Question

Evaluate the integral: ∫(cos x/cos3x) dx

Answer 1

Let \(I = \int \frac{\cos x}{\cos 3x} dx\)

\(= \int \frac{\cos x}{(4\cos^3x - 3\cos x)} dx\)    \([\cos 3A = 4\cos^3A - 3\cos A]\)

\(= \int \frac 1{4\cos^2x - 3}dx\)

Dividing numerator and denominator by cos²x

⇒ \(I = \int \frac{\sec^2 x}{4 -3\sec^2 x} dx\)

\(= \int \frac{\sec^2x}{4 - 3 (1 + \tan^2 x)}dx\)

\(= \int \frac{\sec^2x}{1 - 3\tan^2 x}dx\)

\(= \int \frac{\sec^2x}{1 - (\sqrt 3 \tan x)^2}dx\)

Let \(\sqrt 3\tan x = t\)

⇒ \(\sqrt 3 \sec^2 x\, dx= dt\)

⇒ \(\sec^2 x \, dx = \frac{dt}{\sqrt 3}\)

\(\therefore I = \frac 1{\sqrt 3} \int \frac{dt}{1^2 - t^2}\)

\(= \frac 1{\sqrt 3} \times \frac 12\, \ln \left|\frac{1 + t}{1 - t}\right| + C\)

\(= \frac 1{2\sqrt 3} \, \ln \left|\frac {1 + \sqrt 3 \tan x}{1 - \sqrt 3\tan x}\right| + C\)

Answer 2

Given as ∫(cos x/cos3x) dx

On dividing the numerator and denominator by cos² x 

On replacing sec²x in denominator by 1 + tan² x

On putting tan x = t and sec² x dx = dt