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) \[\frac{1}{2}[{{\sec }^{2}}x+\tan x]+c\]
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 \[\int_{{}}^{{}}{{{\tan }^{-1}}t\,.\,t\,dt}=\frac{x{{\tan }^{2}}x}{2}-\frac{1}{2}t+\frac{1}{2}{{\tan }^{-1}}t\] \[=\frac{x({{\sec }^{2}}x-1)}{2}-\frac{1}{2}\tan x+\frac{1}{2}x=\frac{1}{2}[x{{\sec }^{2}}x-\tan x]+c\].
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