Question

Value of ∫ Cos^2 (x) Sin^2 (x)dx.

(a) \(\frac{1}{8} [x-\frac{Cos(2x)}{2}]\)

(b) \(\frac{1}{4} [x-\frac{Cos(2x)}{2}]\)

(c) \(\frac{1}{8} [x-\frac{Sin(2x)}{2}]\)

(d) \(\frac{1}{4} [x-\frac{Sin(2x)}{2}]\)

This question was posed to me in examination.

This question is from Improper Integrals topic in portion Integral Calculus of Engineering Mathematics

Answer 1

Correct answer is (c) \(\frac{1}{8} [x-\frac{Sin(2x)}{2}]\)

To explain I would say: Add constant automatically

Given,f(x)=\(\int Cos^2 (x) Sin^2 (x)dx=\frac{1}{4} \int Sin^2 (2x) dx=\frac{1}{4} \int \frac{[1-Cos(2x)]}{2} dx=\frac{1}{8} [x-\frac{Sin(2x)}{2}]\)