Q98 / 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 I equal to?

  • A
    $\frac{\pi}{4}+\ln 2$
  • B
    $\frac{\pi}{4}-\ln 2$
  • Correct answer
    C
    $\frac{\pi}{4}-\frac{\ln 2}{2}$
  • D
    $\frac{\pi}{4}+\frac{\ln 2}{2}$

Solution

Given that $f(x)=\sin x$ and $g(x) \sin x+\cos x+1$ Let $I=\int_0^{\frac{\pi}{2}} \frac{f(x)}{g(x)} d x=\int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x+\cos x+1} d x$.....(i) $\begin{aligned} & I=\int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)+1} d x \\ & I=\int_0^{\frac{\pi}{2}} \frac{\cos x}{\cos x+\sin x+1} d x \quad ....(ii) \end{aligned}$ Adding equation (i) and (ii), we get $2 I=\int_0^{\frac{\pi}{2}} \frac{\sin x+\cos x}{\sin x+\cos x+1} d x$ $\begin{aligned} & =\int_0^{\frac{\pi}{2}}\left(1-\frac{1}{\sin x+\cos x+1}\right) d x \\ & =\int_0^{\frac{\pi}{2}} 1 d n-\int_0^{\frac{\pi}{2}} \frac{1}{\sin x+\cos x+1} d x \end{aligned}$ from solution of previous question $\begin{aligned} & 2 \mathrm{I}=\int_0^{\frac{\pi}{2}} 1 \mathrm{dx}-\ln 2=\frac{\pi}{2}-\ln 2 \\ & \therefore \mathrm{I}=\frac{\pi}{4}-\frac{\ln 2}{2} \end{aligned}$

Hence, the correct answer is option: 'C'.

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