Calculus II: Power Reduction and Integration by Parts Flashcards

Algorithmic Power Reduction for Secant and Tangent

• The lecture introduces an algorithmic approach to finding the integrals of high-powered trigonometric functions, specifically secant and tangent, using power reduction formulas.

• This method is more efficient than the manual method of turning powers of secant into tangents (or vice versa) repeatedly, especially as the exponent $n$ increases. It is described as the logic a computer program would follow to solve these integrals.

• Power Reduction Formula for Secant:

  • The formula for a power of secant ($n$) is:     $\int \sec^n(x)\,dx = \frac{1}{n-1} \sec^{n-2}(x) \tan(x) + \frac{n-2}{n-1} \int \sec^{n-2}(x)\,dx$

  • This formula reduces the power of the integral by 2 in each iteration until the remaining integral is either the integral of $\sec(x)$ (if $n$ is odd) or $\sec^0(x) = 1$ (if $n$ is even).

• Example: Evaluating $\int \sec^{3}(x)\,dx$:

  • Identify $n = 3$.

  • Applying the formula: $\frac{1}{3-1} \sec^{3-2}(x) \tan(x) + \frac{3-2}{3-1} \int \sec^{3-2}(x)\,dx$

  • Simplified: $\frac{1}{2} \sec(x) \tan(x) + \frac{1}{2} \int \sec(x)\,dx$

  • Result: The anti-derivative of $\sec(x)$ is $\ln|\sec(x) + \tan(x)|$.

  • Final Solution: $\frac{1}{2} \sec(x) \tan(x) + \frac{1}{2} \ln|\sec(x) + \tan(x)| + C$

• Example: Evaluating $\int \sec^5(x)\,dx$:

  • Identify $n = 5$.

  • First reduction: $\frac{1}{4} \sec^3(x) \tan(x) + \frac{3}{4} \int \sec^3(x)\,dx$

  • We then substitute the previously found value for $\int \sec^3(x)\,dx$ into this expression:     $\frac{1}{4} \sec^3(x) \tan(x) + \frac{3}{4} \left( \frac{1}{2} \sec(x) \tan(x) + \frac{1}{2} \ln|\sec(x) + \tan(x)| \right) + C$

  • Final simplified form (distributing the coefficient): $\frac{1}{4} \sec^3(x) \tan(x) + \frac{3}{8} \sec(x) \tan(x) + \frac{3}{8} \ln|\sec(x) + \tan(x)| + C$

• Power Reduction Formula for Tangent:

  • This formula contains a subtraction rather than an addition between terms because converting tangents to secants (via $\tan^2(x) = \sec^2(x) - 1$) introduces a negative sign.

  • The formula is:     $\int \tan^n(x)\,dx = \frac{\tan^{n-1}(x)}{n-1} - \int \tan^{n-2}(x)\,dx$

  • This formula works for all $n$ as long as the functions are defined on their domain; it is applied repeatedly until the remaining integral becomes $\int \tan(x)\,dx$ or $\int 1\,dx$.

• Example: Evaluating $\int \tan^4(x)\,dx$:

  • Identify $n = 4$.

  • First reduction: $\frac{1}{3} \tan^3(x) - \int \tan^2(x)\,dx$

  • Apply formula again for $n = 2$:     $\frac{1}{3} \tan^3(x) - \left( \frac{\tan^1(x)}{1} - \int \tan^0(x)\,dx \right)$     $\frac{1}{3} \tan^3(x) - \tan(x) + \int 1\,dx$

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

Integration by Parts (IBP) Foundations

• The IBP Formula:   $\int u\,dv = uv - \int v\,du$

• Strategy for Choosing $u$ and $dv$:

  • Choose $u$ such that its derivative $\frac{du}{dx}$ is simpler than $u$ itself.

  • Choose $dv$ such that its anti-derivative $v$ is easy to calculate.

  • The lecturer notes that for polynomials multiplied by exponentials (e.g., $x^4 e^x$), the polynomial is nearly always chosen as $u$ to eventually reduce the power to zero through repeated integration.

• Example: Integration of $\int x^4 e^x\,dx$:

  • Let $u = x^4$, $du = 4x^3\,dx$.

  • Let $dv = e^x \,dx$, $v = e^x$.

  • Applying IBP: $x^4 e^x - \int 4x^3 e^x\,dx$

  • This requires doing integration by parts three more times (total of four iterations) to fully evaluate until the power of $x$ becomes 0.

• Example: Integration with Nested Logarithms $\int (\ln(x))^4\,dx$:

  • Note: The exponent 4 is outside the natural log, meaning it cannot be brought to the front as a coefficient.

  • Let $u = (\ln(x))^4$. Use the chain rule for the derivative: $du = 4(\ln(x))^3 \times \frac{1}{x}\,dx$.

  • Let $dv = dx$, so $v = x$.

  • Result of first step: $x(\ln(x))^4 - \int x \left( \frac{4(\ln(x))^3}{x} \right)\,dx = x(\ln(x))^4 - \int 4(\ln(x))^3\,dx$

  • This process must be repeated until the the logarithm factor is eliminated.

Advanced Integration by Parts Techniques

• Linear Arguments in Logs ($\int \ln(3r - 4)\,dr$):

  • This can be solved two ways:

  1. Direct IBP: Set $u = \ln(3r - 4)$ and $dv = dr$. This leads to a complicated algebraic integral $\int \frac{3r}{3r-4}\,dr$. This requires an algebraic trick: rewrite $3r$ as $(3r - 4) + 4$ to split the fraction into $1 + \frac{4}{3r-4}$.

