Question

Integration by Substitution?

Show that:
`int_0^(pi/2)sin^2(x)dx=int_0^(pi/2)cos^2(x)dx`

Answer 1

`int_0^(pi/2) cos^2(x)dx = int_0^(pi/2) sin^2(x)dx`

`int_0^(pi/2) cos^2(x)dx - int_0^(pi/2) sin^2(x)dx = 0`

using the linearity of the integral:

`int_0^(pi/2) (cos^2(x)dx - sin^2(x))dx = 0`

and the trigonometric identity: `cos(2alpha) = cos^2alpha -sin^2alpha`

`int_0^(pi/2) cos(2x)dx = 0`

In fact:

`int_0^(pi/2) cos(2x)dx = 1/2 [sin(2x)]_0^(pi/2) = 0`

which proves the point.

Answer 2

`int_0^(pi/2)sin^2xdx=int_0^(pi/2)cos^2xdx iff`

`int_0^(pi/2)cos^2xdx-int_0^(pi/2)sin^2xdx=0 iff`

`int_0^(pi/2)(cos^2x-sin^2x)dx=0iff`

`int_0^(pi/2)cos2xdx=0 iff`

Let's substitude `u=2x =>du=2dx=>dx=(du)/2` :

`int_0^picosu(du)/2=0iff`

`1/2[sinu]_0^pi=0iff(sinpi-sin0)=0iff`

`0=0` which is true, so the first statement is true.

Answer 3

Kindly, refer to the Explanation.

Explanation:

Using the well-known Result : `int_0^af(x)dx=int_0^af(a-x)dx,`

we have, `int_0^(pi/2)sin^2xdx=int_0^(pi/2)sin^2(pi/2-x)dx,`

`=int_0^(pi/2)cos^2xdx.`

Hence, the Proof.