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 . &nbsp a = − 1 16 B . &nbsp a = 1 8 C . &nbsp a = 1 16 D . &nbsp a = − 1 8

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,

Discussion Forum : Indefinite Integration

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

Options:

A .

a = \frac{- 1}{16}

B .

a = \frac{1}{8}

C .

a = \frac{1}{16}

D .

a = \frac{- 1}{8}

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}

Was this answer helpful ?

Next Question

Discussion Forum : Indefinite Integration

Submit Your Solution hear: