Q97 / 120 · NDA 2024 (Phase 2) · Mathematics
Paragraph:
Let $\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{f(x)}{g(x)} d x$, where $f(x)=\sin x$ and $g(x)=\sin x+\cos x+1$.
Question:
What is $\int_0^{\frac{\pi}{2}} \frac{d x}{g(x)}$ equal to?
Question: What is $\int_0^{\frac{\pi}{2}} \frac{d x}{g(x)}$ equal to?
- A$\frac{\ln 2}{2}$
- B$\frac{\ln 2}{4}$
- Correct answerC$\ln 2$
- D$2 \ln 2$
Solution
Given that $\mathrm{g}(\mathrm{x})=\sin \mathrm{x}+\cos \mathrm{x}+1$
$\begin{aligned}
& \int_0^{\frac{\pi}{2}} \frac{d x}{g(x)}=\int_0^{\frac{\pi}{2}} \frac{1}{\sin x+\cos x+1} d x \\
& =\int_0^{\frac{\pi}{2}} \frac{1}{\frac{2 \tan \frac{\pi}{2}}{1+\tan ^2 \frac{\pi}{2}}+\frac{1-\tan ^2 \frac{\pi}{2}}{1+\tan ^2 \frac{\pi}{2}}+1} d x
\end{aligned}$
$\begin{aligned}
& =\int_0^{\frac{\pi}{2}} \frac{1+\tan ^2 \frac{x}{2}}{2 \tan \frac{x}{2}+1-\tan ^2 \frac{x}{2}+1+\tan ^2 \frac{x}{2}} d x \\
& =\int_0^{\frac{\pi}{2}} \frac{\sec ^2 \frac{x}{2}}{2\left(1+\tan \frac{x}{2}\right)} d x \\
& \text { let } 1+\tan \frac{x}{2}=t \Rightarrow \frac{1}{2} \sec ^2 \frac{x}{2} d x=d t
\end{aligned}$
when $\mathrm{x}=\pi / 2$ then $\mathrm{t}=2$, and when $\mathrm{x}=0$ then $\mathrm{t}=1$ $=\int_1^2 \frac{1}{\mathrm{t}} \mathrm{dt}=[\ln \mathrm{t}]_1^2=\ln 2-\ln 1=\ln 2$
Hence, the correct answer is option: 'C'.