Question

The integral $\int \frac{{\sin }^{2}x{\cos }^{2}x}{{({\sin }^{5}x+{\cos }^{3}x{\sin }^{2}x+x{\cos }^{2}x+{\cos }^{5}x)}^2}dx$ is equal to:

• (A) \(\frac{ 1}{1+ \cot^{3}x} + C\)
• (B) \(\frac{ -1}{1+ \cot^{3}x} + C\)
• (C) \(\frac{ 1}{ 3\big(1+ \cot^{3}x\big) } + C\)
• (D) \(\frac{ -1}{ 3\big(1+ \cot^{3}x\big) } + C\)

Answer 1

The Correct Option is D

Explanation:
$\int \frac{{\sin }^{2}x{\cos }^{2}xdx}{{({\sin }^{5}x+{\cos }^{3}x{\sin }^{2}x+{\sin }^{3}x{\cos }^{2}x+{\cos }^{5}x)}^2}$$\int \frac{{\sin }^{2}x\cdot {\cos }^{2}xdx}{\{{({\sin }^{2}x({\sin }^{3}x+{\cos }^{3}x)+{\cos }^{2}x({\sin }^{3}x+{\cos }^{3}x)\}}^2}$$\int \frac{{\sin }^{2}x\cdot {\cos }^{2}xdx}{{\left\{({\sin }^{2}x+{\cos }^{2}x)({\sin }^{3}x+{\cos }^{3}x)\right\}}^2}$$\int \frac{{\sin }^{2}x{\cos }^{2}xdx}{{({\sin }^{3}x+{\cos }^{3}x)}^2}$Divide by ${\cos }^{3}x$ in numerator and denominator we get$=\int \frac{{\sec }^{2}x\cdot {\tan }^{2}x}{{({\tan }^{3}x+1)}^2}dx$Let $1+{\tan }^{3}x=t$$3{\tan }^{2}x{\sec }^{2}xdx=dt$$=\frac{1}{3}\int \frac{dt}{t^2}=-\frac{1}{3}\frac{1}{t}+C$$=-\frac{1}{3(1+{\tan }^{3}x)}+C$Hence, the correct option is (D).