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) \[-\frac{{{\cos }^{3}}x}{3\left( 1+{{\cot }^{3}}x \right)}+c\]
Correct Answer: B
Solution :
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\[I=-\frac{1}{3(1+{{\tan }^{3}}x)}+c\]
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