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Let \(n \geq 2\) be a natural number and \(0<\theta<\frac{\pi}{2} . \text {Then,} \int \frac{\left(\sin ^{n} \theta-\sin \theta\right)^{\frac{1}{n}} \cos \theta}{\sin ^{n+1} \theta} d \theta \text { is equal to }\)
(where \(C\) is a constant of integration)( )
A. \(\frac{n}{n^{2}-1}\left(1-\frac{1}{\sin ^{n+1} \theta}\right)^{\frac{n+1}{n}}+C\)
B. \(\frac{n}{n^{2}-1}\left(1+\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{n+1}{n}}+C\)
C. \(\frac{n}{n^{2}-1}\left(1-\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{n+1}{n}}+C\)
D. \(\frac{n}{n^{2}+1}\left(1-\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{n+1}{n}}+C\)
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Let \(n \geq 2\) be a natural number and \(0<\theta<\frac{\pi}{2} . \text {Then,} \int \frac{\left(\sin ^{n} \theta-\sin \theta\right)^{\frac{1}{n}} \cos \theta}{\sin ^{n+1} \theta} d \theta \text { is equal to }\)
(where \(C\) is a constant of integration)( )
A. \(\frac{n}{n^{2}-1}\left(1-\frac{1}{\sin ^{n+1} \theta}\right)^{\frac{n+1}{n}}+C\)
B. \(\frac{n}{n^{2}-1}\left(1+\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{n+1}{n}}+C\)
C. \(\frac{n}{n^{2}-1}\left(1-\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{n+1}{n}}+C\)
D. \(\frac{n}{n^{2}+1}\left(1-\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{n+1}{n}}+C\)

Answer

C
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Solution

\(\text { Let } I=\int \frac{\left(\sin ^{n} \theta-\sin \theta\right)^{1 / n} \cos \theta}{\sin ^{n+1} \theta} d \theta\)
\(\text { Put } \sin \theta=t \Rightarrow \cos \theta d \theta=d t\)
\(\therefore \quad I=\int \frac{\left(t^{n}-t\right)^{1 / n}}{t^{n+1}} d t\)
\(=\int \frac{\left[t^{n}\left(1-\frac{t}{t^{n}}\right)\right]^{1 / n}}{t^{n+1}} d t\)
\(=\int \frac{t\left(1-1 / t^{n-1}\right)^{1 / n}}{t^{n+1}} d t=\int \frac{\left(1-1 / t^{n-1}\right)^{1 / n}}{t^{n}} d t\)
\(\text { Put } \quad 1-\frac{1}{t^{n-1}}=u\)
\(\text { or } \quad 1-t^{-(n-1)}=u \Rightarrow \frac{(n-1)}{t^{n}} d t=d u\)
\(\Rightarrow \quad \frac{d t}{t^{n}}=\frac{d u}{n-1}\)
Analysis
\(\Rightarrow I=\int \frac{u^{1 / n} d u}{n-1}=\frac{u^{\frac{1}{n}+1}}{(n-1)\left(\frac{1}{n}+1\right)}+C\)
\(=\frac{n\left(1-\frac{1}{t^{n-1}}\right)^{\frac{n+1}{n}}}{(n-1)(n+1)}+C\)
\(=\frac{n\left(1-\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{n+1}{n}}}{n^{2}-1}+C\)
\(\left[\because u=1-\frac{1}{t^{n-1}} \text { and } t=\sin \theta\right]\)
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