Q112 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics
$\int \frac{2-\sin x}{2 \cos x+3} d x=$
- A$\frac{2}{\sqrt{5}} \tan ^{-1}\left(\frac{1}{\sqrt{3}} \tan \frac{x}{2}\right)-\log \sqrt{2 \cos x+3}+c$
- Correct answerB$\frac{4}{\sqrt{5}} \tan ^{-1}\left(\frac{1}{\sqrt{5}} \tan \frac{x}{2}\right)+\log \sqrt{2 \cos x+3}+c$
- C$\frac{3}{\sqrt{5}} \tan ^{-1}\left(\frac{1}{\sqrt{5}} \tan \frac{x}{2}\right)+\log \sqrt{2 \cos x+3}+c$
- D$\frac{1}{\sqrt{5}} \tan ^{-1}\left(\frac{1}{\sqrt{5}} \tan \frac{x}{2}\right)-\log \sqrt{2 \cos x-3}+c$
Solution
$\mathrm{I}=\int \frac{2-\sin x}{2 \cos x+3} d x$
$\begin{aligned} & \mathrm{I}=\int \frac{2}{2 \cos x+3} d x+\int \frac{-\sin x}{2 \cos x+3} d x \\ & \mathrm{I}=\mathrm{I}_1+\mathrm{I}_2 \\ & \mathrm{I}_1=\int \frac{2}{2 \cos x+3} d x=2 \int \frac{\sec ^2 \frac{x}{2}}{\tan ^2 \frac{x}{2}+5} d x \\ & \text { Let } \frac{\tan \frac{x}{2}}{\sqrt{5}}=u \Rightarrow d \mathrm{u}=\frac{\sec ^2\left(\frac{x}{2}\right)}{2 \sqrt{5}} d x \\ & \Rightarrow \mathrm{I}_1=2 \int \frac{2 \sqrt{5}}{5 u^2+5} d u \\ & \mathrm{I}_1=\frac{4}{\sqrt{5}} \int \frac{1}{u^2+1} d u=\frac{4}{\sqrt{5}} \tan ^{-1} u+\mathrm{C}_1 \\ & \mathrm{I}_1=\frac{4}{\sqrt{5}} \tan ^{-1}\left(\frac{1}{\sqrt{5}} \tan \frac{x}{2}\right)+\mathrm{C}_1 \\ & \mathrm{I}_2=\int \frac{-\sin x}{2 \cos x+3} \qquad [\because \int \frac{f^{\prime}(x)}{f(x)}=\log f(x)+C] \\ & =\frac{1}{2} \log (2 \cos x+3)+\mathrm{C}_2 \\ & \therefore \mathrm{I}=\frac{4}{\sqrt{5}} \tan ^{-1}\left(\frac{1}{\sqrt{5}} \tan \frac{x}{2}\right)+\log \sqrt{2 \cos x+3}+\mathrm{C}\end{aligned}$
Hence, the correct answer is option: 'B'.