Mathematics - Indefinite Integration Question with Solution | TestHub

To find $f(x)$, we need to evaluate the integral $\int \frac{1-7{\cos }^{2}x}{{\sin }^{7}x{\cos }^{2}x}dx$.

First, split the integrand:

$$\int \left(\frac{1}{{\sin }^{7}x{\cos }^{2}x}-\frac{7{\cos }^{2}x}{{\sin }^{7}x{\cos }^{2}x}\right)dx$$

$$=\int \left(\frac{{\sec }^{2}x}{{\sin }^{7}x}-\frac{7}{{\sin }^{7}x}\right)dx$$

$$=\int \frac{{\sec }^{2}x}{{\sin }^{7}x}dx-7\int \frac{1}{{\sin }^{7}x}dx$$

Let's use integration by parts for the first term, $\int \frac{{\sec }^{2}x}{{\sin }^{7}x}dx$.

Let $u=\frac{1}{{\sin }^{7}x}=(\sin x{)}^{-7}$ and $dv={\sec }^{2}xdx$.

Then $du=-7(\sin x{)}^{-8}\cos xdx=-7\frac{\cos x}{{\sin }^{8}x}dx$ and $v=\tan x$.

Applying the integration by parts formula $\int udv=uv-\int vdu$:

$$\int \frac{{\sec }^{2}x}{{\sin }^{7}x}dx=\frac{\tan x}{{\sin }^{7}x}-\int \tan x\left(-7\frac{\cos x}{{\sin }^{8}x}\right)dx$$

$$=\frac{\tan x}{{\sin }^{7}x}+7\int \frac{\sin x}{\cos x}\frac{\cos x}{{\sin }^{8}x}dx$$

$$=\frac{\tan x}{{\sin }^{7}x}+7\int \frac{1}{{\sin }^{7}x}dx$$

Substitute this back into the original integral:

$$\int \frac{1-7{\cos }^{2}x}{{\sin }^{7}x{\cos }^{2}x}dx=\left(\frac{\tan x}{{\sin }^{7}x}+7\int \frac{1}{{\sin }^{7}x}dx\right)-7\int \frac{1}{{\sin }^{7}x}dx+C$$

$$=\frac{\tan x}{{\sin }^{7}x}+C$$

Comparing this with the given form $\frac{f(x)}{(\sin x{)}^7}+C$, we find that $f(x)=\tan x$.

The final answer is $\boxed{Tanx}$.