Question

\(\int \frac{\mathrm{d} x}{3 \cos 2 x+5}\) equals

• A. \(\frac{1}{2}\mathrm{tan}^{-1}(\mathrm{tan}x)+\mathrm{c}\), where c is the constant of integration.
• B. \(\frac{1}{2}\mathrm{tan}^{-1}(\frac{\mathrm{tan}x}{2})+\mathrm{c}\), where c is the constant of integration.
• C. \(\frac{1}{4}\mathrm{tan}^{-1}(\frac{1}{2}\mathrm{tan}x)+\mathrm{c}\), where c is the constant of integration.
• D. \(\frac{1}{4}\mathrm{tan}^{-1}(\mathrm{tan}x)+\mathrm{c}\), where c is the constant of integration.

Hint:

Write cos 2x in terms of tan x and substitute t = tan x.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Use \(\cos2x=\frac{1-\tan^2x}{1+\tan^2x}\) and the substitution \(t=\tan x\), so \(dx=\frac{dt}{1+t^2}\).

Step 2: Simplify the denominator
\(3\cos2x+5=\frac{3(1-t^2)+5(1+t^2)}{1+t^2}=\frac{8+2t^2}{1+t^2}\).

Step 3: Integral in t
\(\int\frac{dt}{2t^2+8}=\frac12\int\frac{dt}{t^2+4}\).

Step 4: Result
\(=\frac12\cdot\frac12\tan^{-1}\frac t2+c=\frac14\tan^{-1}\left(\frac{\tan x}{2}\right)+c\), option (C).

Final Answer:
The integral equals \(\frac14\tan^{-1}\left(\frac12\tan x\right)+c\). \[ \boxed{\text{(C) }\dfrac{1}{4}\tan^{-1}\left(\dfrac{1}{2}\tan x\right)+\text{c}} \]