$\int_0^{\pi / 2} \frac{d x}{5+4 \sin x}=A \tan ^{-1} B$, then $A+B=$

$\int_0^{\pi / 2} \frac{d x}{5+4 \sin x}=A \tan ^{-1} B$, then $A+B=$
  1. $\frac{2}{3}$
  2. 1
  3. 2
  4. $\frac{1}{3}$

Solution

Let $I=\int_0^{\pi / 2} \frac{d x}{5+4 \sin x}$ Put $\tan \frac{\mathrm{x}}{2}=\mathrm{t} \Rightarrow \sec ^2 \frac{\mathrm{x}}{2}\left(\frac{1}{2}\right) \mathrm{dx}=\mathrm{dt}$ $\therefore \mathrm{dx}=\frac{2 \mathrm{dt}}{1+\mathrm{t}^2} \text { and } \sin \mathrm{x}=\frac{2 \mathrm{t}}{1+\mathrm{t}^2}$ When $\mathrm{x}=0, \mathrm{t}=0$ and when $\mathrm{x}=\frac{\pi}{2}, \mathrm{t}=1$ $\begin{aligned} & \therefore=\int_0^1 \frac{1}{5+4\left(\frac{2 \mathrm{t}}{1+\mathrm{t}^2}\right)} \times \frac{2 \mathrm{dt}}{1+\mathrm{t}^2}=2 \int_0^1 \frac{\mathrm{dt}}{5+5 \mathrm{t}^2+8 \mathrm{t}}=\frac{2}{5} \int_0^1 \frac{\mathrm{dt}}{\mathrm{t}^2+\frac{8}{5} \mathrm{t}+1} \\ & =\frac{2}{5} \int_0^1 \frac{\mathrm{dt}}{\mathrm{t}^2+\frac{8}{5} \mathrm{t}+\frac{16}{25}+\frac{9}{25}}=\frac{2}{5} \int_0^1 \frac{\mathrm{dt}}{\left(\mathrm{t}+\frac{4}{5}\right)^2+\left(\frac{3}{5}\right)^2} \end{aligned}$ $\begin{aligned} & \therefore I=\frac{2}{5} \times \frac{1}{\left(\frac{3}{5}\right)}\left[\tan ^{-1}\left[\frac{\mathrm{t}+\frac{4}{5}}{\left(\frac{3}{5}\right)}\right]\right]_0^1=\frac{2}{3}\left[\tan ^{-1}\left(\frac{5 \mathrm{t}+4}{3}\right)\right]_0^1 \\ & =\frac{2}{3}\left[\tan ^{-1} 3-\tan ^{-1} \frac{4}{3}\right]=\frac{2}{3} \tan ^{-1}\left[\frac{3-\left(\frac{4}{3}\right)}{1+3\left(\frac{4}{3}\right)}\right] \\ & =\frac{2}{3} \tan ^{-1}\left(\frac{5}{3} \times \frac{1}{5}\right)=\frac{2}{3} \tan ^{-1}\left(\frac{1}{3}\right) \end{aligned}$ Comparing with given data, we get $\mathrm{A}=\frac{2}{3}, \mathrm{~B}=\frac{1}{3} \Rightarrow \mathrm{A}+\mathrm{B}=1$

Asked in: MHT CET 2021 (20 Sep Shift 1)

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