int sin^{frac{-1}{2}}x cos^{frac{-7}{2}}x , d | vedclass

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Question

(C) Let $I = \int \sin^{\frac{-1}{2}}x \cos^{\frac{-7}{2}}x \, dx$. We can rewrite the integrand as: $I = \int \frac{1}{\sin^{\frac{1}{2}}x \cos^{\fra

Vedclass · Accepted Answer

$2 	an^{\frac{1}{2}}x + \frac{2}{5} 	an^{\frac{5}{2}}x + C$

Vedclass · Answer

(C) Let $I = \int \sin^{\frac{-1}{2}}x \cos^{\frac{-7}{2}}x \, dx$. We can rewrite the integrand as: $I = \int \frac{1}{\sin^{\frac{1}{2}}x \cos^{\frac{7}{2}}x} \, dx = \int \frac{1}{\left(\frac{\sin x}{\cos x}ight)^{\frac{1}{2}} \cos^4 x} \, dx$. $I = \int \frac{\sec^4 x}{	an^{\frac{1}{2}}x} \, dx = \int \frac{\sec^2 x \cdot \sec^2 x}{	an^{\frac{1}{2}}x} \, dx$. Since $\sec^2 x = 1 + 	an^2 x$,we have: $I = \int \frac{(1 + 	an^2 x) \sec^2 x}{	an^{\frac{1}{2}}x} \, dx$. Let $u = 	an x$,then $du = \sec^2 x \, dx$. $I = \int \frac{1 + u^2}{u^{\frac{1}{2}}} \, du = \int (u^{-\frac{1}{2}} + u^{\frac{3}{2}}) \, du$. Integrating with respect to $u$: $I = \frac{u^{\frac{1}{2}}}{\frac{1}{2}} + \frac{u^{\frac{5}{2}}}{\frac{5}{2}} + C = 2u^{\frac{1}{2}} + \frac{2}{5}u^{\frac{5}{2}} + C$. Substituting $u = 	an x$ back: $I = 2	an^{\frac{1}{2}}x + \frac{2}{5}	an^{\frac{5}{2}}x + C$.

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

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