Question

Solve $\int e^{\log x}.\cos xdx$

• A. $x\sin x-cosx+c$
• B. $\frac{x}{2}\sin x+\cos x+c$
• C. $x\sin x+cosx+c$
• D. $x\sin x+cos2x+c$

Answer 1

The correct option is C $x\sin x+cosx+c$

We know, $e^{logx}=x$

$\int e^{logx}cosxdx=\int xcosxdx$

We know, $\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$

Now by integrating the functions by parts, taking $x$ as $u$ and $cosx$ as $v$, we get

$\int xcosxdx=x\int cosxdx-\int 1(\int cosx)dx$

$=xsinx+cosx+c$

$\int e^{logx}cosxdx=x\cdot sinx+cosx+c\cdot$