MHT CET · Maths · Indefinite Integration
\(\int \frac{1}{3-2 \cos 2 x} \mathrm{~d} x=\) (where \(C\) is constant of integration.)
- A \(\frac{2}{5} \tan ^{-1}(5 \tan x)+C\)
- B \(\frac{1}{\sqrt{5}} \tan ^{-1}(\sqrt{5} \tan x)+C\)
- C \(\frac{2}{\sqrt{5}} \tan ^{-1}(\sqrt{5} \tan x)+C\)
- D \(\frac{1}{5} \tan ^{-1}(5 \tan x)+C\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{\sqrt{5}} \tan ^{-1}(\sqrt{5} \tan x)+C\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \int \frac{\mathrm{d} x}{3-2 \cos 2 x}=\int \frac{\mathrm{d} x}{3-2 \times \frac{1-\tan ^2 x}{1+\tan ^2 x}}=\int \frac{\sec ^2 x \mathrm{~d} x}{5 \tan ^2 x+1} \\ & =\int \frac{\mathrm{d} t}{5 t^2+1}[\text { let } \tan x=t] \\ & =\int \frac{\mathrm{d} t}{(\sqrt{5} t)^2+1^2}=\frac{1}{\sqrt{5}} \tan ^{-1}\left(\frac{\sqrt{5} t}{1}\right)+C \\ & =\frac{1}{\sqrt{5}} \tan ^{-1}(\sqrt{5} \tan x)+C\end{aligned}\)
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