Calculus 9e - Chapter 7.1 Integration by Parts Complete Exercises Study Guide

Core Principles of Integration by Parts

• Fundamental Formula for Indefinite Integrals: Integration by parts is derived from the product rule for differentiation. For differentiable functions $u(x)$ and $v(x)$, the integration by parts formula is:   $\int u\,dv = uv - \int v\,du$

• Definite Integral Formula: When evaluating over a bounded interval $[a, b]$, the formula incorporates evaluation limits:   $\int_a^b u\,dv = [u(x)v(x)]_a^b - \int_a^b v\,du = u(b)v(b) - u(a)v(a) - \int_a^b v\,du$

• Strategy for Selecting $u$ and $dv$ (LIATE Rule): A common heuristic for choosing $u$ prioritizes functions in the following order:

  1. Logarithmic functions (e.g., $\ln(x)$)

  2. Inverse trigonometric functions (e.g., $\arcsin(x)$, $\arctan(x)$)

  3. Algebraic functions (e.g., $x^n$, $\sqrt{x}$)

  4. Trigonometric functions (e.g., $\sin(x)$, $\cos(x)$)

  5. Exponential functions (e.g., $e^x$, $3^x$)

Guided Choice Exercises (Problems 1–4)

• Exercise 1: Evaluate $\int x e^{2x}\,dx$ given $u = x$ and $dv = e^{2x}\,dx$.

  • Differentiating $u$ yields $du = dx$.

  • Integrating $dv$ yields $v = \frac{1}{2} e^{2x}$.

  • Applying the formula:     $\int x e^{2x}\,dx = \frac{1}{2} x e^{2x} - \int \frac{1}{2} e^{2x}\,dx = \frac{1}{2} x e^{2x} - \frac{1}{4} e^{2x} + C = \frac{1}{4} e^{2x} (2x - 1) + C$

• Exercise 2: Evaluate $\int \sqrt{x} \ln(x)\,dx$ given $u = \ln(x)$ and $dv = \sqrt{x}\,dx$.

  • Differentiating $u$ yields du = \frac{1}{x}\,dx$.\n - Integrating dv = x^{1/2}\,dx$yields$v = \frac{2}{3} x^{3/2}.\n - Applying the formula:\n    \int \sqrt{x} \ln(x)\,dx = \frac{2}{3} x^{3/2} \ln(x) - \int \frac{2}{3} x^{3/2} \cdot \frac{1}{x}\,dx = \frac{2}{3} x^{3/2} \ln(x) - \frac{2}{3} \int x^{1/2}\,dx = \frac{2}{3} x^{3/2} \ln(x) - \frac{4}{9} x^{3/2} + C\n\n- **Exercise 3:** Evaluate \int x \cos(4x)\,dx$given$u = x$and$dv = \cos(4x)\,dx.\n - Differentiating u$yields$du = dx$.

  • Integrating $dv$ yields $v = \frac{1}{4} \sin(4x)$.

  • Applying the formula:     $\int x \cos(4x)\,dx = \frac{1}{4} x \sin(4x) - \int \frac{1}{4} \sin(4x)\,dx = \frac{1}{4} x \sin(4x) + \frac{1}{16} \cos(4x) + C$

• Exercise 4: Evaluate $\int \sin^{-1}(x)\,dx$ given $u = \sin^{-1}(x)$ and $dv = dx$.

  • Differentiating $u$ yields du = \frac{1}{\sqrt{1 - x^2}}\,dx$.\n - Integrating dv$yields$v = x$.

  • Applying the formula:     $\int \sin^{-1}(x)\,dx = x \sin^{-1}(x) - \int \frac{x}{\sqrt{1 - x^2}}\,dx = x \sin^{-1}(x) + \sqrt{1 - x^2} + C$

Standard Indefinite and Definite Integrals (Problems 5–42)

• Exercise 5: $\int t e^{2t}\,dt$

  • Let $u = t \implies du = dt$ and $dv = e^{2t}\,dt \implies v = \frac{1}{2} e^{2t}$.

  • Result: $\frac{1}{2} t e^{2t} - \frac{1}{4} e^{2t} + C$

• Exercise 6: $\int y e^{-y}\,dy$

  • Let $u = y \implies du = dy$ and $dv = e^{-y}\,dy \implies v = -e^{-y}$.

  • Result: $-y e^{-y} - e^{-y} + C = -e^{-y} (y + 1) + C$

• Exercise 7: $\int x \sin(10x)\,dx$

  • Let $u = x \implies du = dx$ and $dv = \sin(10x)\,dx \implies v = -\frac{1}{10} \cos(10x)$.

  • Result: $- \frac{1}{10} x \cos(10x) + \frac{1}{100} \sin(10x) + C$

• Exercise 8: $\int (\pi - x) \cos(\pi x)\,dx$

  • Let $u = \pi - x \implies du = -dx$ and $dv = \cos(\pi x)\,dx \implies v = \frac{1}{\pi} \sin(\pi x)$.

  • Result: $\frac{\pi - x}{\pi} \sin(\pi x) - \frac{1}{\pi^2} \cos(\pi x) + C$

• Exercise 9: $\int w \ln(w)\,dw$

  • Let $u = \ln(w) \implies du = \frac{1}{w}\,dw$ and $dv = w\,dw \implies v = \frac{1}{2} w^2$.

