Let n ge 2 be a natural number and 0

Question

(A) Let $I = \int \frac{(\sin^n 	heta - \sin 	heta)^{\frac{1}{n}} \cos 	heta}{\sin^{n+1} 	heta} d	heta$.Substitute $\sin 	heta = t$,then $\cos \

Vedclass · Accepted Answer

$\frac{n}{n^2 - 1} \left( 1 - \frac{1}{\sin^{n-1} 	heta} ight)^{\frac{n+1}{n}} + C$

Vedclass · Answer

(A) Let $I = \int \frac{(\sin^n 	heta - \sin 	heta)^{\frac{1}{n}} \cos 	heta}{\sin^{n+1} 	heta} d	heta$.Substitute $\sin 	heta = t$,then $\cos 	heta d	heta = dt$.The integral becomes $I = \int \frac{(t^n - t)^{\frac{1}{n}}}{t^{n+1}} dt$.Factor out $t^n$ from the numerator: $I = \int \frac{(t^n(1 - t^{1-n}))^{\frac{1}{n}}}{t^{n+1}} dt = \int \frac{t(1 - t^{1-n})^{\frac{1}{n}}}{t^{n+1}} dt = \int \frac{(1 - t^{1-n})^{\frac{1}{n}}}{t^n} dt$.Let $z = 1 - t^{1-n}$. Then $dz = -(1-n)t^{-n} dt = (n-1)t^{-n} dt$.Thus,$t^{-n} dt = \frac{dz}{n-1}$.Substituting into the integral: $I = \frac{1}{n-1} \int z^{\frac{1}{n}} dz$.Integrating with respect to $z$: $I = \frac{1}{n-1} \cdot \frac{z^{\frac{1}{n} + 1}}{\frac{1}{n} + 1} + C = \frac{1}{n-1} \cdot \frac{z^{\frac{n+1}{n}}}{\frac{n+1}{n}} + C = \frac{n}{(n-1)(n+1)} z^{\frac{n+1}{n}} + C$.$I = \frac{n}{n^2 - 1} (1 - t^{1-n})^{\frac{n+1}{n}} + C$.Substituting $t = \sin 	heta$ back: $I = \frac{n}{n^2 - 1} (1 - \frac{1}{\sin^{n-1} 	heta})^{\frac{n+1}{n}} + C$.

Let $n \ge 2$ be a natural number and $0 < \theta < \frac{\pi}{2}$. Then $\int \frac{(\sin^n \theta - \sin \theta)^{\frac{1}{n}} \cos \theta}{\sin^{n+1} \theta} d\theta$ is equal to

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