Question

Evaluate \[ \int \frac{dx}{10\cos^2x+\cos2x+5} \]

• A. \(\dfrac14\tan^{-1}\!\left(\dfrac{\tan x}{2}\right)+c\)
• B. \(\dfrac18\tan^{-1}\!\left(\dfrac{\tan x}{2}\right)+c\)
• C. \(\dfrac14\tan^{-1}\!\left(\dfrac{\tan x}{5}\right)+c\)
• D. \(\dfrac18\tan^{-1}\!\left(\dfrac{\tan x}{5}\right)+c\)

Hint:

For integrals containing both \[ \cos^2x \] and \[ \cos2x, \] first use \[ \boxed{\cos2x=2\cos^2x-1,} \] then apply the substitution \[ \boxed{t=\tan x.} \] This converts the integral into the standard form \[ \int\frac{dt}{t^2+a^2}. \]

Answer 1

The Correct Option is B

Step 1: Simplify the denominator. Using the identity \[ \cos2x=2\cos^2x-1, \] the denominator becomes \[ 10\cos^2x+\cos2x+5 = 10\cos^2x+2\cos^2x-1+5 = 12\cos^2x+4. \] Hence, \[ I = \int\frac{dx}{12\cos^2x+4} = \frac14\int\frac{dx}{3\cos^2x+1}. \]

Step 2: Use the substitution \(t=\tan x\). Let \[ t=\tan x. \] Then, \[ dt=\sec^2x\,dx, \] \[ dx=\frac{dt}{1+t^2}, \] and \[ \cos^2x=\frac1{1+t^2}. \] Therefore, \[ I = \frac14 \int \frac{dt}{3+(1+t^2)} = \frac14 \int \frac{dt}{t^2+4}. \]

Step 3: Integrate using the standard formula. Using \[ \int\frac{dx}{x^2+a^2} = \frac1a\tan^{-1}\!\left(\frac xa\right)+c, \] with \[ a=2, \] we obtain \[ I = \frac14\cdot\frac12 \tan^{-1}\!\left(\frac t2\right)+c. \] Substituting \[ t=\tan x, \] gives \[ \boxed{ I = \frac18 \tan^{-1}\!\left(\frac{\tan x}{2}\right)+c. } \] Hence, the correct option is \(\boxed{(B)}\).