Question

If $\int \frac{cosx-sinx}{8-sin2x}dx=\frac{1}{p}log\left[\frac{3+sinx+cosx}{3-sinx-cosx}\right]+c$ then $p=$ .......

• A. $6$
• B. $1$
• C. $3$
• D. $12$

Answer 1

Answer: (A)

We have,
$\int \frac{\cos x-\sin x}{8-\sin 2x}=\frac{1}{p}\log \left[\frac{3+\sin x+\cos x}{3-\sin x-\cos x}\right]+C$
Now, $\int \frac{\cos x-\sin x}{8-\sin 2x}dx$
$=\int \frac{\cos x-\sin x}{9-(1+2\sin x\cos x)}dx$
$=\int \frac{\cos x-\sin x}{9-\left({\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x\right)}dx$
$=\int \frac{\cos x-\sin x}{(3{)}^2-(\cos x+\sin x{)}^2}dx$
put $\cos x+\sin x=t$
$(-\sin x+\cos x)dx=dt$
$=\int \frac{dt}{(3{)}^2-(t{)}^2}=\frac{1}{2(3)}\log \left|\frac{3+t}{3-t}\right|+C$
$=\frac{1}{6}\log \left|\frac{3+\sin x+\cos x}{3-\sin x-\cos x}\right|+C$
$\therefore p=6$