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#### Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Multiple Choice Questions Question 7 Maths Textbook Solution.

$\frac{1}{16}$

Given:

$\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x=a \cos 8 x+C$

Hint

Using $\int \sin x d x$

Explanation:

Let $I=\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x$

$=\int \frac{2 \cos ^{2} 4 x-1+1}{\frac{\sin 2 x}{\cos 2 x}-\frac{\cos 2 x}{\sin 2 x}} d x \quad\left\[\because 2 \cos ^{2} x=1+\cos 2 x ; \tan x=\frac{\sin x}{\cos x} ; \cot x=\frac{\cos x}{\sin x}\right]$

$=\int \frac{2 \cos ^{2} 4 x \cdot \sin 2 x \cos 2 x}{\sin ^{2} 2 x-\cos ^{2} 2 x} d x$

$=\int \frac{\cos ^{2} 4 x \sin 4 x}{-\cos 4 x} d x$                            $\left[\because \cos ^{2} x-\sin ^{2} x=\cos 2 x\right]$

$=-\int \cos 4 x \sin 4 x d x$

$=-\frac{1}{2} \int \sin 8 x d x$                                    $[\because \sin 2 x=2 \sin x \cos x]$

$=-\frac{1}{2} \int \sin 8 x d x$                                    $[\because \sin 2 x=2 \sin x \cos x]$

\begin{aligned} &=-\frac{1}{2}\left(-\frac{\cos 8 x}{8}\right)+C \\ &I=\frac{\cos 8 x}{16}+C \end{aligned}

According to given,

\begin{aligned} &I=a \cos 8 x+C \\ &\because a \cos 8 x+C=\frac{1}{16} \cos 8 x+C \\ &\therefore a=\frac{1}{16} \end{aligned}