Question

Let $n \ge 2$ be a natural number and $0 < \theta < p/2$.
Then $\int \frac{\left(\sin^{n} \theta - \sin\theta\right)^{\frac{1}{n}} \cos \theta}{\sin^{n+1} \theta} d \theta $ 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 1

Answer: (C)

$\int \frac{\left(\sin ^{n} \theta-\sin \theta\right)^{1 / n} \cos \theta}{\sin ^{n+1} \theta} d \theta$
$=\int \frac{\sin \theta\left(1-\frac{1}{\sin ^{n-1} \theta}\right)^{1 / n}}{\sin ^{n+1} \theta} d \theta$
Put $1-\frac{1}{\sin ^{n-1} \theta}=t$
So $\frac{(n-1)}{\sin ^{n} \theta} \cos \theta d \theta=d t$
Now $\frac{1}{n-1} \int(t)^{1 / n} d t$
$=\frac{1}{(n-1)} \frac{(t)^{\frac{1}{n}+1}}{\frac{1}{n}+1}+C$
$\frac{1}{(n-1)}\left(1-\frac{1}{\sin ^{n-1} \theta}\right)^{\frac{1}{n}+1}+C$