If ∫ ( 1-5 cos ^2 x sin ^5 x cos ^2 x ) d x=f(x)+ C, where C is the constant of integration, then f ( π 6 )-f ( π 4 ) is equal to

$$\int \frac{1 - 5\cos^2 x}{\sin^5 x \cos^2 x} dx = \int \frac{dx}{\sin^5 x \cos^2 x} - 5\int \frac{dx}{\sin^5 x}$$∫sin5xcos2x1−5cos2x​dx=∫sin5xcos2xdx​−5∫sin5xdx​

$$= \int \frac{\sec^2 x \, dx}{\sin^5 x} - 5\int \frac{dx}{\sin^5 x}$$=∫sin5xsec2xdx​−5∫sin5xdx​

By IBP:

$$= \frac{\tan x}{\sin^5 x} - \int \left(-\frac{5}{\sin^6 x}\right) \cos x \cdot \tan x \, dx - 5\int \frac{dx}{\sin^5 x}$$=sin5xtanx​−∫(−sin6x5​)cosx⋅tanxdx−5∫sin5xdx​

$$= \frac{\tan x}{\sin^5 x} + C$$=sin5xtanx​+C

So $f(x) = \frac{\tan x}{\sin^5 x}$f(x)=sin5xtanx​

$$f\left(\frac{\pi}{6}\right) - f\left(\frac{\pi}{4}\right) = \frac{2^5}{\sqrt{3}} - (\sqrt{2})^5 = \frac{32}{\sqrt{3}} - 4\sqrt{2} = \frac{4}{\sqrt{3}}(8 - \sqrt{6})$$f(6π​)−f(4π​)=3​25​−(2​)5=3​32​−42​=3​4​(8−6​)