∫ 1 − 7 c o s 2 x s i n 7 x c o s 2 x d x = f ( x ) ( s i n x ) 7 + C , then f(x) is equal to Options: A . &nbsp sin x B . &nbsp cos x C . &nbsp tan x D . &nbsp cot x

Answer: Option C : C ∫ 1 − 7 c o s 2 x s i n 7 x c o s 2 x d x = ∫ ( s e c 2 x s i n 7 x − 7 s i n 7 x ) d x = ∫ s e c 2 x s i n 7 x d x − ∫ 7 s i n 7 d x = I 1 + I 2 N o w , I 1 = ∫ s e c 2 x s i n 7 d x = t a n x s i n 7 x + 7 ∫ t a n x c o s x s i n 8 x = t a n x s i n 7 x − I 2 ∴ I 1 + I 2 = t a n x s i n 7 x + C ⇒ f ( x ) = t a n x

∫1−7 cos2xsin7xcos2xdx=f(x)(sinx)7+C, then f(x) is equ

Discussion Forum : Indefinite Integration

Question -

$\int \frac{1 - 7 c o s^{2} x}{s i n^{7} x c o s^{2} x} d x = \frac{f (x)}{(s i n x)^{7}} + C$ , then f(x) is equal to

Options:

A . sin x

B . cos x

C . tan x

D . cot x

Answer: Option C
:
C

\int \frac{1 - 7 c o s^{2} x}{s i n^{7} x c o s^{2} x} d x = \int (\frac{s e c^{2} x}{s i n^{7} x} - \frac{7}{s i n^{7} x}) d x = \int \frac{s e c^{2} x}{s i n^{7} x} d x - \int \frac{7}{s i n^{7}} d x = I_{1} + I_{2} N o w, I_{1} = \int \frac{s e c^{2} x}{s i n^{7}} d x = \frac{t a n x}{s i n^{7} x} + 7 \int \frac{t a n x c o s x}{s i n^{8} x} = \frac{t a n x}{s i n^{7} x} - I_{2} ∴ I_{1} + I_{2} = \frac{t a n x}{s i n^{7} x} + C \Rightarrow f (x) = t a n x

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