int cos ^{frac{-3}{7}} x cdot sin ^{frac{-11} | vedclass

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Question

(C) Let $I = \int \cos ^{-\frac{3}{7}} x \cdot \sin ^{-\frac{11}{7}} x \, dx$. We can rewrite the integrand as: $I = \int \frac{\sin ^{-\frac{11}{7}}

Vedclass · Accepted Answer

$-\frac{7}{4} 	an ^{\frac{-4}{7}} x + c$,where $c$ is a constant of integration.

Vedclass · Answer

(C) Let $I = \int \cos ^{-\frac{3}{7}} x \cdot \sin ^{-\frac{11}{7}} x \, dx$. We can rewrite the integrand as: $I = \int \frac{\sin ^{-\frac{11}{7}} x}{\cos ^{-\frac{3}{7}} x} \, dx = \int \frac{\sin ^{-\frac{11}{7}} x}{\cos ^{-\frac{11}{7}} x \cdot \cos ^{-\frac{3}{7} + \frac{11}{7}} x} \, dx$ $I = \int \frac{\sin ^{-\frac{11}{7}} x}{\cos ^{-\frac{11}{7}} x} \cdot \frac{1}{\cos ^{\frac{8}{7}} x} \, dx = \int 	an ^{-\frac{11}{7}} x \cdot \sec ^{\frac{8}{7}} x \, dx$ Wait,let us simplify differently: $I = \int \cos ^{-\frac{3}{7}} x \cdot \sin ^{-\frac{11}{7}} x \, dx = \int \frac{1}{\cos ^{\frac{3}{7}} x \cdot \sin ^{\frac{11}{7}} x} \, dx$ Divide numerator and denominator by $\cos ^{\frac{14}{7}} x = \cos ^2 x$: $I = \int \frac{\sec ^2 x}{\frac{\cos ^{\frac{3}{7}} x \cdot \sin ^{\frac{11}{7}} x}{\cos ^2 x}} \, dx = \int \frac{\sec ^2 x}{	an ^{\frac{11}{7}} x} \, dx = \int 	an ^{-\frac{11}{7}} x \cdot \sec ^2 x \, dx$ Using the substitution $u = 	an x$,then $du = \sec ^2 x \, dx$: $I = \int u^{-\frac{11}{7}} \, du = \frac{u^{-\frac{11}{7} + 1}}{-\frac{11}{7} + 1} + c = \frac{u^{-\frac{4}{7}}}{-\frac{4}{7}} + c = -\frac{7}{4} 	an ^{-\frac{4}{7}} x + c$.

$\int \cos ^{\frac{-3}{7}} x \cdot \sin ^{\frac{-11}{7}} x \, dx =$

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