Q.26. The integral \int \frac{\sin^{2}x\cos^{2}x}{\left(\sin^{5}x+\cos^{3}x\sin^{2}x+\sin^{3}x\cos^{2}x+\cos^{5}x\right)^{2}} \,dx is equal to
A. \frac{1}{3\left(1+\tan^{3}x\right)}+c , where c is a constant of integration.
B. \frac{-1}{3\left(1+\tan^{3}x\right)}+c , where c is a constant of integration.
C. \frac{1}{1+\cot^{3}x}+c , where c is a constant of integration.
D. \frac{-1}{1+\cos^{3}x}+c , where c is a constant of integration.
Answer: B. \frac{-1}{3\left(1+\tan^{3}x\right)}+c , where c is a constant of integration.
Explanation :-
Let
I = \int \frac{\sin^{2}x\cos^{2}x}{\left(\sin^{5}x+\cos^{3}x\sin^{2}x+\sin^{3}x\cos^{2}x+\cos^{5}x\right)^{2}} \,dx = \int \frac{\sin^{2}x\cos^{2}x}{\left(\sin^{5}x+\sin^{3}x\cos^{2}x+\cos^{3}x\sin^{2}x+\cos^{5}x\right)^{2}} \,dx = \int \frac{\sin^{2}x\cos^{2}x}{\left[\sin^{3}x\left(\sin^{2}x+\cos^{2}x\right)+\cos^{3}x\left(\sin^{2}x+\cos^{2}x\right)\right]^{2}} \,dx = \int \frac{\sin^{2}x\cos^{2}x}{\left(\sin^{3}x+\cos^{3}x\right)^{2}} \,dx =\int \frac{sec^{2}xtan^{2}x}{(1+tan^{3}x)^{2}}dx…. ( Dividing numerator and denominator by cos6x)
1+tan3 x = t
Differentiating w.r.t x,we get
3tan2 x sec2x dx = dt
tan2x sec2x dx = \frac{1}{3}
=\frac{1}{3}\int \frac{1}{t^{2}}dt =\frac{-1}{3t}+c= =\frac{-1}{3(1+tan^{3}x)}+c