Calculus: Indefinite Integration by Substitution and Trigonometric Functions

7.3 Integration by Substitution

• General Principle:

  • Some indefinite integrals, such as $\int \frac{1}{(2x+1)^6} dx$ and $\int x \sqrt{2x-1} dx$, cannot be evaluated directly using basic direct integration theorems.

  • In these instances, the method of integration by substitution (Theorem 7.4) is used.

• Theorem 7.4 (Integration by Substitution):

  • Let $u = g(x)$ be a differentiable function.

  • Then: $\int f(g(x)) g'(x) dx = \int f(u) du$

• Proof of Theorem 7.4:

  1. Suppose $F(u)$ is a primitive function of $f(u)$, where $u = g(x)$.

  2. This implies $\frac{d}{du} [F(u)] = f(u)$ (Equation 1).

  3. And $\int f(u) du = F(u) + C$ (Equation 2).

  4. By the chain rule, we have:

     • $\frac{d}{dx} F(u) = \frac{d}{du} F(u) \cdot \frac{du}{dx}$

     • $\frac{d}{dx} F(u) = f(u) \cdot \frac{du}{dx}$ (By Equation 1)

     • $\frac{d}{dx} F(u) = f(g(x)) g'(x)$

  5. By the definition of the indefinite integral:

     • $\int f(g(x)) g'(x) dx = F(u) + C$

     • $\int f(g(x)) g'(x) dx = \int f(u) du$ (By Equation 2)

• Example: Finding $\int (3x+2)^7 dx$ via Theorem 7.4:

  • Let $u = g(x) = 3x+2$ and $f(x) = x^7$.

  • Then $f(g(x)) = (3x+2)^7$ and $g'(x) = 3$.

  • $\int (3x+2)^7 dx = \int \frac{1}{3} (3x+2)^7 \cdot 3 dx$

  • $= \frac{1}{3} \int u^7 du$

  • $= \frac{1}{3} \cdot \frac{u^8}{8} + C$

  • $= \frac{1}{24} (3x+2)^8 + C$

• Standard Steps for Substitution:

  • Step 1: Choose a suitable substitution $u = g(x)$. Note that $du = g'(x) dx$. Rewrite $dx$ in terms of $du$.

  • Step 2: Rewrite the entire integral in terms of $x$ as an integral in terms of $u$.

  • Step 3: Integrate with respect to $u$.

  • Step 4: Substitute back to write the result in terms of the original variable $x$.

Practical Examples of Integration by Substitution

• Example 7.7: Simple Linear Substitution:

  • (a) Find $\int \sqrt{2x+1} dx$:

    • Let $u = 2x+1$, which means $du = 2 dx$, or $dx = \frac{1}{2} du$.

    • $\int \sqrt{u} (\frac{1}{2} du) = \frac{1}{2} \int u^{\frac{1}{2}} du$

    • $= \frac{1}{2} [\frac{u^{\frac{3}{2}}}{\frac{3}{2}}] + C = \frac{1}{3} u^{\frac{3}{2}} + C$

    • Result: $\frac{1}{3} (2x+1)^{\frac{3}{2}} + C$

  • (b) Find $\int x(1-x^2)^3 dx$:

    • Let $u = 1-x^2$, so $du = -2x dx$, or $x dx = -\frac{1}{2} du$.

    • $\int u^3 (-\frac{1}{2} du) = -\frac{1}{2} \int u^3 du$

    • $= -\frac{1}{2} [\frac{u^4}{4}] + C = -\frac{1}{8} u^4 + C$

    • Result: $-\frac{1}{8} (1-x^2)^4 + C$

• Example 7.8: Substitution with Variable Rearrangement:

  • Find $\int x^2 \sqrt{x+2} dx$.

  • Method 1:

    • Let $u = x+2$, so $du = dx$ and $x = u-2$.

    • $\int (u-2)^2 u^{\frac{1}{2}} du = \int (u^2 - 4u + 4) u^{\frac{1}{2}} du$

    • $= \int (u^{\frac{5}{2}} - 4u^{\frac{3}{2}} + 4u^{\frac{1}{2}}) du$

    • $= \frac{2}{7} u^{\frac{7}{2}} - \frac{8}{5} u^{\frac{5}{2}} + \frac{8}{3} u^{\frac{3}{2}} + C$

    • Result: $\frac{2}{7}(x+2)^{\frac{7}{2}} - \frac{8}{5}(x+2)^{\frac{5}{2}} + \frac{8}{3}(x+2)^{\frac{3}{2}} + C$

  • Method 2:

    • Let $u = \sqrt{x+2}$, so $u^2 = x+2$, $x = u^2 - 2$, and $dx = 2u du$.

    • $\int (u^2 - 2)^2 \cdot u \cdot 2u du = 2 \int (u^4 - 4u^2 + 4) u^2 du$

    • $= 2 \int (u^6 - 4u^4 + 4u^2) du = 2(\frac{u^7}{7} - \frac{4u^5}{5} + \frac{4u^3}{3}) + C$

    • Substituting back $u = \sqrt{x+2}$ yields the same result: $\frac{2}{7}(x+2)^{\frac{7}{2}} - \frac{8}{5}(x+2)^{\frac{5}{2}} + \frac{8}{3}(x+2)^{\frac{3}{2}} + C$

Advanced Substitution Techniques

• Implicit Variable Substitution:

  • Substitution can be performed without explicitly introducing $u$ by manipulating the differential $dx$.

