Examples

A and T factors

Compute the integral: ${\displaystyle \int x\cos xdx}$

Solution

(1) Choose $u$ and $v^{\prime }$:

Set $u(x)=x$ because $x$ simplifies when differentiated.

(By the trick: $x$ is Algebraic, i.e. more “$u$”, and $\cos x$ is Trig, more “$v$”.)

Remaining factor must be $v^{\prime }=\cos x$.

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(2) Compute $u^{\prime }$ and $v$:

$$\begin{array}{cc}u=x& \gg \gg u^{\prime }=1\\ & \\ v^{\prime }=\cos x& \gg \gg v=\sin x\end{array}$$

Key chart:

$$\begin{array}{cc}u=x& v^{\prime }=\cos x\\ u^{\prime }=1& v=\sin x\end{array}\begin{array}{c}⟶\int u\cdot v^{\prime }\text{original}\\ ⟶\int u^{\prime }\cdot v\text{final}\end{array}$$

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(3) Evaluate IBP formula:

$$\begin{array}{c}\int u^{\prime }vdx=uv-\int u{v}^{\prime }dx\\ \\ \gg \gg \int x\cos xdx=x\sin x-\int 1\cdot \sin xdx\\ \\ \gg \gg x\sin x+\cos x+C\end{array}$$

Why IBP?

The rationale of IBP is that $\int 1\cdot \sin xdx$ is easier to compute than $\int x\cos xdx$.

Hidden A

Compute the integral:

$$\int \ln xdx$$

Solution

(1) Choose $u=\ln x$:

Because Log is farthest left in LIATE.

Therefore we must choose $v^{\prime }(x)=1$.

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(2) Compute $u^{\prime }$ and $v$:

We have $u^{\prime }=\frac{1}{x}$ and $v=x$. Obtain chart:

$$\begin{array}{cc}u=\ln x& v^{\prime }=1\\ u^{\prime }=1/x& v=x\end{array}\begin{array}{c}⟶\int u\cdot v^{\prime }\text{original}\\ ⟶\int u^{\prime }\cdot v\text{final}\end{array}$$

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(3) Evaluate IBP formula:

$$\begin{array}{c}\int u^{\prime }vdx=uv-\int u{v}^{\prime }dx\\ \\ \gg \gg \int \ln x\cdot 1dx=x\ln x-\int \frac{1}{x}\cdot xdx\end{array}$$

Perform new integration:

$$-\int \frac{1}{x}\cdot xdx\gg \gg -\int 1dx\gg \gg -x+C$$

Final answer is: $x\ln x-x+C$