Integral Calculus Question 476
Question: $ \int_{{}}^{{}}{{{\tan }^{3}}}2x\sec 2x\ dx= $
[IIT 1977]
Options:
A) $ \frac{1}{6}{{\sec }^{3}}2x-\frac{1}{2}\sec 2x+c $
B) $ \frac{1}{6}{{\sec }^{3}}2x+\frac{1}{2}\sec 2x+c $
C) $ \frac{1}{9}{{\sec }^{2}}2x-\frac{1}{3}\sec 2x+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{{{\tan }^{3}}2x\sec 2x,dx}=\int_{{}}^{{}}{({{\sec }^{2}}2x-1)\sec 2x\tan 2x,dx} $
$ \int_{{}}^{{}}{({{\sec }^{3}}2x\tan 2x-\sec 2x\tan 2x)dx} $
$ =\int_{{}}^{{}}{{{\sec }^{3}}2x\tan 2x,dx}-\int_{{}}^{{}}{\sec 2x\tan 2x,dx} $ ..?(i)
Now, we take $ \int_{{}}^{{}}{{{\sec }^{3}}2x\tan 2x,dx} $
Put $ \sec 2x=t\Rightarrow \sec 2x\tan 2x=\frac{dt}{2}, $ then it reduces to
$ \frac{1}{2}\int_{{}}^{{}}{t^{2}dt}=\frac{t^{3}}{6}=\frac{{{\sec }^{3}}2x}{6} $
From (i), $ \int_{{}}^{{}}{{{\sec }^{3}}2x\tan 2x,dx}-\int_{{}}^{{}}{\sec 2x\tan 2x,dx} $
$ =\frac{{{\sec }^{3}}2x}{6}-\frac{\sec 2x}{2}+c. $
Trick : Let $ \sec 2x=t, $ then $ \sec 2x\tan 2x,dx=\frac{1}{2}dt $
\ $ \frac{1}{2}\int_{{}}^{{}}{(t^{2}-1),dt}=\frac{1}{6}t^{3}-\frac{1}{2}t+c=\frac{1}{6}{{\sec }^{3}}2x-\frac{1}{2}\sec 2x+c $ .
BETA
