§5.4 Integration by parts (continued)

Using Integration by Parts Multiple Times

Evaluating Definite Integrals Using Integration by Parts

When

Summary

Applying integration by parts twice to the integral $\int t^2e^t\,dt$ yields which antiderivative?

Evaluating $\int x^2\sin(x)\,dx$ requires by parts twice, and the final antiderivative is

For $\int e^z\sin(z)\,dz$, applying parts twice and solving for the unknown integral gives

Which combination of techniques evaluates $\int x^5\cos(x^3)\,dx$?

Evaluating $\int s^2e^{3s}\,ds$ by parts twice gives

Using $\frac{t^2}{1+t^2} = 1 - \frac{1}{1+t^2}$, the integral $\int t\arctan(t)\,dt$ equals

The integral $\int e^{2x}\cos(e^{2x})\,dx$ differs from $\int e^{2x}\cos(e^x)\,dx$ because

Among $\int x^3\sin(x^4)\,dx$ and $\int x^7\sin(x^4)\,dx$, the correct classifications are

For $F(x) = \int_0^x te^{-2t}\,dt$, by parts gives $F(x) = -\frac{1}{2}xe^{-2x} - \frac{1}{4}e^{-2x} + \frac{1}{4}$. Compute $F(1)$, rounded to four decimals.

Using the antiderivative $-t\cos(t) + \sin(t)$ from the by parts computation, evaluate $\int_0^{\pi/2}t\sin(t)\,dt$.

Using the antiderivative $\frac{1}{4}t^4\ln(t) - \frac{1}{16}t^4$, evaluate $\int_1^e t^3\ln(t)\,dt$ as a decimal, rounded to four places.

When integration by parts reproduces the original integral on the other side, collecting like integrals and solving algebraically is a valid way to finish.