Integral Calculus Question 249
Question: $ \int_{{}}^{{}}{x\sin x{{\sec }^{3}}x,dx=} $
Options:
A) $ \frac{1}{2}[{{\sec }^{2}}x-\tan x]+c $
B) $ \frac{1}{2}[x{{\sec }^{2}}x-\tan x]+c $
C) $ \frac{1}{2}[x{{\sec }^{2}}x+\tan x]+c $
D) $ \frac{1}{2}[{{\sec }^{2}}x+\tan x]+c $
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Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{x\sin x{{\sec }^{3}}x,dx}=\int_{{}}^{{}}{x\sin x\frac{1}{{{\cos }^{3}}x},dx} $ $ =\int_{{}}^{{}}{x\tan x,.,{{\sec }^{2}}x,dx} $ Now put $ \tan x=t\Rightarrow {{\sec }^{2}}x,dx=dt $ and $ x={{\tan }^{-1}}t, $ then it reduces to $ \int_{{}}^{{}}{{{\tan }^{-1}}t,.,t,dt}=\frac{x{{\tan }^{2}}x}{2}-\frac{1}{2}t+\frac{1}{2}{{\tan }^{-1}}t $ $ =\frac{x({{\sec }^{2}}x-1)}{2}-\frac{1}{2}\tan x+\frac{1}{2}x=\frac{1}{2}[x{{\sec }^{2}}x-\tan x]+c $ .
BETA
