Question
If $\int_{}^{}\ \left( \frac{1 - 5\cos^{2}x}{\sin^{5}x\cos^{2}x} \right)dx = f(x) + C$ where $C$ is the constant of integration, then $f\left( \frac{\pi}{6} \right) - f\left( \frac{\pi}{4} \right)$ is equal to
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Solution
$\int_{}^{}\ \frac{dx}{\sin^{5}\cos^{2}x} - 5\int_{}^{}\ \frac{dx}{\sin^{5}x}$
$$\int\frac{\sec^{2}xdx}{\sin^{5}x} - 5\int\frac{dx}{\sin^{5}x} $$By IBP
$${= \frac{tanx}{\sin^{5}x} - \int - \frac{5}{\sin^{6}x} \cdot cosx \cdot tanxdx - 5\int\frac{dx}{\sin^{5}x} }{= \frac{tanx}{\sin^{5}x} + c }{f(x) = \frac{tanx}{\sin^{5}x} }{f\left( \frac{\pi}{6} \right) - f\left( \frac{\pi}{4} \right) = \frac{2^{5}}{\sqrt{3}} - (\sqrt{2})^{5} = 4\sqrt{2} - \frac{32}{\sqrt{3}} }{= \frac{32}{\sqrt{3}} - 4\sqrt{2} }{= \frac{4}{\sqrt{3}}(8 - \sqrt{6})}$$
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