Question

Evalaute $\int \frac{\cos 4x-\cos 2x}{\sin 4x-\sin 2x}dx$

• A. $-\tan x+c$
• B. $-\tan 3x+c$
• C. $\frac{1}{3}\log \left(\cos 3x\right)+c$
• D. $\frac{1}{3}\log \left(\tan 3x\right)+c$

Answer 1

The correct option is C $\frac{1}{3}\log \left(\cos 3x\right)+c$

$\int \left(\frac{cos4x-cos2x}{sin4x-sin2x}\right)dx\int \left(\frac{-2sin\left(\frac{4x+2x}{2}\right)sin\left(\frac{4x-2x}{2}\right)}{2cos\left(\frac{4x+2x}{2}\right)sin\left(\frac{4x-2x}{2}\right)}\right)dx-\int \left(\frac{sin3x}{cos3x}\right)dxletcos3x=t\therefore -sin3x\times 3dx=dt-\int \left(\frac{1}{-3t}\right)dt=\left(\frac{1}{3}\right)log\left|t\right|+c=\left(\frac{1}{3}\right)log\left|cos3x\right|+c$