§5.4 Integration by parts

Introduction

Reversing the Product Rule: Integration by Parts

Some Subtleties with Integration by Parts

Integration by parts reverses the product rule and states that $\int u\,dv$ equals

Evaluate $\int x\ln(x)\,dx$ using parts.

Evaluate $\int\ln(z)\,dz$ by parts with $dv = dz$.

Applying integration by parts to this product of a power and a trig function, evaluate $\int x\cos(x)\,dx$.

Evaluate $\int 4x\sin(3x)\,dx$ by parts.

Evaluate $\int z\sec^2(z)\,dz$, using the known antiderivative of tangent in the remainder.

Evaluate $\int\arctan(x)\,dx$ by parts with $u = \arctan(x)$ and $dv = dx$.

Using $z = t^2$ first, $\int t^3\sin(t^2)\,dt$ transforms into $\frac{1}{2}\int z\sin(z)\,dz$ and evaluates to

Using $z = s^3$ so that $dz = 3s^2\,ds$, the integral $\int s^5 e^{s^3}\,ds$ becomes $\frac{1}{3}\int z e^z\,dz$ and equals

Writing $e^{2t} = e^t \cdot e^t$ and substituting $z = e^t$, the integral $\int e^{2t}\cos(e^t)\,dt$ evaluates to

Using the antiderivative $\frac{1}{2}x^2\ln(x) - \frac{1}{4}x^2$ from the by parts computation, evaluate $\int_0^1 x\ln(x)\,dx$ as a decimal.

Using the antiderivative $-x\cos(x) + \sin(x)$, evaluate the definite integral $\int_0^{\pi}x\sin(x)\,dx$, rounded to three decimals.

For $\int te^{-t}\,dt$, choosing $u = t$ and $dv = e^{-t}\,dt$ leads to $-te^{-t} - e^{-t} + C$.