Solve this following

Question:

Let $\mathrm{n} \geq 2$ be a natural number and $0<\theta<\pi / 2$.

Then $\int \frac{\left(\sin ^{\mathrm{n}} \theta-\sin \theta\right)^{\frac{1}{n}} \cos \theta}{\sin ^{\mathrm{n}+1} \theta} \mathrm{d} \theta$ is equal to :

(Where $\mathrm{C}$ is a constant of integration)

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

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

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

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


Correct Option: , 3

Solution:

$\int \frac{\left(\sin ^{n} \theta-\sin \theta\right)^{1 / n} \cos \theta}{\sin ^{n+1} \theta} d \theta$

$=\int \frac{\sin \theta\left(1-\frac{1}{\sin ^{\mathrm{n}-1} \theta}\right)^{1 / \mathrm{n}}}{\sin ^{\mathrm{n}+1} \theta} \mathrm{d} \theta$

Put $1-\frac{1}{\sin ^{n-1} \theta}=\mathrm{t}$

So $\frac{(n-1)}{\sin ^{n} \theta} \cos \theta d \theta=d t$

Now $\frac{1}{n-1} \int(t)^{1 / n} d t$

$=\frac{1}{(n-1)} \frac{(t)^{\frac{1}{n}+1}}{\frac{1}{n}+1}+C$

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

 

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