If -cos 8 x-1-tan 2 x- 2 x d x-a cos 8 x-C- then

The correct option is C a=1/16∫cos 8x+1/tan 2x cot 2xdx =∫2 cos^2 4x/sin^2 2x cos^2 2x.sin 2x cos 2x dx = ∫ sin 4x cos 4x dx= 1/2∫ sin 8x dx =1/16cos 8x+ ...

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Standard XII

Mathematics

Integration of Trigonometric Functions

If ∫cos 8 x+1...

Question

If $\int \frac{c o s 8 x + 1}{t a n 2 x - c o t 2 x} d x = a c o s 8 x + C$ , then

a = \frac{- 1}{16}

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a = \frac{1}{8}

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a = \frac{1}{16}

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a = \frac{- 1}{8}

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Solution

The correct option is C $a = \frac{1}{16}$
$\int \frac{c o s 8 x + 1}{t a n 2 x - c o t 2 x} d x = \int \frac{2 c o s^{2} 4 x}{s i n^{2} 2 x - c o s^{2} 2 x} . s i n 2 x c o s 2 x d x = - \int s i n 4 x c o s 4 x d x = - \frac{1}{2} \int s i n 8 x d x = \frac{1}{16} c o s 8 x + C ∴ a = \frac{1}{16}$

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If ∫cos 8x+1tan 2x−cot 2xdx=a cos 8x+C, then

The correct option is C a=116∫cos 8x+1tan 2x−cot 2xdx=∫2 cos2 4xsin2 2x−cos2 2x.sin 2x cos 2x dx=−∫sin 4x cos 4x dx=−12∫sin 8x dx=116cos 8x+C∴a=116

If $\int \frac{c o s 8 x + 1}{t a n 2 x - c o t 2 x} d x = a c o s 8 x + C$ , then