∫x sinx sec3x dx=∫x sinx .1cos3x dx =∫x tanx sec2x dx integration by parts, let u=x and v=tanx sec2x =x ∫sec xtanx secx dx−∫sec xtanx secx dxdx =xsec2x2−∫sec2x2dx =xsec2x2−tanx2+c

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$\int x \sin x s e c^{3} x d x =$

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$\frac{1}{2} [x s e c^{2} x + \tan x] + c$

$\frac{1}{2} [x s e c^{2} x - \tan x] + c$

$\frac{1}{2} [s e c^{2} x - \tan x] + c$

$\frac{1}{2} [s e c^{2} x + \tan x] + c$

answer is B.

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Detailed Solution

$\begin{array}{l} \int x \sin x s e c^{3} x d x = \int x \sin x . \frac{1}{\cos^{3} x} d x \\ = \int x \tan x s e c^{2} x d x \\ (i n t e g r a t i o n b y p a r t s, l e t u = x a n d v = \tan x s e c^{2} x) \\ = x \int s e c x (t a n x s e c x) d x - \int [s e c x (\tan x s e c x) d x] d x \\ = x \frac{s e c^{2} x}{2} - \int \frac{s e c^{2} x}{2} d x \\ = \frac{x s e c^{2} x}{2} - \frac{\tan x}{2} + c \end{array}$