Question

Evaluate:$\int \left[\sin (\log x)+\cos (\log x)\right]dx$

• A. $x\cos (\log x)+C$
• B. $\cos (\log x)+C$
• C. $x\sin (\log x)+C$
• D. $\sin (\log x)+C$

Answer 1

The correct option is C

$x\sin (\log x)+C$

Evaluating the integral

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

Substituting $t=\log x$then $x=e^t$ and putting the value of $dx$ after differentiating w.r.r.t $x$

$dx=e^{t}dt$

$$\begin{gathered} \Rightarrow \int \left[\sin t+\cos t\right]e^{t}dt \\ \Rightarrow \int e^t\sin \left(t\right)dt+\int e^t\cos \left(t\right)dt \\ \Rightarrow \int e^t\sin \left(t\right)dt+\left(e^t\int \cos \left(t\right)dt-\int \left[\frac{d}{dt}{e}^t\int \cos \left(t\right)dt\right]dt\right)\left[\text{Integrating by parts}\right] \\ \Rightarrow \int e^t\sin \left(t\right)dt+\left(e^t\sin \left(t\right)-\int \sin \left(t\right)dt\right)+C\left[\text{C is an integrating constant}\right] \\ \Rightarrow e^t\sin \left(t\right)+C \\ \Rightarrow x\sin \left(\log x\right)+C\left[\text{substituting back}t=\log x\&x=e^t\right] \end{gathered}$$

Hence, option C is the correct answer.