JEE Main On 08 April 2018 Question 22
Question: The integral $ \int _{{}}^{{}}{\frac{{{\sin }^{2}}x{{\cos }^{2}}x}{{{({{\sin }^{5}}x+{{\cos }^{3}}x{{\sin }^{2}}x+{{\sin }^{3}}x{{\cos }^{2}}x+{{\cos }^{5}}x)}^{2}}}dx} $ is equal to: [JEE Main Online 08-04-2018]
Options:
A) $ \frac{1}{1+{{\cot }^{3}}x}+C $
B) $ \frac{-1}{1+{{\cot }^{3}}x}+C $
C) $ \frac{1}{3(1+{{\tan }^{3}}x)}+C $
D) $ \frac{-1}{3(1+{{\tan }^{3}}x)}+C $ (where C is a constant of integration)
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int _{{}}^{{}}{\frac{{{\sin }^{2}}x.{{\cos }^{2}}x}{{{({{\sin }^{5}}x+{{\cos }^{3}}x-{{\sin }^{2}}x-{{\sin }^{3}}x.{{\cos }^{2}}x+{{\cos }^{5}}x)}^{2}}}dx} $
$ \int _{{}}^{{}}{\frac{{{\sin }^{2}}x.{{\cos }^{2}}x}{{{(({{\sin }^{3}}x+{{\cos }^{3}}x)(sin^{2}x+{{\cos }^{2}}x))}^{2}}}dx} $
$ \int _{{}}^{{}}{\frac{{{\tan }^{2}}x.{{\sec }^{2}}x}{{{({{\tan }^{3}}x+1)}^{2}}}} $
Put $ {{\tan }^{3}}x+1=t $
$ \frac{1}{3}\int _{{}}^{{}}{\frac{dt}{t^{2}}=-\frac{1}{3t}+C\Rightarrow -\frac{1}{3({{\tan }^{3}}x+1)}+C} $
