Q54 / 79 · NEST (NISER) 2026 · Mathematics
$\displaystyle\int_{\frac{\pi}{2}}^{\pi} \dfrac{\sin x - x\cos x}{x(x + \sin x)}\, dx$ equals
- Correct answerA$\log\left(1 + \dfrac{2}{\pi}\right)$
- B$1 + \dfrac{\pi}{2}$
- C$\log\left(1 + \dfrac{\pi}{2}\right)$
- D$1 + \dfrac{2}{\pi}$
Solution
Let $I = \displaystyle\int_{\dfrac{\pi}{2}}^{\pi} \dfrac{\sin x - x\cos x}{x(x + \sin x)}\, dx$
Adding and subtracting $x$ in the numerator, we get:
$I = \displaystyle\int_{\dfrac{\pi}{2}}^{\pi} \dfrac{x + \sin x - x - x\cos x}{x(x + \sin x)}\, dx$
$I = \displaystyle\int_{\dfrac{\pi}{2}}^{\pi} \dfrac{(x + \sin x) - x(1 + \cos x)}{x(x + \sin x)}\, dx$
Separating the terms, we obtain:
$I = \displaystyle\int_{\dfrac{\pi}{2}}^{\pi} \left( \dfrac{x + \sin x}{x(x + \sin x)} - \dfrac{x(1 + \cos x)}{x(x + \sin x)} \right)\, dx$
$I = \displaystyle\int_{\dfrac{\pi}{2}}^{\pi} \left( \dfrac{1}{x} - \dfrac{1 + \cos x}{x + \sin x} \right)\, dx$
Integrating each term with respect to $x$:
$I = \left[ \log x - \log(x + \sin x) \right]_{\dfrac{\pi}{2}}^{\pi}$
$I = \left[ \log\left(\dfrac{x}{x + \sin x}\right) \right]_{\dfrac{\pi}{2}}^{\pi}$
Substituting the upper and lower limits:
$I = \log\left(\dfrac{\pi}{\pi + \sin \pi}\right) - \log\left(\dfrac{\dfrac{\pi}{2}}{\dfrac{\pi}{2} + \sin \dfrac{\pi}{2}}\right)$
Since $\sin \pi = 0$ and $\sin \dfrac{\pi}{2} = 1$:
$I = \log\left(\dfrac{\pi}{\pi + 0}\right) - \log\left(\dfrac{\dfrac{\pi}{2}}{\dfrac{\pi}{2} + 1}\right)$
$I = \log(1) - \log\left(\dfrac{\pi}{\pi + 2}\right)$
$I = 0 - \log\left(\dfrac{\pi}{\pi + 2}\right)$
$I = \log\left(\dfrac{\pi + 2}{\pi}\right) = \log\left(1 + \dfrac{2}{\pi}\right)$
Answer: $\log\left(1 + \dfrac{2}{\pi}\right)$
Adding and subtracting $x$ in the numerator, we get:
$I = \displaystyle\int_{\dfrac{\pi}{2}}^{\pi} \dfrac{x + \sin x - x - x\cos x}{x(x + \sin x)}\, dx$
$I = \displaystyle\int_{\dfrac{\pi}{2}}^{\pi} \dfrac{(x + \sin x) - x(1 + \cos x)}{x(x + \sin x)}\, dx$
Separating the terms, we obtain:
$I = \displaystyle\int_{\dfrac{\pi}{2}}^{\pi} \left( \dfrac{x + \sin x}{x(x + \sin x)} - \dfrac{x(1 + \cos x)}{x(x + \sin x)} \right)\, dx$
$I = \displaystyle\int_{\dfrac{\pi}{2}}^{\pi} \left( \dfrac{1}{x} - \dfrac{1 + \cos x}{x + \sin x} \right)\, dx$
Integrating each term with respect to $x$:
$I = \left[ \log x - \log(x + \sin x) \right]_{\dfrac{\pi}{2}}^{\pi}$
$I = \left[ \log\left(\dfrac{x}{x + \sin x}\right) \right]_{\dfrac{\pi}{2}}^{\pi}$
Substituting the upper and lower limits:
$I = \log\left(\dfrac{\pi}{\pi + \sin \pi}\right) - \log\left(\dfrac{\dfrac{\pi}{2}}{\dfrac{\pi}{2} + \sin \dfrac{\pi}{2}}\right)$
Since $\sin \pi = 0$ and $\sin \dfrac{\pi}{2} = 1$:
$I = \log\left(\dfrac{\pi}{\pi + 0}\right) - \log\left(\dfrac{\dfrac{\pi}{2}}{\dfrac{\pi}{2} + 1}\right)$
$I = \log(1) - \log\left(\dfrac{\pi}{\pi + 2}\right)$
$I = 0 - \log\left(\dfrac{\pi}{\pi + 2}\right)$
$I = \log\left(\dfrac{\pi + 2}{\pi}\right) = \log\left(1 + \dfrac{2}{\pi}\right)$
Answer: $\log\left(1 + \dfrac{2}{\pi}\right)$
Hence, the correct answer is option: 'A'.