  • Result: $\frac{1}{2} w^2 \ln(w) - \frac{1}{4} w^2 + C$

• Exercise 10: $\int \frac{\ln(x)}{x^2}\,dx$

  • Let $u = \ln(x) \implies du = \frac{1}{x}\,dx$ and $dv = x^{-2}\,dx \implies v = -x^{-1}$.

  • Result: $- \frac{\ln(x)}{x} - \frac{1}{x} + C$

• Exercise 11: $\int (x^2 + 2x) \cos(x)\,dx$

  • Requires integration by parts twice:

  1. First pass: $u = x^2 + 2x \implies du = (2x + 2)\,dx$; $dv = \cos(x)\,dx \implies v = \sin(x)$.

  2. Second pass on $\int (2x + 2) \sin(x)\,dx$: $u = 2x + 2 \implies du = 2\,dx$; $dv = \sin(x)\,dx \implies v = -\cos(x)$.

  • Result: $(x^2 + 2x - 2) \sin(x) + (2x + 2) \cos(x) + C$

• Exercise 12: $\int t^2 \sin(\beta t)\,dt$

  • Requires integration by parts twice.

  • Result: $- \frac{t^2}{\beta} \cos(\beta t) + \frac{2t}{\beta^2} \sin(\beta t) + \frac{2}{\beta^3} \cos(\beta t) + C$

• Exercise 13: $\int \cos^{-1}(x)\,dx$

  • Let $u = \cos^{-1}(x) \implies du = -\frac{1}{\sqrt{1 - x^2}}\,dx$ and dv = dx \implies v = x$.\n - Result: x \cos^{-1}(x) - \sqrt{1 - x^2} + C\n\n- **Exercise 14:** \int \ln(\sqrt{x})\,dx\n - Simplify first using logarithm properties: \ln(\sqrt{x}) = \frac{1}{2} \ln(x).\n - Result: \frac{1}{2} x \ln(x) - \frac{1}{2} x + C\n\n- **Exercise 15:** \int t^4 \ln(t)\,dt\n - Let u = \ln(t) \implies du = \frac{1}{t}\,dt$and$dv = t^4\,dt \implies v = \frac{1}{5} t^5$.

  • Result: $\frac{1}{5} t^5 \ln(t) - \frac{1}{25} t^5 + C$

• Exercise 16: $\int \tan^{-1}(2y)\,dy$

  • Let $u = \tan^{-1}(2y) \implies du = \frac{2}{1 + 4y^2}\,dy$ and dv = dy \implies v = y$.\n - Result: y \tan^{-1}(2y) - \frac{1}{4} \ln(1 + 4y^2) + C\n\n- **Exercise 17:** \int t \csc^2(t)\,dt\n - Let u = t \implies du = dt$and$dv = \csc^2(t)\,dt \implies v = -\cot(t).\n - Result: -t \cot(t) + \ln|\sin(t)| + C\n\n- **Exercise 18:** \int x \cosh(ax)\,dx\n - Let u = x \implies du = dx$and$dv = \cosh(ax)\,dx \implies v = \frac{1}{a} \sinh(ax).\n - Result: \frac{x}{a} \sinh(ax) - \frac{1}{a^2} \cosh(ax) + C\n\n- **Exercise 19:** \int (\ln(x))^2\,dx\n - Let u = (\ln(x))^2 \implies du = \frac{2 \ln(x)}{x}\,dx$and$dv = dx \implies v = x$.

  • Result: $x (\ln(x))^2 - 2x \ln(x) + 2x + C$

• Exercise 20: $\int \frac{z}{10^z}\,dz$

  • Rewrite integrand as $z 10^{-z}$. Let $u = z \implies du = dz$ and $dv = 10^{-z}\,dz \implies v = -\frac{10^{-z}}{\ln(10)}$ .

  • Result: $- \frac{z 10^{-z}}{\ln(10)} - \frac{10^{-z}}{(\ln(10))^2} + C$

• Exercise 21: $\int e^{3x} \cos(x)\,dx$

  • Cyclic Integration by Parts: Applying integration by parts twice returns a constant multiple of the original integral $I$.

  • Let $u = e^{3x}, dv = \cos(x)\,dx \implies I = e^{3x} \sin(x) - 3 \int e^{3x} \sin(x)\,dx$

  • For the second integral, let $u = e^{3x}, dv = \sin(x)\,dx \implies \int e^{3x} \sin(x)\,dx = -e^{3x} \cos(x) + 3I$

  • Solving for $I$:     $I = e^{3x} \sin(x) + 3e^{3x} \cos(x) - 9I \implies 10I = e^{3x} (\sin(x) + 3\cos(x))$

  • Result: $\frac{e^{3x}}{10} (\sin(x) + 3 \cos(x)) + C$

• Exercise 22: $\int e^x \sin(\pi x)\,dx$

  • Cyclic Integration by Parts.

  • Result: $\frac{e^x}{1 + \pi^2} (\sin(\pi x) - \pi \cos(\pi x)) + C$

• Exercise 23: $\int e^{2\theta} \sin(3\theta)\,d\theta$

  • Cyclic Integration by Parts.

