Question #24fd5

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Answer 1 · 2017-04-20T07:15:14

$pi/4.$

Respected Eddie has suggested the Method to solve the

Problem.

Here is another way to Solve it.

For that, we need the following useful Result :

Result : $int_0^a f(x)dx=int_0^af(a-x)dx; a>0, and, f" is cont. on "[0,a].$

So, $I=int_0^(pi/2)sin^2xdx......(1)$

$rArr I=int_0^(pi/2)sin^2(pi/2-x)dx=int_0^(pi/2)cos^2xdx......(2)$

$:., (1)+(2) rArr I+I=2I=int_0^(pi/2)(sin^2x+cos^2x)dx, i.e.,$

$2I=int_0^(pi/2)(1)dx=[x]_0^(pi/2)=pi/2$

$:. I=pi/4.$

Enjoy Maths.!