Q51 / 100 · BITSAT 2023 (Memory Based Paper 1) · Mathematics
If $\int \frac{\mathrm{e}^{\mathrm{x}}(1+\sin \mathrm{x}) \mathrm{dx}}{1+\cos \mathrm{x}}=\mathrm{e}^{\mathrm{x}} \mathrm{f}(\mathrm{x})+\mathrm{C}$, then $\mathrm{f}(\mathrm{x})$ is equal to
- A$\sin \frac{x}{2}$
- B$\cos \frac{x}{2}$
- Correct answerC$\tan \frac{x}{2}$
- D$\log \frac{x}{2}$
Solution
$\int e^x \frac{(1+\sin x)}{(1+\cos x)} d x$
$=\int e^x\left[\frac{1}{2} \sec ^2 \frac{x}{2}+\tan \frac{x}{2}\right] d x$
$=\frac{1}{2} \int e^x \sec ^2 \frac{x}{2} d x+\int e^x \tan \frac{x}{2} d x$
$=\mathrm{e}^{\mathrm{x}} \tan \frac{\mathrm{x}}{2}+\mathrm{C}$
But $\mathrm{I}=\mathrm{e}^{\mathrm{x}} \mathrm{f}(\mathrm{x})+\mathrm{C}$ (given)
$\therefore \mathrm{f}(\mathrm{x})=\tan \frac{\mathrm{x}}{2}$
$=\int e^x\left[\frac{1}{2} \sec ^2 \frac{x}{2}+\tan \frac{x}{2}\right] d x$
$=\frac{1}{2} \int e^x \sec ^2 \frac{x}{2} d x+\int e^x \tan \frac{x}{2} d x$
$=\mathrm{e}^{\mathrm{x}} \tan \frac{\mathrm{x}}{2}+\mathrm{C}$
But $\mathrm{I}=\mathrm{e}^{\mathrm{x}} \mathrm{f}(\mathrm{x})+\mathrm{C}$ (given)
$\therefore \mathrm{f}(\mathrm{x})=\tan \frac{\mathrm{x}}{2}$
Hence, the correct answer is option: 'C'.