  • Result: $\frac{e^{2\theta}}{13} (2 \sin(3\theta) - 3 \cos(3\theta)) + C$

• Exercise 24: $\int e^{-\theta} \cos(2\theta)\,d\theta$

  • Cyclic Integration by Parts.

  • Result: $\frac{e^{-\theta}}{5} (-\cos(2\theta) + 2 \sin(2\theta)) + C$

• Exercise 25: $\int z^3 e^z\,dz$

  • Tabular integration with repeated differentiation of $z^3$ down to 0.

  • Result: $e^z (z^3 - 3z^2 + 6z - 6) + C$

• Exercise 26: $\int (\arcsin(x))^2\,dx$

  • Let $u = (\arcsin(x))^2 \implies du = \frac{2 \arcsin(x)}{\sqrt{1 - x^2}}\,dx$ and dv = dx \implies v = x$.\n - Result: x (\arcsin(x))^2 + 2\sqrt{1 - x^2} \arcsin(x) - 2x + C\n\n- **Exercise 27:** \int (1 + x^2) e^{3x}\,dx\n - Tabular integration with u = 1 + x^2$and$dv = e^{3x}\,dx$.

  • Result: $e^{3x} \left( \frac{1}{3} x^2 - \frac{2}{9} x + \frac{11}{27} \right) + C$

• Exercise 28: $\int_0^{1/2} \theta \sin(3\pi \theta)\,d\theta$

  • Let $u = \theta, dv = \sin(3\pi \theta)\,d\theta \implies du = d\theta, v = -\frac{1}{3\pi} \cos(3\pi \theta)$.

  • Evaluate boundary terms and remaining integral:     $\left[ - \frac{\theta}{3\pi} \cos(3\pi \theta) \right]_0^{1/2} + \frac{1}{3\pi} \int_0^{1/2} \cos(3\pi \theta)\,d\theta = 0 + \left[ \frac{1}{9\pi^2} \sin(3\pi \theta) \right]_0^{1/2} = -\frac{1}{9\pi^2}$

  • Result: $-\frac{1}{9\pi^2}$

• Exercise 29: $\int_0^1 x 3^x\,dx$

  • Let $u = x, dv = 3^x\,dx \implies du = dx, v = \frac{3^x}{\ln(3)}$.

  • Result: $\frac{3}{\ln(3)} - \frac{2}{(\ln(3))^2}$

• Exercise 30: $\int_0^1 \frac{x e^x}{(1 + x)^2}\,dx$

  • Let $u = x e^x \implies du = e^x (1 + x)\,dx$ and $dv = (1 + x)^{-2}\,dx \implies v = -\frac{1}{1 + x}$.

  • Evaluating formula:     $\left[ -\frac{x e^x}{1 + x} \right]_0^1 + \int_0^1 e^x\,dx = -\frac{e}{2} + (e - 1) = \frac{e}{2} - 1$

  • Result: $\frac{e}{2} - 1$

• Exercise 31: $\int_0^2 y \sinh(y)\,dy$

  • Let $u = y, dv = \sinh(y)\,dy \implies du = dy, v = \cosh(y)$.

  • Result: $2 \cosh(2) - \sinh(2)$

• Exercise 32: $\int_1^2 w^2 \ln(w)\,dw$

  • Let u = \ln(w), dv = w^2\,dw \implies du = \frac{1}{w}\,dw, v = \frac{1}{3} w^3$.\n - Result: \frac{8}{3} \ln(2) - \frac{7}{9}\n\n- **Exercise 33:** \int_1^5 \frac{\ln(R)}{R^2}\,dR\n - Let u = \ln(R), dv = R^{-2}\,dR \implies du = \frac{1}{R}\,dR, v = -R^{-1}$.

  • Result: $\frac{4 - \ln(5)}{5}$

• Exercise 34: $\int_0^{2\pi} t^2 \sin(2t)\,dt$

  • Integration by parts twice over $[0, 2\pi]$.

  • Result: $-2\pi^2$

• Exercise 35: $\int_0^\pi x \sin(x) \cos(x)\,dx$

  • Rewrite integrand using double-angle identity: $\sin(x) \cos(x) = \frac{1}{2} \sin(2x)$.

  • Integral becomes $\frac{1}{2} \int_0^\pi x \sin(2x)\,dx$.

