# Integration By Parts

## Related Calculator: Integral (Antiderivative) Calculator with Steps

It is easy to compute integral int e^xdx but how to handle integrals like int xe^xdx?
In general if you have under integral product of functions that can be easily integrated separately then you should use integration by parts.

Formula for Integration by Parts. int udv=uv-int vdu.

Proof.

Using Product Rule we have that (f(x)g(x))'=f'(x)g(x)+f(x)g'(x).

Integrating both sides gives: int (f(x)g(x))'dx=int (f'(x)g(x)+f(x)g'(x))dx.

Which can be rewritten as f(x)g(x)=int f'(x)g(x)dx+int f(x)g'(x)dx or int f(x)g'(x)dx=f(x)g(x)-int f'(x)g(x)dx.

If we take u=f(x) and v=g(x) then du=f'(x)dx and dv=g'(x)dx, and above formula can be rewritten using Substitution Rule as int udv=uv-int vdu .

As can be seen Integration by Parts corresponds to the Product Rule (just like Substitution Rule corresponds to the Chain Rule).

In fact every differentiation rule has a corresponding integration rule, because these processes are inverse of each other.

Example 1. Evaluate int xe^xdx.

Let u=x and dv=e^xdx then du=(x)'dx=dx and v=int e^xdx=e^x.

So, int xe^xdx=int overbrace (x)^u overbrace (e^xdx)^(dv)=overbrace (x)^u overbrace (e^x)^v-int overbrace (e^x)^v overbrace(dx)^(du)=xe^x-e^x+C.

Note, that it is very important to choose appropriate u and v because wrong choice will only complicate integral.

For example, assume that instead of choosing u=x and dv=e^xdx in the above example, we choose u=e^x and v=xdx. Then du=(e^x)'dx=e^xdx and v=int xdx=x^2/2.

So, int xe^xdx=int overbrace (e^x)^u overbrace (xdx)^(dv)=overbrace (e^x)^u overbrace (1/2 x^2)^v-int overbrace (1/2 x^2)^v overbrace(e^xdx)^(du).

Although this equation is true, but integral int 1/2 x^2 e^xdx is more difficult to evaluate than the integral we started with.

Example 2. Evaluate int x^2 cos(x)dx.

Notice that x^2 becomes simpler when differentiated. Therefore, let u=x^2 and dv=cos(x)dx then du=(x^2)'dx=2xdx and v=int cos(x)dx=sin(x).

So, int x^2cos(x)dx=x^2sin(x)-int 2xsin(x)dx=x^2sin(x)-2int xsin(x)dx .

We obtained simpler integral, but still it is not obvious. Therefore, we apply integration by parts once more to integral int xsin(x)dx:

let u=x and dv=sin(x)dx then du=(x)'dx=dx and v=int sin(x)dx=-cos(x).

So, int overbrace (x)^u overbrace (sin(x)dx)^(dv)=overbrace (x)^u overbrace (-cos(x))^v-int overbrace (-cos(x))^(v) overbrace (dx)^(du)=-xcos(x)+int cos(x)dx=

=-xcos(x)+sin(x)+C .

Finally, int x^2 cos(x)dx=x^2sin(x)-2int xsin(x)dx=x^2sin(x)-2(-xcos(x)+sin(x)+C)=

=x^2sin(x)+2xcos(x)-2sin(x)+C_1 where C_1=-2C.

Example 2 showed that in some cases we need to apply integration by parts more than once.

Example 3. Evaluate int ln(x)dx.

Here, there is only one choice for u and v, namely, u=ln(x) and dv=dx, so du=(ln(x))'dx=1/x dx and v=int dx=x.

Therefore, int overbrace (ln(x))^u overbrace (dx)^(dv)=overbrace (ln(x))^u overbrace (x)^v-int overbrace (x)^v overbrace (1/xdx)^(du)=

=xln(x)-int dx=xln(x)-x+C.

Example 4. Evaluate int e^xsin(x)dx .

Neither e^x, nor sin(x) become simpler when differentiated, but we try choosing u=e^x and v=sin(x)dx.

Then du=(e^x)'dx=e^xdx and v=int sin(x)dx=-cos(x)dx.

