If int frac{d x}{sqrt[3]{sin ^{11} x cos x}}= | vedclass

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Question

(A) Let $I = \int \frac{d x}{\sqrt[3]{\sin ^{11} x \cdot \cos x}}$.Divide the numerator and denominator by $\cos^{\frac{12}{3}} x = \cos^4 x$:$I = \in

Vedclass · Accepted Answer

$f(x)=	an ^{\frac{-8}{3}} x, g(x)=	an ^{\frac{-2}{3}} x$,(where $c$ is a constant of integration)

Vedclass · Answer

(A) Let $I = \int \frac{d x}{\sqrt[3]{\sin ^{11} x \cdot \cos x}}$.Divide the numerator and denominator by $\cos^{\frac{12}{3}} x = \cos^4 x$:$I = \int \frac{\sec^4 x}{	an^{\frac{11}{3}} x} d x$.Using $\sec^4 x = (1 + 	an^2 x) \sec^2 x$,we get:$I = \int \frac{(1 + 	an^2 x) \sec^2 x}{	an^{\frac{11}{3}} x} d x$.Let $	an x = t$,so $\sec^2 x d x = d t$.$I = \int \frac{1 + t^2}{t^{\frac{11}{3}}} d t = \int (t^{-\frac{11}{3}} + t^{-\frac{5}{3}}) d t$.Integrating term by term:$I = \frac{t^{-\frac{8}{3}}}{-\frac{8}{3}} + \frac{t^{-\frac{2}{3}}}{-\frac{2}{3}} + c = -\left(\frac{3}{8} t^{-\frac{8}{3}} + \frac{3}{2} t^{-\frac{2}{3}}ight) + c$.Substituting $t = 	an x$:$I = -\left(\frac{3}{8} 	an^{-\frac{8}{3}} x + \frac{3}{2} 	an^{-\frac{2}{3}} xight) + c$.Comparing with the given form,we get $f(x) = 	an^{-\frac{8}{3}} x$ and $g(x) = 	an^{-\frac{2}{3}} x$.

If $\int \frac{d x}{\sqrt[3]{\sin ^{11} x \cos x}}=-\left(\frac{3}{8} f(x)+\frac{3}{2} g(x)\right)+c$,then:

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