  • Example 7.9: Find $\int \frac{1}{(2x+1)^6} dx$.

    • Note that $d(2x+1) = 2 dx$, so $dx = \frac{1}{2} d(2x+1)$.

    • $\int (2x+1)^{-6} \frac{1}{2} d(2x+1) = \frac{1}{2} \cdot \frac{(2x+1)^{-5}}{-5} + C = -\frac{1}{10} (2x+1)^{-5} + C$

  • Example 7.10 (a): Find $\int x^4 (x^5-2)^9 dx$.

    • Note that $d(x^5-2) = 5x^4 dx$, so $x^4 dx = \frac{1}{5} d(x^5-2)$.

    • $\int (x^5-2)^9 \frac{1}{5} d(x^5-2) = \frac{1}{5} [\frac{(x^5-2)^{10}}{10}] + C = \frac{1}{50}(x^5-2)^{10} + C$

  • Example 7.10 (b): Find $\int \frac{2x-1}{x^2-x+3} dx$.

    • Note that $d(x^2-x+3) = (2x-1) dx$.

    • $\int \frac{1}{x^2-x+3} d(x^2-x+3) = \ln |x^2-x+3| + C$

• Exponential and Logarithmic Substitution (Example 7.11):

  • (a) $\int e^{4x-2} dx = \int e^{4x-2} \frac{1}{4} d(4x-2) = \frac{1}{4} e^{4x-2} + C$

  • (b) $\int x e^{x^2} dx = \int e^{x^2} \frac{1}{2} d(x^2) = \frac{1}{2} e^{x^2} + C$

  • (c) $\int \frac{e^x}{e^x+1} dx = \int \frac{1}{e^x+1} d(e^x+1) = \ln(e^x+1) + C$ (Note: absolute value not needed as e^x+1 > 0)

  • (d) $\int \frac{\ln x}{x} dx = \int \ln x \, d(\ln x) = \frac{(\ln x)^2}{2} + C$

7.4 Integration of Trigonometric Functions

• Fundamental Formulas (Theorem 7.5):

  • $\int \cos x \, dx = \sin x + C$

  • $\int \sin x \, dx = -\cos x + C$

  • $\int \sec^2 x \, dx = \tan x + C$

• Essential Trigonometric Relations:

  • (i) $\sec A = \frac{1}{\cos A}$

  • (ii) $\tan A = \frac{\sin A}{\cos A}$

  • (iii) $\sin^2 A + \cos^2 A = 1$

  • (iv) $1 + \tan^2 A = \sec^2 A$

• Double Angle Formulas:

  • (i) $\sin 2A = 2 \sin A \cos A$

  • (ii) $\cos 2A = 2 \cos^2 A - 1 \implies \cos^2 A = \frac{1}{2}(1 + \cos 2A)$

  • (iii) $\cos 2A = 1 - 2 \sin^2 A \implies \sin^2 A = \frac{1}{2}(1 - \cos 2A)$

• Trigonometric Integrals via Identities (Example 7.12 & 7.13):

  • Find $\int \tan^2 x \, dx$:

    • $\int (\sec^2 x - 1) dx = \tan x - x + C$

  • Find $\int \frac{\tan x}{\sin 2x} dx$:

    • Using $\sin 2x = 2 \sin x \cos x$ and $\tan x = \frac{\sin x}{\cos x}$.

    • $\int \frac{\frac{\sin x}{\cos x}}{2 \sin x \cos x} dx = \int \frac{1}{2 \cos^2 x} dx = \frac{1}{2} \int \sec^2 x \, dx = \frac{1}{2} \tan x + C$

Trigonometric Integration by Substitution

• Example 7.14 & 7.15 (Functional Composition):

  • (a) $\int \sin(3x+2) dx = \int \sin(3x+2) \frac{1}{3} d(3x+2) = -\frac{1}{3} \cos(3x+2) + C$

  • (b) $\int \cos^2 x \, dx = \int \frac{1}{2} (1 + \cos 2x) dx = \frac{1}{2} x + \frac{1}{2} \int \cos 2x \, dx$

    • $= \frac{1}{2} x + \frac{1}{2} \int \cos 2x \, \frac{1}{2} d(2x) = \frac{1}{2} x + \frac{1}{4} \sin 2x + C$

  • (c) $\int x \cos(x^2) dx = \int \cos(x^2) \frac{1}{2} d(x^2) = \frac{1}{2} \sin(x^2) + C$

  • (d) $\int \frac{2 \cos x}{(3 + \sin x)^2} dx = \int \frac{2}{(3 + \sin x)^2} d(3 + \sin x) = 2 [\frac{(3+\sin x)^{-1}}{-1}] + C = -\frac{2}{3+\sin x} + C$

Integration Using Product-to-Sum Formulas

• Formulas for products where $m$ and $n$ are constants:

  • (i) $\sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)]$

  • (ii) $\cos A \sin B = \frac{1}{2} [\sin(A + B) - \sin(A - B)]$

  • (iii) $\cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)]$

  • (iv) $\sin A \sin B = -\frac{1}{2} [\cos(A + B) - \cos(A - B)]$

• Example 7.16: Find $\int \sin 2x \cos 4x \, dx$.

