$\int \frac{\mathrm{d} x}{3-2 \cos 2 x}=\frac{\tan ^{-1}(\mathrm{f}(x))}{\sqrt{5}}+\mathrm{c}$, (where c is…

$\int \frac{\mathrm{d} x}{3-2 \cos 2 x}=\frac{\tan ^{-1}(\mathrm{f}(x))}{\sqrt{5}}+\mathrm{c}$, (where c is constant of integration), then $f(\pi / 4)$ has the value
  1. $-\sqrt{5}$
  2. $\sqrt{5}$
  3. $\frac{2}{\sqrt{5}}$
  4. $\frac{1}{\sqrt{5}}$

Solution

$\begin{aligned} & \text { Let } \mathrm{I}=\int \frac{\mathrm{d} x}{3-2 \cos 2 x} \\ & \text { Put } \tan x=\mathrm{t} \\ \therefore \quad & x=\tan ^{-1} \mathrm{t}\end{aligned}$ $\begin{aligned} & \therefore \quad \mathrm{d} x=\frac{\mathrm{dt}}{1+\mathrm{t}^2} \\ & \quad \begin{aligned} & \cos 2 x=\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2} \\ \therefore \quad \mathrm{I} & =\int \frac{\frac{\mathrm{dt}}{1+\mathrm{t}^2}}{3-2\left(\frac{1-\mathrm{t}^2}{1+\mathrm{t}^2}\right)} \\ & =\int \frac{\mathrm{dt}}{3\left(1+\mathrm{t}^2\right)-2\left(1-\mathrm{t}^2\right)} \\ & =\int \frac{\mathrm{dt}}{3+3 \mathrm{t}^2-2+2 \mathrm{t}^2} \\ & =\int \frac{\mathrm{dt}}{(\sqrt{5} \mathrm{t})^2+(1)^2} \\ & =\frac{1}{\sqrt{5}} \tan ^{-1}(\sqrt{5} \mathrm{t})+\mathrm{c}\end{aligned}\end{aligned}$ $\therefore \quad I=\frac{1}{\sqrt{5}} \tan ^{-1} \sqrt{5} \tan x+c$
Comparing with $\frac{\tan ^{-1} \mathrm{f}(x)}{\sqrt{5}}$, we get $\begin{aligned} & \mathrm{f}(x)=\sqrt{5} \tan x \\ \therefore \quad & \mathrm{f}\left(\frac{\pi}{4}\right)=\sqrt{5} \tan \frac{\pi}{4}=\sqrt{5} \end{aligned}$

Asked in: MHT CET 2024 (02 May Shift 1)

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