Integration by Substitution?

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

3 Answers
Jul 25, 2017

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.

Jul 25, 2017

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.

Jul 25, 2017

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.