Question

$\int$ sin x d(cos x) is equal to

• A. $-\frac{1}{4}cos2x+c$
• B. $\frac{1}{4}sin2x-\frac{x}{2}+c$
• C. $2sin2x+c$
• D. $sinx+cosx+c$

Answer 1

The correct option is A $-\frac{1}{4}cos2x+c$

Consider the given equation:

$:\int \sin xd\left(\cos x\right)dx$

$=\int \sin \left(x\right)\cos \left(x\right)dx$

$=\int \frac{\sin \left(2x\right)}{2}dx$

$=\frac{1}{2}\int \sin \left(2x\right)dx$

$\text{put}\text{then}v=2x$

$\text{so}\text{that}dv=2dx$

$=\frac{1}{4}\int \sin \left(v\right).\left(2\right)dx$

$=\frac{1}{4}\int \sin \left(v\right)dx$

$=-\frac{1}{4}\cos \left(v\right)+c$

$=-\frac{1}{4}\cos \left(2x\right)+c$

$Hence,thisistheanswer.$