If ∫ 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 Options: A . &nbspa=−116 B . &nbspa=18 C . &nbspa=116 D . &nbspa=−18

Answer: Option C : C ∫ c o s 8 x + 1 t a n 2 x − c o t 2 x d x = ∫ 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 = − ∫ s i n 4 x c o s 4 x d x = − 1 2 ∫ s i n 8 x d x = 1 16 c o s 8 x + C ∴ a = 1 16

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

Question

\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

Options:

A . a=−116

B . a=18

C . a=116

D . a=−18

Answer: Option C
:
C

\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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