Q96 / 180 · COMEDK 2015 · Mathematics

If $\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x=a \cos 8 x+c$, then $a=$

  • A
    $-\frac{1}{16}$
  • B
    $\frac{1}{8}$
  • Correct answer
    C
    $\frac{1}{16}$
  • D
    $-\frac{1}{8}$

Solution

$\mathrm{LHS}=\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x$ $=\int \frac{2 \cos ^{2} 4 x}{\left(\frac{\sin ^{2} 2 x-\cos ^{2} 2 x}{\sin 2 x \cos 2 x}\right)} d x$ $\left[\because \cos 2 x=2 \cos ^{2} x-1\right]$ $=-\int \frac{\cos ^{2} 4 x(2 \sin 2 x \cos 2 x)}{\left(\cos ^{2} 2 x-\sin ^{2} 2 x\right)} d x$ $=-\int \frac{\cos ^{2} 4 x \times \sin 4 x}{\cos 4 x} d x$ $[\because \sin 2 x=2 \sin x \cos x]$ $=-\frac{1}{2} \int 2 \sin 4 x \cos 4 x d x$ $=-\frac{1}{2} \int \sin 8 x d x=\frac{1}{2} \times \frac{\cos 8 x}{8}+c$ Now, $\frac{1}{2} \frac{\cos 8 x}{8}+c=a \cos 8 x+c$ $\therefore \quad a=\frac{1}{16}$

Hence, the correct answer is option: 'C'.

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