Previous Year NEET Question- Definite Integral

1. Evaluate ∫0π/4 sin2x.cos2x dx. (NEET 2019)

We know that sin2x = 2sinxcosx. Therefore,

∫0π/4 sin2x.cos2x dx = ∫0π/4 2sinx.cosx.cos2x dx = ∫0π/4 2sinx.cos2x.cosx dx

Now, let u = sinx and v = cosx. Then, du = cosx dx and dv = -sinx dx.

So, ∫0π/4 2sinxcos2xcosx dx = ∫u dv = uv|0π/4 - ∫v du = uv|0π/4 - ∫v du

We know that ∫sin2x dx = -cos2x/2 + C. So,

∫0π/4 sin2xcos2x dx = (1/2)∫0π/4 sin4x dx = -(1/8)cos4x|0π/4 = -(1/8)[cosπ - cos0] = -(1/8)[-1 - 1] = 1/4