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

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\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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