If n geq 2 is a natural number and 0

Question

(D) Let $I = \int \frac{(\cos ^n 	heta-\cos 	heta)^{\frac{1}{n}}}{\cos ^{n+1} 	heta} \sin 	heta d 	heta$. Factor out $\cos^n 	heta$ from the num

Vedclass · Accepted Answer

$\frac{n}{1-n^2}(1-\cos ^{(1-n)} 	heta)^{\frac{n+1}{n}}+c$

Vedclass · Answer

(D) Let $I = \int \frac{(\cos ^n 	heta-\cos 	heta)^{\frac{1}{n}}}{\cos ^{n+1} 	heta} \sin 	heta d 	heta$. Factor out $\cos^n 	heta$ from the numerator: $I = \int \frac{(\cos^n 	heta (1 - \cos^{1-n} 	heta))^{\frac{1}{n}}}{\cos^{n+1} 	heta} \sin 	heta d 	heta$ $I = \int \frac{\cos 	heta (1 - \cos^{1-n} 	heta)^{\frac{1}{n}}}{\cos^{n+1} 	heta} \sin 	heta d 	heta = \int \frac{(1 - \cos^{1-n} 	heta)^{\frac{1}{n}}}{\cos^n 	heta} \sin 	heta d 	heta$. Let $t = 1 - \cos^{1-n} 	heta$. Then $dt = -(1-n) \cos^{-n} 	heta (-\sin 	heta) d 	heta = (1-n) \cos^{-n} 	heta \sin 	heta d 	heta$. Thus,$\frac{dt}{1-n} = \frac{\sin 	heta}{\cos^n 	heta} d 	heta$. Substituting into the integral: $I = \int \frac{t^{\frac{1}{n}}}{1-n} dt = \frac{1}{1-n} \cdot \frac{t^{\frac{1}{n}+1}}{\frac{1}{n}+1} + c = \frac{1}{1-n} \cdot \frac{t^{\frac{n+1}{n}}}{\frac{n+1}{n}} + c$ $I = \frac{n}{(1-n)(n+1)} t^{\frac{n+1}{n}} + c = \frac{n}{1-n^2} (1 - \cos^{1-n} 	heta)^{\frac{n+1}{n}} + c$.

If $n \geq 2$ is a natural number and $0 < \theta < \frac{\pi}{2}$,then $\int \frac{(\cos ^n \theta-\cos \theta)^{1 / n}}{\cos ^{n+1} \theta} \sin \theta d \theta =$

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