Introduction

Welcome to this guide on Integration by Parts of the FiveHive Calculus BC course. This article will guide you through what is integration by parts and how to perform this technique.

Deriving Integration By Parts Formula

Start from the product rule of differentiation:

$\frac{d}{dx}(uv)=u^{\prime }v+v^{\prime }u$

If we multiply both side by $dx$ on both sides and take the indefinite integral, we have:

$\int d(uv)=\int u^{\prime }vdx+\int v^{\prime }udx$

Evaluate the integral, we have 

$\int udv=uv-\int vdu$

Here we introduced a new variable that $v=vdx$ and $u=u^{\prime }dx$, you can see them as completely new variable and have no relation with the original $u$ and $v$

Integrating Using Integration By Parts

Indefinite Integral

The formula for integration by parts is simple:

$\int udv=uv-\int vdu$

Consider this integral:

$\int x{e}^{x}dx$

Let $u=x$, $dv=e^{x}dx$, thus $du=dx$, $v=e^x$, by integration of parts, we have 

$\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=x{e}^x-e^x+C$

The key of integration by parts is to find the correct $dv$, it must be easy to integrate, here are some expressions you should consider choosing as $dv$:

Let's take a look at another another example:

$\int x\cos xdx$

Let $u=x$ and $dv=\cos xdx$, thus $du=dx$ and $v=\sin x$, we can apply integration by parts:

$\int x\cos xdx=x\sin x-\int \sin xdx=x\sin x-\cos x+C$

Applying Integration By Parts more than once

Sometimes we need to apply integration by parts more than once, for example:

$\int e^x\sin xdx$

Let $u=\sin x$, $dv=e^{x}dx$, thus $du=\cos xdx$, $v=e^x$, by integration by parts:

$\int e^x\sin xdx=e^x\sin x-\int e^x\cos xdx$

Here we arrived at a new integral of $\int e^x\cos xdx$, which again can be evaluated by integration by parts:

Let $u=\sin x$, $dv=e^{x}dx$, thus $du=-\cos xdx$ and $v=e^x$, by integration by parts:

$\int e^x\cos xdx=e^x\cos x+\int e^x\sin xdx$

Here we see a problem, it seems that we need to evaluate our original integral to get an expression for our original integral, but this can be easily bypassed,

notice that

$\int e^x\sin xdx=e^x\sin x-e^x\cos x-\int e^x\sin xdx$

We can treat our original integral as an unknown value and solve this equation, thus:

$2\int e^x\sin xdx=e^x\sin x-e^x\cos x+C$

$\int e^x\sin xdx=\frac{1}{2}{e}^x(\sin x-\cos x)+C$

Inverse Trigonometric Function

Integration by parts can be used to calculate the indefinite integral of inverse trig function:

$\int \arcsin xdx$

We immediately see a $udv$ structure, let $u=\arcsin x$ and $dv=dx$, thus $du=\frac{1}{\sqrt{1-x^2}}dx$, $v=x$:

$\int \arcsin xdx$

$=x\arcsin x-\int \frac{x}{\sqrt{1-x^2}}dx$

The last integral can be evaluate with a u-substitution, thus we arrive at our final example:

$\int \arcsin xdx=x\arcsin x+\sqrt{1-x^2}+C$

Definite Integral

For definite integral, the formula for integration by parts turn to:

${\int }_a^{b}udv=uv{|}_a^b-{\int }_a^{b}vdu$

Consider this integral:

${\int }_0^{\frac{1}{2}}\arcsin xdx$

Previously we derived that

${\int }_0^{\frac{1}{2}}\arcsin xdx$

$=x\arcsin x{|}_0^{\frac{1}{2}}-{\int }_0^{\frac{1}{2}}\frac{x}{\sqrt{1-x^2}}dx$

$=\frac{1}{2}\arcsin \frac{1}{2}+\sqrt{1-x^2}{|}_0^{\frac{1}{2}}$

$=\frac{\pi }{12}+\frac{\sqrt{3}}{2}-1$

Practice