Question

Evaluate \[ \int \frac{\tan 2x}{\cos^4 x}\,dx \]

• A. \(\tan^2x-\log\!\left(1-\tan^2x\right)^2+c\)
• B. \(-\tan^2x-\log\!\left(1-\tan^2x\right)^2+c\)
• C. \(-\tan^2x+\log\!\left(1-\tan^2x\right)+c\)
• D. \(\tan^2x+\log\!\left(1-\tan^2x\right)+c\)

Hint:

For integrals involving \[ \tan x,\qquad \sec x, \] use the substitution \[ \boxed{t=\tan x.} \] Also remember \[ \boxed{\tan2x=\frac{2\tan x}{1-\tan^2x}}. \]

Answer 1

The Correct Option is B

Step 1: Express the integrand in terms of \(\tan x\). Using \[ \tan2x=\frac{2\tan x}{1-\tan^2x}, \] and \[ \sec^4x=\frac{1}{\cos^4x}, \] the integral becomes \[ I = \int \frac{2\tan x\,\sec^4x}{1-\tan^2x}\,dx. \]

Step 2: Use substitution. Let \[ t=\tan x. \] Then, \[ dt=\sec^2x\,dx, \] and \[ \sec^4x\,dx = (1+t^2)\,dt. \] Hence, \[ I = \int \frac{2t(1+t^2)}{1-t^2}\,dt. \]

Step 3: Simplify the integrand. Performing algebraic simplification, \[ \frac{2t(1+t^2)}{1-t^2} = -2t+\frac{4t}{1-t^2}. \] Therefore, \[ I = \int \left( -2t+\frac{4t}{1-t^2} \right)dt. \]

Step 4: Integrate. Now, \[ \int -2t\,dt=-t^2, \] and \[ \int\frac{4t}{1-t^2}\,dt = -2\log|1-t^2|. \] Thus, \[ I = -t^2 - 2\log|1-t^2| +c. \] Replacing \[ t=\tan x, \] we get \[ I = -\tan^2x - \log\!\left(1-\tan^2x\right)^2 +c. \] Hence, \[ \boxed{ \int \frac{\tan2x}{\cos^4x}\,dx = -\tan^2x-\log\!\left(1-\tan^2x\right)^2+c. } \] Therefore, the correct option is \(\boxed{(B)}\).