  • Result: $-\frac{\pi}{4}$

• Exercise 36: $\int_1^{\sqrt{3}} \arctan(1/x)\,dx$

  • Let $u = \arctan(1/x) \implies du = -\frac{1}{x^2 + 1}\,dx$ and dv = dx \implies v = x$.\n - Result: \frac{\pi \sqrt{3}}{6} - \frac{\pi}{4} + \frac{1}{2} \ln(2)\n\n- **Exercise 37:** \int_1^5 \frac{M}{e^M}\,dM\n - Let u = M, dv = e^{-M}\,dM \implies du = dM, v = -e^{-M}.\n - Result: \frac{2}{e} - \frac{6}{e^5}\n\n- **Exercise 38:** \int_1^2 \frac{(\ln(x))^2}{x^3}\,dx\n - Let u = (\ln(x))^2, dv = x^{-3}\,dx \implies du = \frac{2 \ln(x)}{x}\,dx, v = -\frac{1}{2x^2}.\n - Result: \frac{3 - 2\ln(2) - 2(\ln(2))^2}{16}\n\n- **Exercise 39:** \int_0^{\pi/3} \sin(x) \ln(\cos(x))\,dx\n - Let u = \ln(\cos(x)) \implies du = -\tan(x)\,dx$and$dv = \sin(x)\,dx \implies v = -\cos(x).\n - Result: \frac{1}{2} (\ln(2) - 1)\n\n- **Exercise 40:** \int_0^1 \frac{r^3}{\sqrt{4 + r^2}}\,dr\n - Rewrite r^3 = r^2 \cdot r$. Let$u = r^2 \implies du = 2r\,dr$and$dv = \frac{r}{\sqrt{4 + r^2}}\,dr \implies v = \sqrt{4 + r^2}.\n - Result: \frac{16 - 7\sqrt{5}}{3}\n\n- **Exercise 41:** \int_0^\pi \cos(x) \sinh(x)\,dx\n - Cyclic Integration by Parts.\n - Result: -\frac{1 + \cosh(\pi)}{2}\n\n- **Exercise 42:** \int_0^t e^s \sin(t - s)\,ds\n - Integration with respect to variable s using cyclic integration by parts.\n - Result: \frac{1}{2} (e^t - \sin(t) - \cos(t))\n\n\n# Substitution Combined with Integration by Parts (Problems 43–48)\n\n- **Exercise 43:** \int e^{\sqrt{x}}\,dx\n - Substitution: Let w = \sqrt{x} \implies x = w^2 \implies dx = 2w\,dw$.

  • Integral transforms to 2 \int w e^w\,dw$.\n - Apply integration by parts (u = w, dv = e^w\,dw$):$2(w e^w - e^w) + C$.

  • Substitute back: $2 e^{\sqrt{x}} (\sqrt{x} - 1) + C$

• Exercise 44: $\int \cos(\ln(x))\,dx$

  • Substitution: Let w = \ln(x) \implies x = e^w \implies dx = e^w\,dw$.\n - Integral transforms to \int e^w \cos(w)\,dw$.

  • Apply cyclic integration by parts: \frac{1}{2} e^w (\cos(w) + \sin(w)) + C$.\n - Substitute back: \frac{1}{2} x (\cos(\ln(x)) + \sin(\ln(x))) + C\n\n- **Exercise 45:** \int_{\sqrt{\pi/2}}^{\sqrt{\pi}} \theta^3 \cos(\theta^2)\,d\theta\n - Substitution: Let x = \theta^2 \implies dx = 2\theta\,d\theta$.

  • Limits change: when $\theta = \sqrt{\pi/2} \implies x = \pi/2$; when \theta = \sqrt{\pi} \implies x = \pi$.\n - Integral transforms to \frac{1}{2} \int_{\pi/2}^\pi x \cos(x)\,dx$.

  • Apply integration by parts ($u = x, dv = \cos(x)\,dx$).

  • Result: $\frac{1}{2} \left( [x \sin(x)]_{\pi/2}^\pi - \int_{\pi/2}^\pi \sin(x)\,dx \right) = -\frac{\pi + 2}{4}$

• Exercise 46: $\int_0^\pi e^{\cos(t)} \sin(2t)\,dt$

  • Rewrite $\sin(2t) = 2 \sin(t) \cos(t)$.

  • Substitution: Let w = \cos(t) \implies dw = -\sin(t)\,dt$.\n - Limits change: t = 0 \implies w = 1$;$t = \pi \implies w = -1$.

  • Integral transforms to 2 \int_{-1}^1 w e^w\,dw$.\n - Apply integration by parts.\n - Result: \frac{4}{e}\n\n- **Exercise 47:** \int x \ln(1 + x)\,dx\n - Substitution: Let w = 1 + x \implies x = w - 1 \implies dx = dw$.

  • Integral transforms to \int (w - 1) \ln(w)\,dw = \int w \ln(w)\,dw - \int \ln(w)\,dw$.\n - Result: \frac{x^2 - 1}{2} \ln(1 + x) - \frac{x^2}{4} + \frac{x}{2} + C\n\n- **Exercise 48:** \int \frac{\arcsin(\ln(x))}{x}\,dx\n - Substitution: Let w = \ln(x) \implies dw = \frac{1}{x}\,dx$.

