Question

$\int$ { $sin\left(logx\right)+cos\left(logx\right)$} dx is equal to

• A. $sin\left(logx\right)+c$
• B. $cos\left(logx\right)+c$
• C. $x.sin\left(logx\right)+c$
• D. $x.cos\left(logx\right)+c$

Answer 1

The correct option is C $x.sin\left(logx\right)+c$

Given,

$\int \sin \left(\log \left(x\right)\right)+\cos \left(\log \left(x\right)\right)dx$

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

using product rule of integration, we get,

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

$=2\times \frac{1}{2}x\sin \left(\log \left(x\right)\right)$

$=x\sin \left(\log \left(x\right)\right)+c$