  2. Initial U-Substitution (Recommended): Let $w = 3r - 4$, then $dw = 3\,dr$ or $dr = \frac{1}{3}\,dw$. The integral becomes $\frac{1}{3} \int \ln(w)\,dw$, which is a standard IBP setup ($u = \ln(w)$, $dv = dw$).

• Cyclical Integrals ($\int e^x \sin(x)\,dx$):

  • Neither a derivative of $e^x$ nor $\sin(x)$ ever reaches zero.

  • First IBP step: Choose $u = \sin(x)$, $dv = e^x\,dx$. Result: $e^x \sin(x) - \int e^x \cos(x)\,dx$.

  • Second IBP step: Perform IBP on $\int e^x \cos(x)\,dx$. Choose $u = \cos(x)$, $dv = e^x\,dx$. Result: $e^x \cos(x) - \int e^x (-\sin(x))\,dx$.

  • Algebraic solve: Substitute back to get $I = e^x \sin(x) - (e^x \cos(x) + I)$, where $I$ is the original integral.

  • Combine terms: $2I = e^x \sin(x) - e^x \cos(x)$.

  • Result: $\int e^x \sin(x)\,dx = \frac{1}{2} (e^x \sin(x) - e^x \cos(x)) + C$.

Definite Integrals with Integration by Parts

• Definite Integral Example: $\int_2^4 t^3 \ln(5t)\,dt$:

  • In this case, choose $u = \ln(5t)$ because the derivative makes the log part disappear: $du = \frac{5}{5t}\,dt = \frac{1}{t}\,dt$.

  • Let $dv = t^3\,dt$, so $v = \frac{t^4}{4}$.

  • IBP evaluation: $\left[ \frac{t^4}{4} \ln(5t) \right]_2^4 - \int_2^4 \frac{t^4}{4} \times \frac{1}{t}\,dt$

  • $\left[ \frac{t^4}{4} \ln(5t) \right]_2^4 - \int_2^4 \frac{t^3}{4}\,dt = \left[ \frac{t^4}{4} \ln(5t) - \frac{t^4}{16} \right]_2^4$

  • Plug in upper bound ($t=4$): $\frac{256}{4} \ln(20) - \frac{256}{16} = 64 \ln(20) - 16$

  • Plug in lower bound ($t=2$): $\frac{16}{4} \ln(10) - \frac{16}{16} = 4 \ln(10) - 1$

  • Subtracting bounds: $(64 \ln(20) - 16) - (4 \ln(10) - 1) = 64 \ln(20) - 4 \ln(10) - 15$

• Definite Integral Example: $\int_1^e 4t^2 \ln(t)\,dt$:

  • Pull out the constant: $4 \int_1^e t^2 \ln(t)\,dt$.

  • Set $u = \ln(t)$, $du = \frac{1}{t}\,dt$.

  • Set $dv = t^2\,dt$, $v = \frac{t^3}{3}$.

  • Evaluating: $4 \left( \left[ \frac{t^3}{3} \ln(t) \right]_1^e - \int_1^e \frac{t^2}{3}\,dt \right) = 4 \left[ \frac{t^3}{3} \ln(t) - \frac{t^3}{9} \right]_1^e$

  • Evaluate at $e$: $\frac{4e^3}{3} \ln(e) - \frac{4e^3}{9} = \frac{4e^3}{3} - \frac{4e^3}{9}$.

  • Evaluate at $1$: $\frac{4(1)^3}{3} \ln(1) - \frac{4(1)^3}{9} = 0 - \frac{4}{9}$.

  • Result: $\frac{12e^3}{9} - \frac{4e^3}{9} - (-\frac{4}{9}) = \frac{8e^3 + 4}{9}$.

Inverse Trigonometric Integrals

• Evaluating $\int \tan^{-1}(x)\,dx$:

  • Treat as single function. Set $u = \tan^{-1}(x)$ and $dv = dx$.

  • Derivatives: $du = \frac{1}{x^2+1}\,dx$, $v = x$.

  • IBP: $x \tan^{-1}(x) - \int \frac{x}{x^2+1}\,dx$.

  • To solve the remaining integral, use U-substitution: $u = x^2 + 1$, $du = 2x\,dx$. This yields $\frac{1}{2} \int \frac{du}{u} = \frac{1}{2} \ln|u|$.

  • Final result: $x \tan^{-1}(x) - \frac{1}{2} \ln(x^2 + 1) + C$.

  • Note: Absolute value bars are dropped for $x^2 + 1$ because that expression is always positive.

Questions & Discussion

• Q: Does $\tan^0(x)$ simplify to 1?

  • A: Yes, anything raised to the power of zero is 1. While $0^0$ is technically undefined, for the purposes of these integrals (power reduction), the function behaves as a constant 1 across the valid domain.

• Q: When do we use these formulas vs. standard U-sub?

  • A: For something like $\tan^6(x)$, these formulas are significantly faster. For combinations of secant and tangent where powers are mixed, one must still check the rules for even/odd powers to decide on the best strategy.

• Q: Are there formulas for combined secant and tangent?

  • A: No single algorithmic formula captures all combinations. You must evaluate the specific case (e.g., if secant is even, if tangent is odd).

• Q: How to provide answers on tests?

  • A: The exact answer (with natural logs and exponents) is preferred over decimal approximations from a calculator unless specifically asked for rounding.