  • Integral transforms to \int \arcsin(w)\,dw$.\n - Apply integration by parts (u = \arcsin(w), dv = dw).\n - Result: \ln(x) \arcsin(\ln(x)) + \sqrt{1 - (\ln(x))^2} + C\n\n\n# Indefinite Integrals and Graphical Verification (Problems 49–52)\n\n- **Instruction:** Evaluate the indefinite integral. Illustrate and verify reasonableness by graphing the function f(x)$and its antiderivative$F(x)$with constant$C = 0.\n\n- **Exercise 49:** \int x e^{-2x}\,dx\n - Antiderivative (C = 0$):$F(x) = -\frac{1}{4} e^{-2x} (2x + 1)\n - Function: f(x) = x e^{-2x}\n - Graphical check: F'(x) = f(x)$.$f(x) = 0$at$x = 0$, where$F(x)$exhibits a local minimum at$(0, -0.25).\n\n- **Exercise 50:** \int x^{3/2} \ln(x)\,dx\n - Antiderivative (C = 0$):$F(x) = \frac{2}{25} x^{5/2} (5 \ln(x) - 2)\n - Function: f(x) = x^{3/2} \ln(x)\n\n- **Exercise 51:** \int x^3 \sqrt{1 + x^2}\,dx\n - Antiderivative (C = 0$):$F(x) = \frac{1}{15} (1 + x^2)^{3/2} (3x^2 - 2)\n - Function: f(x) = x^3 \sqrt{1 + x^2}\n\n- **Exercise 52:** \int x^2 \sin(2x)\,dx\n - Antiderivative (C = 0$):$F(x) = \frac{1}{4} \left( (1 - 2x^2) \cos(2x) + 2x \sin(2x) \right)\n - Function: f(x) = x^2 \sin(2x)\n\n\n# Derivation and Application of Reduction Formulas (Problems 53–62)\n\n- **Exercise 53:**\n - **Part (a):** Use the reduction formula \int \sin^n(x)\,dx = -\frac{1}{n} \sin^{n-1}(x) \cos(x) + \frac{n-1}{n} \int \sin^{n-2}(x)\,dx$for$n = 2:\n    \int \sin^2(x)\,dx = -\frac{1}{2} \sin(x) \cos(x) + \frac{1}{2} \int 1\,dx = \frac{x}{2} - \frac{\sin(2x)}{4} + C\n - **Part (b):** Evaluate \int \sin^4(x)\,dx$using part (a) and reduction for$n = 4:\n    \int \sin^4(x)\,dx = -\frac{1}{4} \sin^3(x) \cos(x) + \frac{3}{4} \left( \frac{x}{2} - \frac{\sin(2x)}{4} \right) + C = \frac{3}{8} x - \frac{1}{4} \sin^3(x) \cos(x) - \frac{3}{16} \sin(2x) + C\n\n- **Exercise 54:**\n - **Part (a):** Prove the reduction formula:\n    \int \cos^n(x)\,dx = \frac{1}{n} \cos^{n-1}(x) \sin(x) + \frac{n-1}{n} \int \cos^{n-2}(x)\,dx\n - *Proof:* Let u = \cos^{n-1}(x) \implies du = -(n-1)\cos^{n-2}(x)\sin(x)\,dx$and$dv = \cos(x)\,dx \implies v = \sin(x).\n - Applying integration by parts:\n      \int \cos^n(x)\,dx = \cos^{n-1}(x) \sin(x) + (n-1) \int \cos^{n-2}(x) \sin^2(x)\,dx\n - Substitute \sin^2(x) = 1 - \cos^2(x):\n      \int \cos^n(x)\,dx = \cos^{n-1}(x) \sin(x) + (n-1) \int \cos^{n-2}(x)\,dx - (n-1) \int \cos^n(x)\,dx\n - Rearranging terms gives n \int \cos^n(x)\,dx = \cos^{n-1}(x) \sin(x) + (n-1) \int \cos^{n-2}(x)\,dx$. Dividing by$n completes the proof.\n - **Part (b):** Evaluate \int \cos^2(x)\,dx = \frac{x}{2} + \frac{\sin(2x)}{4} + C\n - **Part (c):** Evaluate \int \cos^4(x)\,dx = \frac{3}{8} x + \frac{1}{4} \cos^3(x) \sin(x) + \frac{3}{16} \sin(2x) + C\n\n- **Exercise 55:**\n - **Part (a):** Show that for integer n \ge 2:\n    \int_0^{\pi/2} \sin^n(x)\,dx = \frac{n-1}{n} \int_0^{\pi/2} \sin^{n-2}(x)\,dx\n - The boundary term \left[ -\frac{1}{n} \sin^{n-1}(x) \cos(x) \right]_0^{\pi/2} = 0$since$\cos(\pi/2) = 0$and$\sin(0) = 0.\n - **Part (b):** Evaluate integrals:\n - \int_0^{\pi/2} \sin^3(x)\,dx = \frac{2}{3} \int_0^{\pi/2} \sin(x)\,dx = \frac{2}{3}\n - \int_0^{\pi/2} \sin^5(x)\,dx = \frac{4}{5} \cdot \frac{2}{3} = \frac{8}{15}\n - **Part (c):** Show that for odd powers of sine (2n+1):\n    \int_0^{\pi/2} \sin^{2n+1}(x)\,dx = \frac{2 \cdot 4 \cdot 6 \cdot \cdots \cdot 2n}{3 \cdot 5 \cdot 7 \cdot \cdots \cdot (2n+1)}\n\n- **Exercise 56:** Prove that for even powers of sine (2n):\n  \int_0^{\pi/2} \sin^{2n}(x)\,dx = \frac{1 \cdot 3 \cdot 5 \cdot \cdots \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \cdots \cdot 2n} \frac{\pi}{2}\n\n- **Exercise 57:** Prove reduction formula for logarithm powers:\n  \int (\ln(x))^n\,dx = x (\ln(x))^n - n \int (\ln(x))^{n-1}\,dx\n - *Proof:* Let u = (\ln(x))^n \implies du = n (\ln(x))^{n-1} \frac{1}{x}\,dx$and$dv = dx \implies v = x$.

