Question

$\int x\sin (-x^2)dx=$

• A. $\frac{1}{2}\sin \left(x^2\right)+c$
• B. $\frac{1}{2}\cos \left(x^2\right)+c$
• C. $\frac{1}{8}\cos \left(x^2\right)+c$
• D. $\frac{1}{2}x\sin x^2+c$

Answer 1

The correct option is B $\frac{1}{2}\cos \left(x^2\right)+c$

$\int xsin(-x^2)dx$
Substitute $x^2=t$.
$\Rightarrow 2xdx=dt$
$xdx=\frac{1}{2}dt$
$\int \frac{1}{2}dtsin(-t)$
$=\frac{-1}{2}\int sintdt$
$=\frac{1}{2}cost+c$
$=\frac{1}{2}cos\left(x^2\right)+c$
$\int xsin(-x^2)dx=\frac{1}{2}cos\left(x^2\right)+c$