Let, $$n \geq 2$$ be a natural number and $$0 \lt \theta \lt \frac{\pi}{2}$$. Then $$\int \frac{(\sin^n\theta - \sin\theta)^{1/n} \cos\theta}{\sin^{n+1}\theta} d\theta$$, is equal to:

We have to evaluate

$$I \;=\;\int \dfrac{(\sin ^n\theta-\sin\theta)^{1/n}\;\cos\theta}{\sin^{\,n+1}\theta}\,d\theta ,\qquad n\ge 2,\;0\lt \theta\lt \dfrac{\pi}{2}.$$

The factor $$\sin ^n\theta-\sin\theta$$ contains a common $$\sin\theta$$, so we write

$$\sin ^n\theta-\sin\theta=\sin\theta\bigl(\sin^{\,n-1}\theta-1\bigr). \quad -(1)$$

Because of the denominator $$\sin^{\,n+1}\theta$$ in the integrand, it is convenient to make the quantity

$$1-\dfrac{1}{\sin^{\,n-1}\theta}$$

appear. Hence we put

$$y \;=\;1-\dfrac{1}{\sin^{\,n-1}\theta}. \quad -(2)$$

We now express the integrand in terms of $$y$$.

First rewrite (1) by multiplying and dividing the bracket by $$\sin^{\,n-1}\theta$$:

$$\sin ^n\theta-\sin\theta=\sin\theta\Bigl[\sin^{\,n-1}\theta-1\Bigr] \;=\;\sin\theta\;\sin^{\,n-1}\theta \Bigl[1-\dfrac{1}{\sin^{\,n-1}\theta}\Bigr] \;=\;\,y\,\sin^{\,n}\theta. \quad -(3)$$

Taking the $$n^{\text{th}}$$ root of (3) we obtain

$$\bigl(\sin ^n\theta-\sin\theta\bigr)^{1/n} \;=\;(y\,\sin^{\,n}\theta)^{1/n} \;=\;y^{1/n}\,\sin\theta. \quad -(4)$$

Substituting (4) into the integrand gives

$$\dfrac{(\sin ^n\theta-\sin\theta)^{1/n}\cos\theta}{\sin^{\,n+1}\theta} \;=\;\dfrac{y^{1/n}\sin\theta\,\cos\theta}{\sin^{\,n+1}\theta} \;=\;y^{1/n}\,\dfrac{\cos\theta}{\sin^{\,n}\theta}. \quad -(5)$$

Next we need $$dy/d\theta$$. From definition (2)

$$y=1-\bigl(\sin\theta\bigr)^{-(n-1)},$$

and using the standard derivative

$$\dfrac{d}{d\theta}\bigl(\sin\theta\bigr)^m =m\bigl(\sin\theta\bigr)^{m-1}\cos\theta,$$

with $$m=-(n-1)$$, we get

$$\dfrac{dy}{d\theta} =-( -(n-1))\,\bigl(\sin\theta\bigr)^{-n}\cos\theta =(n-1)\dfrac{\cos\theta}{\sin^{\,n}\theta}. \quad -(6)$$

Equation (6) shows that

$$\dfrac{\cos\theta}{\sin^{\,n}\theta} =\dfrac{1}{n-1}\,\dfrac{dy}{d\theta}. \quad -(7)$$

Insert (7) into (5):

$$\dfrac{(\sin ^n\theta-\sin\theta)^{1/n}\cos\theta}{\sin^{\,n+1}\theta} \;=\;y^{1/n}\,\dfrac{1}{n-1}\,\dfrac{dy}{d\theta} \;=\;\dfrac{1}{n-1}\,y^{1/n}\,dy/d\theta. \quad -(8)$$

Now the integral becomes

$$I=\int\dfrac{1}{n-1}\,y^{1/n}\,\dfrac{dy}{d\theta}\,d\theta \;=\;\dfrac{1}{n-1}\int y^{1/n}\,dy. \quad -(9)$$

We integrate $$y^{1/n}$$ using the power rule

$$\int y^{k}\,dy=\dfrac{y^{k+1}}{k+1}+C,$$

with $$k=\dfrac{1}{n}$$, so

$$\int y^{1/n}\,dy=\dfrac{y^{1/n+1}}{1/n+1}+C =\dfrac{n}{n+1}\,y^{(n+1)/n}+C. \quad -(10)$$

Insert (10) into (9):

$$I=\dfrac{1}{n-1}\,\dfrac{n}{n+1}\,y^{(n+1)/n}+C =\dfrac{n}{(n-1)(n+1)}\,y^{(n+1)/n}+C =\dfrac{n}{n^{2}-1}\, \Bigl(1-\dfrac{1}{\sin^{\,n-1}\theta}\Bigr)^{\!(n+1)/n}+C. \quad -(11)$$

The expression in (11) is exactly Option C.

Hence, the correct answer is Option 3.