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Question

Question: How do you integrate \[\smallint {\sin ^2}(2x)dx\] ?...

How do you integrate sin2(2x)dx\smallint {\sin ^2}(2x)dx ?

Explanation

Solution

In this question we have to evaluate the given integral, therefore, 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 solution:
Given integral
I=sin2(2x)dxI = \smallint {\sin ^2}(2x)dx
As per our prior knowledge of trigonometric functions, we know that,
sin2x=(1cos2x2){\sin ^2}x = \left( {\dfrac{{1 - \cos 2x}}{2}} \right) so, after substituting this value integral we have,
I=(1cos2(2x)2)dxI = \int {\left( {\dfrac{{1 - \cos 2(2x)}}{2}} \right)dx}
I=12(1cos2(2x))dxI = \dfrac{1}{2}\int {(1 - \cos 2(2x))dx}
Here , Put 2x=t2x = t
taking derivative on both sides with respect to x we get
d(2x)dx=d(t)dx\dfrac{{d(2x)}}{{dx}} = \dfrac{{d(t)}}{{dx}}
2=dtdx\Rightarrow 2 = \dfrac{{dt}}{{dx}} or 2dx=dt\Rightarrow 2dx = dt
Therefore, we get the transformed equation as
I=14(1cos2t)dtI = \dfrac{1}{4}\int {(1 - \cos 2t)dt}
Now as we know that the integration of a constant is xx and cos(nx)dx=sin(nx)n+c\int {\cos (nx)dx = \dfrac{{\sin (nx)}}{n}} + c hence, using this property in above integral we get
I=14[tsin2t2]I = \dfrac{1}{4}\left[ {t - \dfrac{{\sin 2t}}{2}} \right]
Substitute the value if tt to get the answer in terms of xx we have
I=14[2xsin4x2]I = \dfrac{1}{4}\left[ {2x - \dfrac{{\sin 4x}}{2}} \right]
Thus, this is the required answer.
So, the correct answer is “ I=14[2xsin4x2]I = \dfrac{1}{4}\left[ {2x - \dfrac{{\sin 4x}}{2}} \right]”.

Note : Whenever we are required to solve these type of questions the key concept is to simplify the inside entries of the integration to the basic level so that the direct integration formulas for the trigonometric terms could be applied.