If the integral 8x+1 2x- 2x,dx = A 8x+k where k is an arbitrary constant, then A is equal …

Question

If the integral $\displaystyle\int \dfrac{\cos 8x+1}{\cot 2x-	an 2x}\,dx = A\cos 8x+k$ where $k$ is an arbitrary constant, then $A$ is equal to:

Accepted Answer

$\cot 2x-	an 2x = \dfrac{\cos 2x}{\sin 2x}-\dfrac{\sin 2x}{\cos 2x} = \dfrac{\cos^22x-\sin^22x}{\sin 2x\cos 2x} = \dfrac{2\cos 4x}{\sin 4x}$

$I = \int \dfrac{2\cos^24x}{\dfrac{2\cos 4x}{\sin4x}}\,dx = \int \dfrac{2\cos^24x\cdot\sin4x}{2\cos4x}\,dx = \dfrac12\int \sin8x\,dx = -\dfrac12\cdot\dfrac{\cos8x}{8}+k$

$= -\dfrac{1}{16}\cos8x+k$

$\Rightarrow A = -\dfrac{1}{16}$

Solutions · 1