• question_answer The integral $\int_{{}}^{{}}{\frac{\sin x{{\cos }^{2}}x}{{{\left( {{\sin }^{3}}x+{{\cos }^{3}}x \right)}^{2}}}dx}$is equal to:   [JEE Main Online Paper ( Held On 12 Apirl  2014 ) A) $\frac{1}{\left( 1+{{\cot }^{3}}x \right)}+c$        B) $-\frac{1}{3\left( 1+{{\cot }^{3}}x \right)}+c$ C)  $-\frac{{{\sin }^{3}}x}{\left( 1+{{\cot }^{3}}x \right)}+c$              D) $-\frac{{{\cos }^{3}}x}{3\left( 1+{{\cot }^{3}}x \right)}+c$

Let $I=\int_{{}}^{{}}{\frac{{{\sin }^{2}}x{{\cos }^{2}}x}{{{({{\sin }^{3}}x+{{\cos }^{3}}x)}^{2}}}dx}$ $I=\int_{{}}^{{}}{{{\left( \frac{\sin x.\cos x}{{{\sin }^{3}}x+{{\cos }^{3}}x)} \right)}^{2}}dx}$ $I=\int_{{}}^{{}}{{{\left( \frac{\sin x.\cancel{\cos }x}{\cancel{{{\cos }^{3}}}x(1+{{\tan }^{3}}x)} \right)}^{2}}dx}$ $=\int_{{}}^{{}}{{{\left( \frac{\sin x.{{\sec }^{2}}x}{(1+{{\tan }^{3}}x)} \right)}^{2}}dx}$ Put$1+{{\tan }^{3}}x=t$ $dt=3{{\tan }^{2}}x{{\sec }^{2}}xdx$or$dx=\frac{dt}{3{{\tan }^{2}}x{{\sec }^{2}}x}$ $\therefore$$I=\int_{{}}^{{}}{\frac{{{\sin }^{2}}x.{{\sec }^{4}}x}{{{t}^{2}}}}\times \frac{dt}{3{{\tan }^{2}}x{{\sec }^{2}}x}$ $=\frac{1}{3}\int_{{}}^{{}}{\frac{\cancel{{{\sin }^{2}}x.{{\sec }^{4}}}x}{{{t}^{2}}}}\times \frac{dt}{\cancel{{{\sin }^{2}}x{{\sec }^{4}}x}}$ $\therefore$$I=\frac{1}{3}\int_{{}}^{{}}{\frac{dt}{{{t}^{2}}}=\frac{1}{3}}\int_{{}}^{{}}{{{t}^{-2}}dt}$ $I=\frac{1}{3}\left[ \frac{{{t}^{-2}}+1}{-2+1} \right]+c=\frac{-1}{3}\left[ \frac{1}{t} \right]+c$ or$I=-\frac{1}{3(1+{{\tan }^{3}}x)}+c$