MCQ
$\int_{}^{} {x\sin x{{\sec }^3}x\,dx = } $
- A$\frac{1}{2}[{\sec ^2}x - \tan x] + c$
- ✓$\frac{1}{2}[x{\sec ^2}x - \tan x] + c$
- C$\frac{1}{2}[x{\sec ^2}x + \tan x] + c$
- D$\frac{1}{2}[{\sec ^2}x + \tan x] + c$
✓
Answer
Correct option: B.
$\frac{1}{2}[x{\sec ^2}x - \tan x] + c$
b
(b)$\int_{}^{} {x\sin x{{\sec }^3}x\,dx} = \int_{}^{} {x\sin x\frac{1}{{{{\cos }^3}x}}\,dx} $
(b)$\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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