• Exercise 58: Prove reduction formula for exponential product:   $\int x^n e^x\,dx = x^n e^x - n \int x^{n-1} e^x\,dx$

  • Proof: Let $u = x^n \implies du = n x^{n-1}\,dx$ and dv = e^x\,dx \implies v = e^x$.\n\n- **Exercise 59:** Prove reduction formula for tangent powers (n eq 1):\n  \int \tan^n(x)\,dx = \frac{\tan^{n-1}(x)}{n-1} - \int \tan^{n-2}(x)\,dx\n - *Proof:* Split \tan^n(x) = \tan^{n-2}(x) (\sec^2(x) - 1) = \tan^{n-2}(x) \sec^2(x) - \tan^{n-2}(x)$. Integrate the first term using substitution$w = \tan(x).\n\n- **Exercise 60:** Prove reduction formula for secant powers (n eq 1):\n  \int \sec^n(x)\,dx = \frac{\tan(x) \sec^{n-2}(x)}{n-1} + \frac{n-2}{n-1} \int \sec^{n-2}(x)\,dx\n - *Proof:* Let u = \sec^{n-2}(x)$and$dv = \sec^2(x)\,dx$. Apply integration by parts and trigonometric identity$\tan^2(x) = \sec^2(x) - 1$.

• Exercise 61: Use Exercise 57 to find $\int (\ln(x))^3\,dx$:   $\int (\ln(x))^3\,dx = x (\ln(x))^3 - 3x (\ln(x))^2 + 6x \ln(x) - 6x + C$

• Exercise 62: Use Exercise 58 to find $\int x^4 e^x\,dx$:   $\int x^4 e^x\,dx = e^x (x^4 - 4x^3 + 12x^2 - 24x + 24) + C$

Geometric Applications: Area Bounded by Curves (Problems 63–66)

• Exercise 63: Bounded area between $y = x^2 \ln(x)$ and $y = 4 \ln(x)$.

  • Intersection points: Set x^2 \ln(x) = 4 \ln(x) \implies (x^2 - 4) \ln(x) = 0 \implies x = 1, x = 2$.\n - On interval [1, 2]$,$4 \ln(x) \ge x^2 \ln(x).\n - Area integral: A = \int_1^2 (4 - x^2) \ln(x)\,dx\n - Apply integration by parts with u = \ln(x)$and$dv = (4 - x^2)\,dx \implies v = 4x - \frac{x^3}{3}.\n - Result: A = \frac{16}{3} \ln(2) - \frac{29}{9}\n\n- **Exercise 64:** Bounded area between y = x^2 e^{-x}$and$y = x e^{-x}.\n - Intersection points: Set x^2 e^{-x} = x e^{-x} \implies x(x - 1) = 0 \implies x = 0, x = 1$.

  • Area integral: $A = \int_0^1 (x - x^2) e^{-x}\,dx = \frac{3}{e} - 1$

• Exercise 65: Find approximate $x$--coordinates of intersection and bounded area for $y = \arcsin\left(\frac{1}{2}x\right)$ and $y = 2 - x^2$.

  • Approximate intersection bounds: $x \approx -1.801$ and $x \approx 1.028$

  • Area: $A \approx \int_{-1.801}^{1.028} \left( 2 - x^2 - \arcsin\left(\frac{1}{2}x\right) \right)\,dx \approx 4.09$

• Exercise 66: Find approximate $x$--coordinates of intersection and bounded area for $y = x \ln(x + 1)$ and $y = 3x - x^2$.

  • Approximate intersection bounds: $x = 0$ and $x \approx 1.638$

  • Area: $A \approx \int_0^{1.638} \left( 3x - x^2 - x \ln(x + 1) \right)\,dx \approx 1.22$

Geometric Applications: Volumes of Solids of Revolution (Problems 67–71)

• Exercise 67: Region bounded by $y = \cos(\pi x / 2)$, $y = 0$, $0 \le x \le 1$ rotated about the $y$-axis.

  • Using cylindrical shells:     $V = \int_0^1 2\pi x \cos(\pi x / 2)\,dx = 4 - \frac{8}{\pi}$

• Exercise 68: Region bounded by $y = e^x$, $y = e^{-x}$, $x = 1$ rotated about the $y$-axis.

  • Using cylindrical shells:     $V = \int_0^1 2\pi x (e^x - e^{-x})\,dx = 2\pi \left( [x e^x - e^x]_0^1 - [-x e^{-x} - e^{-x}]_0^1 \right) = \frac{4\pi}{e}$

