Integral Calculus Question 9
Question: Let $f(x)=\int \frac{x^2}{\left(1+x^2\right)\left(1+\sqrt{1+x^2}\right)} d x$ and $ f(0)=0 $ , then the value of $ f(1) $ be
[AMU 2005]
Options:
A) $ \log (1+\sqrt{2}) $
B) $ \log (1+\sqrt{2})-\frac{\pi }{4} $
C) $ \log (1+\sqrt{2})+\frac{\pi }{2} $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ f(x)=\int{\frac{x^{2}dx}{(1+x^{2})( 1+\sqrt{1+x^{2}} )}} $
Let $ x=\tan \theta ,dx={{\sec }^{2}}\theta ,d\theta $ $ =(1+x^{2}).d\theta $
$ f(x)=\int{\frac{x^{2}dx}{(1+x^{2})( 1+\sqrt{1+x^{2}} )}}=\int{\frac{{{\tan }^{2}}\theta {{\sec }^{2}}\theta d\theta }{{{\sec }^{2}}\theta (1+\sec \theta )}} $
$ =\int{\frac{{{\tan }^{2}}\theta d\theta }{1+\sec \theta }\theta }=\int{\frac{{{\sin }^{2}}\theta ,d\theta }{\cos \theta (1+\cos \theta )}}=\int{\frac{1-{{\cos }^{2}}\theta ,d\theta }{\cos \theta (1+\cos \theta )}} $
$ =\int{\frac{(1-\cos \theta )d.\theta }{\cos \theta }} $ $ =\int{\sec \theta d\theta -\int{d\theta }} $
$ =\log (x+\sqrt{1+x^{2}})-{{\tan }^{-1}}x+c $
$ f(0)=\log (0+\sqrt{1+0}-{{\tan }^{-1}}(0)+c $
$ 0=\log 1-0+c $
$ c=0 $ $ f(1)=\log (1+\sqrt{1+1^{2}})-{{\tan }^{-1}}(1)=\log (1+\sqrt{2})-\frac{\pi }{4} $ .
BETA
