Question

$\int \frac{1-7co{s}^{2}x}{si{n}^{7}xco{s}^{2}x}dx=\frac{f\left(x\right)}{(sinx{)}^7}+C$, then f(x) is equal to

• A. sin x
• B. cos x
• C. tan x
• D. cot x

Answer 1

The correct option is C tan x

$\int \frac{1-7co{s}^{2}x}{si{n}^{7}xco{s}^{2}x}dx=\int (\frac{se{c}^{2}x}{si{n}^{7}x}-\frac{7}{si{n}^{7}x})dx=\int \frac{se{c}^{2}x}{si{n}^{7}x}dx-\int \frac{7}{si{n}^7}dx=I_1+I_{2}Now,I_1=\int \frac{se{c}^{2}x}{si{n}^7}dx=\frac{tanx}{si{n}^{7}x}+7\int \frac{tanxcosx}{si{n}^{8}x}=\frac{tanx}{si{n}^{7}x}-I_2\therefore I_1+I_2=\frac{tanx}{si{n}^{7}x}+C\Rightarrow f\left(x\right)=tanx$