  • $\sin 2x \cos 4x = \frac{1}{2} [\sin(2x + 4x) + \sin(2x - 4x)] = \frac{1}{2} [\sin 6x - \sin 2x]$.

  • $\frac{1}{2} \int (\sin 6x - \sin 2x) dx = \frac{1}{2} [-\frac{1}{6} \cos 6x + \frac{1}{2} \cos 2x] + C = -\frac{1}{12} \cos 6x + \frac{1}{4} \cos 2x + C$

Integrands in the form $\int \sin^m x \cos^n x \, dx$

• Rules for Substitution:

  • (i) When $m$ (power of sin) is odd: Substitute $u = \cos x$ (i.e., $\sin x \, dx = -d(\cos x)$).

  • (ii) When $n$ (power of cos) is odd: Substitute $u = \sin x$ (i.e., $\cos x \, dx = d(\sin x)$).

  • (iii) When both $m$ and $n$ are even: Use double angle formulas to reduce powers.

  • (iv) When both are odd: Either substitution works, but picking the one with the higher power for $u$ is usually easier.

• Example 7.17 (Case $n$ is odd): $\int \sin^2 x \cos^3 x \, dx$

  • $\int \sin^2 x \cos^2 x \, d(\sin x) = \int \sin^2 x (1 - \sin^2 x) d(\sin x)$

  • $= \int (\sin^2 x - \sin^4 x) d(\sin x) = \frac{1}{3} \sin^3 x - \frac{1}{5} \sin^5 x + C$

• Example 7.18 (Both odd): $\int \sin^3 2\theta \cos^5 2\theta \, d\theta$

  • Tip: Since the power of $\sin 2\theta$ is smaller, use $\sin 2\theta \, d\theta = -\frac{1}{2} d(\cos 2\theta)$.

  • $\int \sin^2 2\theta \cos^5 2\theta \sin 2\theta \, d\theta = \int (1 - \cos^2 2\theta) \cos^5 2\theta [-\frac{1}{2} d(\cos 2\theta)]$

  • $= -\frac{1}{2} \int (\cos^5 2\theta - \cos^7 2\theta) d(\cos 2\theta) = -\frac{1}{2} [\frac{1}{6} \cos^6 2\theta - \frac{1}{8} \cos^8 2\theta] + C$

  • Result: $\frac{1}{16} \cos^8 2\theta - \frac{1}{12} \cos^6 2\theta + C$

• Example 7.19 (Both even): $\int \sin^2 x \cos^2 x \, dx$

  • $\sin^2 x \cos^2 x = (\sin x \cos x)^2 = (\frac{1}{2} \sin 2x)^2 = \frac{1}{4} \sin^2 2x$

  • $= \frac{1}{4} [\frac{1}{2} (1 - \cos 4x)] = \frac{1}{8} (1 - \cos 4x)$

  • $\int \frac{1}{8} (1 - \cos 4x) dx = \frac{1}{8} [x - \frac{1}{4} \sin 4x] + C = \frac{1}{8}x - \frac{1}{32} \sin 4x + C$

Integrands in the form $\tan^m x \sec^n x$

• Condition: If $n$ (power of sec) is even.

• Substitution: Use $u = \tan x$, implying $\sec^2 x \, dx = d(\tan x)$.

• Example 7.20: Find $\int \tan^2 x \sec^4 x \, dx$.

  • $\int \tan^2 x \sec^2 x \cdot \sec^2 x \, dx = \int \tan^2 x (\tan^2 x + 1) d(\tan x)$

  • $= \int (\tan^4 x + \tan^2 x) d(\tan x) = \frac{1}{5} \tan^5 x + \frac{1}{3} \tan^3 x + C$

Questions & Discussion

• Tips for Students:

  • When performing substitution, every expression in $x$ must be replaced by an expression in $u$. If an integral contains both variables (e.g., $\int x u \, du$), the substitution is incomplete.

  • For $\int \frac{e^x}{e^x+1} dx$, the absolute value sign in $\ln$ is not strictly required because $e^x+1$ is always positive.

  • When dealing with $\sin^m x \cos^n x$ where both are even, always use double angle formulas to reduce the powers before integrating.

• Think Further:

  • Check correctness of $\int \sin 2x \, dx = \int \sin 2x \, \frac{1}{2} d(2x) = \frac{1}{2} [-\cos 2x] + C$.

  • Alternatively, $\int \sin 2x \, dx = \int 2 \sin x \cos x \, dx = \int 2 \sin x \, d(\sin x) = \sin^2 x + C$.

  • Are these the same? Yes, because $\sin^2 x = \frac{1 - \cos 2x}{2} = -\frac{1}{2} \cos 2x + \frac{1}{2}$. The constant discrepancy is absorbed into $C$.