• Exercise 69: Region bounded by $y = e^{-x}$, $y = 0$, $x = -1$, $x = 0$ rotated about x = 1$.\n - Shell radius is 1 - x$, shell height is$e^{-x}.\n - V = \int_{-1}^0 2\pi (1 - x) e^{-x}\,dx = 2\pi e\n\n- **Exercise 70:** Region bounded by y = e^x$,$x = 0$,$y = 3$rotated about the$x-axis.\n - Slicing method (washers):\n    V = \pi \int_0^{\ln(3)} (3^2 - (e^x)^2)\,dx = \pi \left[ 9x - \frac{1}{2} e^{2x} \right]0^{\ln(3)} = \pi (9 \ln(3) - 4)\n\n- **Exercise 71:** Volume generated by rotating region bounded by y = \ln(x)$,$y = 0$, and$x = 2:\n - **Part (a):** About the y-axis (Cylindrical Shells):\n    V = \int_1^2 2\pi x \ln(x)\,dx = \pi \left( 4 \ln(2) - \frac{3}{2} \right)\n - **Part (b):** About the x-axis (Disks):\n    V = \int_1^2 \pi (\ln(x))^2\,dx = 2\pi (\ln(2) - 1)^2\n\n\n# Advanced Calculus Applications and Functional Identities (Problems 72–78)\n\n- **Exercise 72 (Average Value of Function):** Calculate the average value of f(x) = x \sec^2(x)$on$[0, \pi/4].\n - Formula: f{\text{ave}} = \frac{1}{\pi/4 - 0} \int_0^{\pi/4} x \sec^2(x)\,dx = \frac{4}{\pi} \int_0^{\pi/4} x \sec^2(x)\,dx\n - Integration by parts (u = x, dv = \sec^2(x)\,dx \implies v = \tan(x)):\n    \int_0^{\pi/4} x \sec^2(x)\,dx = [x \tan(x)]0^{\pi/4} - \int_0^{\pi/4} \tan(x)\,dx = \frac{\pi}{4} - \frac{1}{2} \ln(2)\n - Average Value: f{\text{ave}} = 1 - \frac{2 \ln(2)}{\pi}\n\n- **Exercise 73 (Optics / Fresnel Function):** The Fresnel function is defined as S(x) = \int_0^x \sin\left(\frac{1}{2}\pi t^2\right)\,dt$. Find$\int S(x)\,dx\n - Apply integration by parts with u = S(x)$and$dv = dx.\n - By the Fundamental Theorem of Calculus, S'(x) = \sin\left(\frac{1}{2}\pi x^2\right).\n - Formula: \int S(x)\,dx = x S(x) - \int x \sin\left(\frac{1}{2}\pi x^2\right)\,dx\n - Integrating remaining term via substitution w = \frac{1}{2}\pi x^2.\n - Result: x S(x) + \frac{1}{\pi} \cos\left(\frac{1}{2}\pi x^2\right) + C\n\n- **Exercise 74 (A Rocket Equation):**\n - Velocity equation: v(t) = -gt - v_e \ln\left(\frac{m - rt}{m}\right)\n - Constants: g = 9.8\,\text{m/s}^2$, initial mass$m = 30\,000\,\text{kg}$, burn rate$r = 160\,\text{kg/s}$, exhaust velocity$v_e = 3000\,\text{m/s}.\n - Position / Height function h(t) = \int_0^t v(s)\,ds via integration by parts:\n    h(t) = - \frac{1}{2} g t^2 + v_e t + \frac{v_e (m - rt)}{r} \ln\left(\frac{m - rt}{m}\right)\n - **Part (a): Height one minute (t = 60\,\text{s}) after liftoff:**\n - m - rt = 30\,000 - 160(60) = 20\,400\,\text{kg}\n - h(60) = -\frac{1}{2}(9.8)(3600) + 3000(60) + \frac{3000(20\,400)}{160} \ln\left(\frac{20\,400}{30\,000}\right) \approx 14\,844\,\text{meters} \approx 14.8\,\text{km}\n - **Part (b): Height after burning 6000\,\text{kg} of fuel:**\n - Burn duration t = \frac{6000}{160} = 37.5\,\text{seconds}$, remaining mass$m - rt = 24\,000\,\text{kg}.\n - h(37.5) \approx 5195\,\text{meters} \approx 5.19\,\text{km}\n\n- **Exercise 75 (Particle Motion / Kinematics):** Straight-line motion with velocity v(t) = t^2 e^{-t}\,\text{m/s}$. Distance traveled in first$t seconds:\n  s(t) = \int_0^t \tau^2 e^{-\tau}\,d\tau = \left[ -e^{-\tau} (\tau^2 + 2\tau + 2) \right]0^t = 2 - e^{-t} (t^2 + 2t + 2)\,\text{meters}\n\n- **Exercise 76 (Second Derivative Identity Proof):** Given f(0) = g(0) = 0$with continuous$f''$and$g'', show:\n  \int_0^a f(x) g''(x)\,dx = f(a) g'(a) - f'(a) g(a) + \int_0^a f''(x) g(x)\,dx\n - *Proof:* Apply integration by parts with u = f(x), dv = g''(x)\,dx \implies \int_0^a f(x) g''(x)\,dx = f(a)g'(a) - f(0)g'(0) - \int_0^a f'(x)g'(x)\,dx$. Apply integration by parts to the second term with$u = g'(x), dv = f'(x)\,dx$. Substitution of$f(0) = 0$and$g(0) = 0 yields the identity.\n\n- **Exercise 77:** Given f(1) = 2$,$f(4) = 7$,$f'(1) = 5$,$f'(4) = 3$, evaluate$\int_1^4 x f''(x)\,dx:\n - Apply integration by parts (u = x, dv = f''(x)\,dx \implies du = dx, v = f'(x)):\n    \int_1^4 x f''(x)\,dx = [x f'(x)]_1^4 - \int_1^4 f'(x)\,dx = (4 f'(4) - 1 f'(1)) - (f(4) - f(1))\n    = (4(3) - 1(5)) - (7 - 2) = 7 - 5 = 2\n\n- **Exercise 78 (Inverse Function Integration):**\n - **Part (a):** Integration by parts identity:\n    \int f(x)\,dx = x f(x) - \int x f'(x)\,dx\n - **Part (b):** If f$and$g$are inverse functions ($g = f^{-1}):\n    \int_a^b f(x)\,dx = b f(b) - a f(a) - \int{f(a)}^{f(b)} g(y)\,dy\n - *Proof:* Make substitution y = f(x) \implies dy = f'(x)\,dx$and$x = g(y)$on$\int_a^b x f'(x)\,dx.\n - **Part (c): Geometric Interpretation:** The large rectangle area b f(b)$minus small rectangle area$a f(a)$forms an L-shaped region decomposed into two orthogonal areas under$y = f(x)$along the$x$-axis and$x = g(y)$along the$y-axis.\n - **Part (d):** Evaluate \int_1^e \ln(x)\,dx using part (b):\n - Here f(x) = \ln(x) \implies g(y) = e^y$. Limits:$a = 1 \implies f(1) = 0$;$b = e \implies f(e) = 1$.

