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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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## QuestionMathsClass 12

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. $$\frac{n}{n^{2}+1}\left(1-\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{n+1}{n}}+C$$

C
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4.6

## 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
$$\Rightarrow I=\int \frac{u^{1 / n} d u}{n-1}=\frac{u^{\frac{1}{n}+1}}{(n-1)\left(\frac{1}{n}+1\right)}+C$$
$$=\frac{n\left(1-\frac{1}{t^{n-1}}\right)^{\frac{n+1}{n}}}{(n-1)(n+1)}+C$$
$$=\frac{n\left(1-\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{n+1}{n}}}{n^{2}-1}+C$$
$$\left[\because u=1-\frac{1}{t^{n-1}} \text { and } t=\sin \theta\right]$$