If int left( frac{1-5 cos^{2}x}{sin^{5}x cos^ | vedclass

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Question

(B) Let $I = \int \left( \frac{1-5 \cos^{2}x}{\sin^{5}x \cos^{2}x} \right) dx = \int \left( \frac{\sec^{2}x}{\sin^{5}x} - \frac{5}{\sin^{5}x} \right)

Vedclass · Accepted Answer

$\frac{4}{\sqrt{3}}(8-\sqrt{6})$

Vedclass · Answer

(B) Let $I = \int \left( \frac{1-5 \cos^{2}x}{\sin^{5}x \cos^{2}x} ight) dx = \int \left( \frac{\sec^{2}x}{\sin^{5}x} - \frac{5}{\sin^{5}x} ight) dx$. Using integration by parts on $\int \frac{\sec^{2}x}{\sin^{5}x} dx$, let $u = \frac{1}{\sin^{5}x}$ and $dv = \sec^{2}x dx$. Then $du = -5 \sin^{-6}x \cos x dx$ and $v = 	an x$. $\int \frac{\sec^{2}x}{\sin^{5}x} dx = \frac{	an x}{\sin^{5}x} - \int 	an x (-5 \sin^{-6}x \cos x) dx = \frac{	an x}{\sin^{5}x} + 5 \int \frac{\sin x / \cos x \cdot \cos x}{\sin^{6}x} dx = \frac{	an x}{\sin^{5}x} + 5 \int \frac{1}{\sin^{5}x} dx$. Substituting this back into $I$: $I = \left( \frac{	an x}{\sin^{5}x} + 5 \int \frac{1}{\sin^{5}x} dx ight) - 5 \int \frac{1}{\sin^{5}x} dx = \frac{	an x}{\sin^{5}x} + C$. Thus, $f(x) = \frac{	an x}{\sin^{5}x} = \frac{\sin x}{\cos x \sin^{5}x} = \frac{1}{\cos x \sin^{4}x}$. $f(\frac{\pi}{6}) = \frac{1}{\cos(\pi/6) \sin^{4}(\pi/6)} = \frac{1}{(\sqrt{3}/2) \cdot (1/2)^{4}} = \frac{1}{(\sqrt{3}/2) \cdot (1/16)} = \frac{32}{\sqrt{3}}$. $f(\frac{\pi}{4}) = \frac{1}{\cos(\pi/4) \sin^{4}(\pi/4)} = \frac{1}{(1/\sqrt{2}) \cdot (1/\sqrt{2})^{4}} = \frac{1}{(1/\sqrt{2}) \cdot (1/4)} = 4\sqrt{2}$. $f(\frac{\pi}{6}) - f(\frac{\pi}{4}) = \frac{32}{\sqrt{3}} - 4\sqrt{2} = \frac{32 - 4\sqrt{6}}{\sqrt{3}} = \frac{4}{\sqrt{3}}(8 - \sqrt{6})$.

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 $f(\frac{\pi}{6}) - f(\frac{\pi}{4})$ is equal to

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