Question

If $\int \frac{cos\,x-sin\,x}{8-sin\,2x} dx = \frac{1}{p} log \left[\frac{3+sin\,x+cos\,x}{3-sin\,x -cos\,x}\right] +c$ then $p= $ .......

• A. $6$
• B. $1$
• C. $3$
• D. $12$

Answer 1

Answer: (A)

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