Q248 / 270 · NDA 2020 · Mathematics
What is $\int\left(e^{\log x}+\sin x\right) \cos x d x$ equal to ?
- A$\sin x+x \cos x+\frac{\sin ^2 x}{2}+c$
- B$\sin x-x \cos x+\frac{\sin ^2 x}{2}+c$
- Correct answerC$x \sin x+\cos x+\frac{\sin ^2 x}{2}+c$
- D$x \sin x-x \cos x+\frac{\sin ^2 x}{2}+c$
Solution
$\begin{aligned} & \text { } \int\left(e^{\log x}+\sin x\right) \cdot \cos x \cdot d x=\int(x+\sin x) \cdot \cos x \cdot d x \\ & =\int(x \cdot \cos x+\sin x \cdot \cos x) \cdot d x\end{aligned}$
$\begin{aligned} & =\int x \cdot \cos x \cdot d x+\int \sin x \cdot \cos x \cdot d x \\ & =x \cdot \int \cos x \cdot d x-\int\left(\frac{d x}{d x} \cdot \int(\cos x) \cdot d x\right) \cdot d x \\ & \quad+\int \sin x \cdot \cos x \cdot d x\end{aligned}$
$\begin{aligned} & =x \cdot \sin x-\int \sin x \cdot d x+\int \sin x \cdot \cos x \cdot d x \\ & =x \cdot \sin x-(-\cos x)+\int \sin x \cdot \cos x \cdot d x\end{aligned}$
$=x \sin x+\cos x+\int \sin x \cdot \cos x \cdot d x$
Now, $\int \sin x \cdot \cos x \cdot d x$
Let $\sin x=z, d z=\cos x \cdot d x$
$\int \sin x \cdot \cos x d x=\int z \cdot d z=\frac{z^2}{2}=\frac{\sin ^2 x}{2}$
$\therefore I=x \cdot \sin x+\cos x+\frac{\sin ^2 x}{2}+C$
Hence, the correct answer is option: 'C'.