Question

The value of ${\int }_{}^{}\frac{{\sin }^{2}x{\cos }^{2}x}{{({\sin }^{3}x+{\cos }^{3}x)}^2}dx$ is

• A. $\frac{1}{3(1+{\tan }^{3}x)}+C$
• B. $-\frac{1}{3(1+{\tan }^{3}x)}+C$
• C. $\frac{1}{1+{\tan }^{3}x}+C$
• D. $-\frac{1}{1+{\tan }^{3}x}+C$

Answer 1

Answer: (B)

Let $I={\int }_{}^{}\frac{{\sin }^{2}x{\cos }^{2}x}{{({\sin }^{3}x+{\cos }^{3}x)}^2}dx$ ${\int }_{}^{}\frac{{\tan }^{2}x{\sec }^{2}x}{{(1+{\tan }^{3}x)}^2}dx$ [dividing numerator and denominator by ${\cos }^{6}x$ ] $=\frac{1}{3}{\int }_{}^{}\frac{3{\tan }^{2}x{\sec }^{2}x}{{(1+{\tan }^{3}x)}^2}dx$ $=\frac{1}{3}{\int }_{}^{}\frac{1}{{(1+{\tan }^{3}x)}^2}d(1+{\tan }^{3}x)$ $=-\frac{1}{3(1+{\tan }^{2}x)}+C$