devarshi-dt-logo

Question:

The integral ∫sin2xcos2x(sin3x+cos3x)²dx is equal to :

1(1+cot2x)+c

⅓(1+tan3x)+c

−cos3x3(1+sin3x)+c

sin3x1+cos3x+c

Solution:

I=∫sin2xcos2x(sin3x+cos3x)²dx
Dividing both denominator and numerator by cos⁶x, we get
=∫tan²xsec²x(1+tan3x)²dx
Substitute 1+tan3x=t ⇒ 3tan²xsec²xdx=dt
The integral becomes
⅓∫dt/t² = ⅓ × 1/t + C = ⅓(1+tan3x)+C