Question

निम्न के मान बताये।
1. | sin^2 2x dx​

Answer 1

We have that

cos(4x)=cos2(2x)−sin2(2x)⇒

cos(4x)=1−sin2(2x)−sin2(2x)⇒

2sin2(2x)=1−cos(4x)⇒

sin2(2x)=12⋅(1−cos(4x))

Hence we have that

∫sin2(2x)dx=∫[12⋅(1−cos(4x))]dx=x2−sin(4x)8+c

Footnote

We used the following trig identities

1) cos(2⋅a)=cos2a−sin2a

2) cos2a=1−sin2a

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