Exercise 3.4 — Integration by Parts

All questions in this exercise are listed below. Click on a question to view its solution.

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Key Concepts

This exercise focuses on integration by parts, a technique used when the integrand is a product of two functions.

Integration by Parts Formula

$\int udv=uv-\int vdu$

LIATE Rule

The LIATE rule guides the selection of $u$ and $dv$:

Choose the function highest on this list as $u$; assign the remaining factor to $dv$.

Repeated Integration by Parts (Tabular Method)

When the integrand contains a polynomial factor $x^n$ multiplied by an exponential or trigonometric function, apply integration by parts repeatedly until the polynomial reduces to a constant.

Example: $\int x^2{e}^{x}dx$

Result: $\int x^2{e}^{x}dx=e^x(x^2-2x+2)+C$

Cyclic Integrals

For integrals like $\int e^x\cos xdx$ or $\int e^x\sin xdx$, applying integration by parts twice returns the original integral $I$ on the right-hand side. Solve algebraically:

$I=e^x\sin x+e^x\cos x-I⟹2I=e^x(\sin x+\cos x)⟹I=\frac{e^x(\sin x+\cos x)}{2}+C$

Integration of Inverse Trigonometric Functions

For integrals like $\int {\sin }^{-1}xdx$, $\int {\cos }^{-1}xdx$, $\int {\tan }^{-1}xdx$: set $u$ = the inverse trig function and $dv=dx$.

Combining Substitution with Integration by Parts

Some integrals require a substitution first to simplify the integrand before applying integration by parts.

Example: $\int \sin (\ln x)dx$ — substitute $t=\ln x$ first, then apply cyclic IBP.

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Important Formulas

Integration by Parts: $\int udv=uv-\int vdu$

Cyclic Integral Result (for $\int e^{ax}\cos (bx)dx$): $\int e^{ax}\cos (bx)dx=\frac{e^{ax}}{a^2+b^2}(a\cos (bx)+b\sin (bx))+C$

Standard Logarithmic Integral: $\int \ln xdx=x\ln x-x+C$

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Summary

This exercise develops proficiency in integration by parts across diverse function types. The key strategy involves selecting $u$ using the LIATE priority rule: logarithmic and inverse trig functions are typically chosen as $u$, while exponentials and trig functions are assigned to $dv$. For integrands with polynomial factors, repeated integration by parts (tabular method) is efficient until the polynomial reduces to a constant. Some questions require combining substitution with integration by parts. The cyclic pattern arises when integration by parts twice yields an equation solvable for the original integral. Inverse trig integrals are handled by setting $u$ as the inverse function and $dv=dx$.