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Question: Solve the integral \(I = \int\limits_0^\pi {{{\sin }^2}x{\text{ }}dx} \)....

Solve the integral I=∫0πsin⁡2x dxI = \int\limits_0^\pi {{{\sin }^2}x{\text{ }}dx} .

Explanation

Solution

Hint – In this question we have to evaluate the given integral so use the trigonometric half angle formula to simplify the trigonometric part inside the integral then use the integration of basic trigonometric terms to reach the answer.

“Complete step-by-step answer:”
Given integral
I=∫0πsin⁡2x dxI = \int\limits_0^\pi {{{\sin }^2}x{\text{ }}dx}
As we know sin⁡2x=(1−cos⁡2x2){\sin ^2}x = \left( {\dfrac{{1 - \cos 2x}}{2}} \right) so, substitute this value in given integral we have,
I=∫0π(1−cos⁡2x2) dxI = \int\limits_0^\pi {\left( {\dfrac{{1 - \cos 2x}}{2}} \right){\text{ }}dx}
I=12∫0π(1−cos⁡2x) dxI = \dfrac{1}{2}\int\limits_0^\pi {\left( {1 - \cos 2x} \right){\text{ }}dx}
Now as we know integration of constant is x and ∫cos⁡nx dx=sin⁡nxn+c\int {\cos nx{\text{ }}dx} = \dfrac{{\sin nx}}{n} + c so, use this property in above integral we have,
I=12[x−sin⁡2x2]0πI = \dfrac{1}{2}\left[ {x - \dfrac{{\sin 2x}}{2}} \right]_0^\pi
Now apply integral limit we have,
I=12[π−sin⁡2π2−(0−sin⁡02)]I = \dfrac{1}{2}\left[ {\pi - \dfrac{{\sin 2\pi }}{2} - \left( {0 - \dfrac{{\sin 0}}{2}} \right)} \right]
Now as we know the value of sin⁡2π\sin 2\pi and sin⁡0\sin 0 is zero so, substitute this value in given integral we have,
I=12[π−0−0]=π2I = \dfrac{1}{2}\left[ {\pi - 0 - 0} \right] = \dfrac{\pi }{2}
So, this is the required value of the integral.
Thus, this is the required answer.

Note – Whenever we face such types of problems the key concept involved is to simplify the inside entity of the integration to the basic level so that the direct integration formula for trigonometric terms could be applied directly. This will help you to get on the right track to reach the answer.