Primitive function/Integration by parts/Section

Theorem

Let

$f,g\colon [a,b]\longrightarrow \mathbb {R}$

denote continuously differentiable functions. Then

${}\int _{a}^{b}f(t)g'(t)\,dt=fg|_{a}^{b}-\int _{a}^{b}f'(t)g(t)\,dt\,.$

Proof

Due to the product rule, the function ${}fg$ is a primitive function for ${}fg'+f'g$. Therefore,

${}\int _{a}^{b}f(t)g'(t)\,dt+\int _{a}^{b}f'(t)g(t)\,dt=\int _{a}^{b}{\left(fg'+f'g\right)}(t)\,dt=fg|_{a}^{b}\,.$

$\Box$

In using integration by parts, two things are to be considered. Firstly, the function to be integrated is usually not in the form ${}fg'$, but just as a product ${}uv$ (if there is no product, then this rule will probably not help, however, sometimes the trivial product ${}1u$ might help). Then for one factor, we have to find a primitive function, and we have to differentiate the other factor. If ${}V$ is a primitive function of ${}v$, then the formula reads

${}\int uv=uV-\int u'V\,.$

Secondly, integration by parts only helps when the integral on the right, i.e. ${}\int _{a}^{b}f'(t)g(t)\,dt$, can be integrated.

Example

We determine a primitive function for the natural logarithm ${}\ln x$, with integration by parts. We write ${}\ln x=1\cdot \ln x$, and we integrate the constant function ${}1$, and we differentiate the logarithm. Then

${}\int _{a}^{b}\ln x\,dx=(x\cdot \ln x)|_{a}^{b}-\int _{a}^{b}x\cdot {\frac {1}{x}}\,dx=(x\cdot \ln x)|_{a}^{b}-\int _{a}^{b}1\,dx=(x\cdot \ln x)|_{a}^{b}-x|_{a}^{b}\,.$

So a primitive function is ${}x\cdot \ln x-x$.

Example

A primitive function for the sine function ${}\sin x$ is ${}-\cos x$. In order to find a primitive function for ${}\sin ^{n}x$, we use integration by parts to get a recursive relation to a power with a smaller exponent. To make this more precise, we work over an interval, the primitive function shall start at ${}0$ and have the value ${}0$ there. For ${}n\geq 2$, with integration by parts, we get

${}{\begin{aligned}\int _{0}^{x}\sin ^{n}t\,dt&=\int _{0}^{x}\sin ^{n-2}t\cdot \sin ^{2}t\,dt\\&=\int _{0}^{x}\sin ^{n-2}t\cdot {\left(1-\cos ^{2}t\right)}\,dt\\&=\int _{0}^{x}\sin ^{n-2}t\,dt-\int _{0}^{x}{\left(\sin ^{n-2}t\cos t\right)}\cos t\,dt\\&=\int _{0}^{x}\sin ^{n-2}t\,dt-{\frac {\sin ^{n-1}t}{n-1}}\cos t|_{0}^{x}-{\frac {1}{n-1}}{\left(\int _{0}^{x}\sin ^{n}t\,dt\right)}.\end{aligned}}$

Multiplication with ${}n-1$ and rearranging yields

${}n\int _{0}^{x}\sin ^{n}t\,dt=(n-1)\int _{0}^{x}\sin ^{n-2}t\,dt-\sin ^{n-1}x\cos x\,.$

In particular, for ${}n=2$, we have

${}\int _{0}^{x}\sin ^{2}t\,dt={\frac {1}{2}}{\left(x-\sin x\cos x\right)}\,.$