Therefore, int e^x sin(x)dx=-e^xcos(x)-int -cos(x)e^xdx=-e^xcos(x)+int e^xcos(x)dx.

Obtained integral is not simpler than original, but it is not more difficult either. So, we apply integration by parts once more: let u=e^x and dv=cos(x)dx then du=(e^x)'dx=e^xdx and v=int cos(x)dx=sin(x).

So, int e^xcos(x)dx=e^xsin(x)-int e^xsin(x)dx.

We obtained initial integral! It seems that we obtained nothing because we arrived at int e^xsin(x)dx , which is where we started.

However, since int e^x sin(x)dx=-e^xcos(x)+int e^xcos(x)dx then

int e^x sin(x)dx=-e^xcos(x)+(e^x sin(x)-int e^xsin(x)dx) .

This can be regarded as equation with the unknown variable int e^x sin(x)dx.

This equation can be rewritten as 2int e^xsin(x)dx=e^x(sin(x)-cos(x)) .

Dividing by 2 and adding constant of integration yields final answer: int e^xsin(x)dx=1/2 e^x(sin(x)-cos(x))+C.

Example 5. Prove Reduction Formula int sin^n(x)dx=-1/n cos(x)sin^(n-1)(x)+(n-1)/n int sin^(n-2)(x)dx where n>=2 is integer.

Let u=sin^(n-1)(x) and dv=sin(x)dx then du=(sin^(n-1))'dx=(n-1)sin^(n-2)(x)(sin(x))'dx=(n-1)sin^(n-2)(x)cos(x)dx and v=int sin(x)dx=-cos(x).

So, int sin^n(x)dx=int sin^(n-1)sin(x)dx=-cos(x)sin^(n-1)(x)-int cos(x)(n-1)sin^(n-2)(x)cos(x)dx=

=-cos(x)sin^(n-1)(x)+(n-1) int sin^(n-2)(x) cos^2(x)dx.

Since cos^2(x)=1-sin^2(x) then (n-1)int sin^(n-2)(x)cos^2(x)dx=(n-1)int sin^(n-2)(x)(1-sin^2(x))dx=

=(n-1)int sin^(n-2)(x)dx-(n-1)int sin^(n-2)sin^2(x)dx=

=(n-1)int sin^(n-2)(x)dx-(n-1)int sin^n(x)dx.

Therefore,

int sin^n(x)dx=-cos(x)sin^(n-1)(x)+(n-1)int sin^(n-2)(x)dx-(n-1) int sin^n(x)dx .

This is an equation with unknown variable int sin^n(x)dx.

It can be rewritten as n int sin^n(x)dx=-cos(x)sin^(n-1)(x)+(n-1)int sin^(n-2)(x)dx.

So, finally we have that int sin^n(x)dx=-1/n cos(x)sin^(n-1)(x)+(n-1)/n int sin^(n-2)(x)dx.

Last question in this section is how to calculate definite integrals with the help of integration by parts. In fact it is very easy, just combine integration by parts with Newton-Leibniz formula: int_a^b udv=uv|_a^b-int_a^b vdu.

Example 6. Calculate int_0^1 tan^(-1)(x)dx.

Let u=tan^(-1)(x) and dv=dx then du=(tan^(-1)(x))'dx=1/(x^2+1)dx and v=int dx=x.

So, int_0^1 tan^(-1)(x)dx=xtan^(-1)(x)|_0^1-int_0^1 x/(x^2+1)dx=

=(1*tan^(-1)(1)-0*tan^(-1)(0))-int_0^1 x/(x^2+1)dx=pi/4-int_0^1 x/(x^2+1)dx.

To calculate int_0^1 x/(x^2+1)dx we use Substitution Rule: let t=x^2+1 then dt=(x^2+1)'dx=2xdx or xdx=1/2 dt.

Since x is changing from 0 to 1 then t is changing from 2*0=0 to 2*1=2.

So, int_0^1 x/(x^2+1)dx=int_1^2 1/t 1/2 dt=1/2 [ln|t|]_1^2=1/2(ln(2)-ln(1))=

=1/2(ln(2)-0)=1/2 ln(2).

Finally we have that int_0^1 tan^(-1)(x)=pi/4-int_0^1 x/(x^2+1)dx=pi/4-1/2ln(2).