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Mathematics Question on integral

If ∫cos⁡ 8x+1tan⁡ 2x−cot⁡ 2xdx=a cos⁡ 8x+c,\int\limits \frac{\cos \, 8x + 1}{ \tan \, 2x - \cot \, 2x} dx = a \, \cos \, 8x + c , then aa =

A

−116 - \frac{1}{16}

B

18 \frac{1}{8}

C

116 \frac{1}{16}

D

−18 - \frac{1}{8}

Answer

116 \frac{1}{16}

Explanation

Solution

L.H.S = ∫cos⁡ 8x+1tan⁡ 2x−cot⁡ 2xdx\int\limits \frac{\cos \, 8x + 1}{ \tan \, 2x - \cot \, 2x} dx
∫2cos⁡24x(sin⁡22x−cos⁡22xsin⁡2xcos⁡2x)dx\int\frac{2 \cos^{2} 4x}{\left(\frac{\sin^{2} 2x -\cos^{2}2x}{\sin 2x\cos 2x}\right)} dx
=−∫cos⁡24x(2sin⁡2xcos⁡2x)(cos⁡22x−sin⁡22x)dx= - \int\frac{\cos^{2} 4x \left(2 \sin 2x \cos 2x\right)}{\left(\cos^{2} 2x - \sin^{2 } 2x \right) }dx
=−∫cos⁡24x×sin⁡ 4xcos⁡ 4xdx= - \int\frac{\cos^{2} 4x \times \sin \, 4x}{\cos\, 4x}dx
=−12∫2 sin⁡ 4x cos 4x dx= - \frac{1}{2} \int 2 \, \sin \, 4x \, cos \, 4x \, dx
=−12∫sin⁡ 8x dx=12×cos⁡ 8x8+c= - \frac{1}{2} \int \sin \, 8x \, dx = \frac{1}{2} \times \frac{\cos \, 8x}{8} + c
Now, 12cos⁡ 8x8+c=a cos 8x+c \frac{1}{2} \frac{\cos \, 8x}{8} +c = a \, cos \, 8x + c
∴  a=116\therefore \:\: a = \frac{1}{16}