  • $\int_1^e \ln(x)\,dx = e \ln(e) - 1 \ln(1) - \int_0^1 e^y\,dy = e - 0 - (e - 1) = 1$

Theoretical Extensions and Special Formulas (Problems 79–81)

• Exercise 79 (Integration Formula from Quotient Rule):

  • Part (a): Derive $\int \frac{u}{v^2}\,dv = -\frac{u}{v} + \int \frac{1}{v}\,du$

  • Proof: From Quotient Rule $\frac{d}{dv}\left(\frac{u}{v}\right) = \frac{1}{v}\frac{du}{dv} - \frac{u}{v^2}$. Rearranging and integrating with respect to $v$ gives the formula.

  • Part (b): Evaluate $\int \frac{\ln(x)}{x^2}\,dx$ using part (a):

  • Set $u = \ln(x)$ and v = x \implies du = \frac{1}{x}\,dx$.\n - Formula gives: \int \frac{\ln(x)}{x^2}\,dx = -\frac{\ln(x)}{x} + \int \frac{1}{x^2}\,dx = -\frac{\ln(x)}{x} - \frac{1}{x} + C\n\n- **Exercise 80 (The Wallis Product Formula for \pi):**\n - Let I_n = \int_0^{\pi/2} \sin^n(x)\,dx$.

  • Part (a): Show $I_{2n+2} \le I_{2n+1} \le I_{2n}$: Since $0 \le \sin(x) \le 1$ on $[0, \pi/2]$, $\sin^{k+1}(x) \le \sin^k(x)$.

  • Part (b): Show $\frac{I_{2n+2}}{I_{2n}} = \frac{2n+1}{2n+2}$ using reduction formula $I_m = \frac{m-1}{m} I_{m-2}$.

  • Part (c): Show $\frac{2n+1}{2n+2} \le \frac{I_{2n+1}}{I_{2n}} \le 1$ and deduce $\lim_{n \to \infty} \frac{I_{2n+1}}{I_{2n}} = 1$ by the Squeeze Theorem.

  • Part (d): Deduce the Wallis Product:     $\frac{\pi}{2} = \lim_{n \to \infty} \left( \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \cdots \cdot \frac{2n}{2n-1} \cdot \frac{2n}{2n+1} \right)$     $\frac{\pi}{2} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \cdots$

  • Part (e): Ratio of width to height for nested rectangles:          

    

        

  • Starting with unit square (area 1), attach rectangles of area 1 alternately beside or on top.

  • Sequence of ratios of width to height reproduces the partial products of the Wallis product.

  • Limit of ratio of width to height as $n \to \infty$ is $\frac{\pi}{2}$.

• Exercise 81 (Proof of Shell Method Volume via Integration by Parts):      

  

  • Slicing formula for solid rotated about $y$-axis:     $V = \pi b^2 d - \pi a^2 c - \int_c^d \pi [g(y)]^2\,dy$

  • Substitution $y = f(x) \implies dy = f'(x)\,dx$ and $g(y) = x$ transforms integral to:     $\int_c^d \pi [g(y)]^2\,dy = \int_a^b \pi x^2 f'(x)\,dx$

  • Apply integration by parts to $\int_a^b x^2 f'(x)\,dx$ ($u = x^2, dv = f'(x)\,dx \implies v = f(x)$):     $\int_a^b x^2 f'(x)\,dx = [x^2 f(x)]_a^b - \int_a^b 2x f(x)\,dx = b^2 d - a^2 c - \int_a^b 2x f(x)\,dx$

  • Substituting back into formula for $V$:     $V = \pi b^2 d - \pi a^2 c - \pi \left( b^2 d - a^2 c - \int_a^b 2x f(x)\,dx \right) = \int_a^b 2\pi x f(x)\,dx$

  • This proves the Cylindrical Shells volume formula using horizontal slicing and integration by parts.