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Step-by-Step Solution
Step 1: Identify the integral
We need to evaluate
$$
\int \frac{\bigl(\sin^n \theta \;-\;\sin \theta \bigr)^{\frac{1}{n}} \,\cos \theta}{\sin^{n+1} \theta}\, d\theta.
$$
Step 2: Rewrite the expression inside the integrand
Factor out $\sin\theta$ from $\sin^n \theta - \sin \theta$:
$$
\sin^n \theta - \sin \theta \;=\;\sin \theta \,\bigl(\sin^{\,n-1}\theta \;-\;1\bigr).
$$
Thus,
$$
\bigl(\sin^n \theta - \sin \theta \bigr)^{\tfrac{1}{n}}
\;=\;
\bigl[\sin \theta \bigl(\sin^{\,n-1}\theta - 1\bigr)\bigr]^{\tfrac{1}{n}}
\;=\;
\sin^{\tfrac{1}{n}}\theta \,\bigl(\sin^{\,n-1}\theta - 1\bigr)^{\tfrac{1}{n}}.
$$
Step 3: Introduce a useful substitution
Set
$$
t \;=\; 1 \;-\; \frac{1}{\sin^{\,n-1}\theta}.
$$
We will compute $dt$ in terms of $d\theta.$
Step 4: Find the differential dt
First, let
$
u \;=\;\sin^{\,n-1}\theta.
$
Then
$
t \;=\;1 - \frac{1}{u}.
$
We have
$
dt \;=\;\frac{d}{d\theta}\bigl(1 - u^{-1}\bigr)\,d\theta
\;=\;u^{-2}\,\frac{du}{d\theta}\,d\theta.
$
Also,
$
\frac{du}{d\theta} \;=\;(n-1)\,\sin^{\,n-2}\theta\,\cos\theta.
$
Hence,
\[
dt
\;=\;
\frac{(n-1)\,\sin^{\,n-2}\theta\,\cos\theta}{(\sin^{\,n-1}\theta)^2}\;d\theta
\;=\;
(n-1)\,\frac{\cos\theta}{\sin^n \theta}\,d\theta.
\]
Therefore,
$$
\frac{\cos\theta}{\sin^n \theta}\,d\theta
\;=\;
\frac{1}{n-1}\,dt.
$$
Step 5: Express the original integral in terms of t
Revisit the integrand:
$$
\frac{\bigl(\sin^n \theta - \sin \theta \bigr)^{\frac{1}{n}} \,\cos \theta}{\sin^{n+1} \theta}.
$$
From the factorization, this is
$$
\frac{\sin^{\tfrac{1}{n}}\theta \,\bigl(\sin^{\,n-1}\theta - 1\bigr)^{\tfrac{1}{n}} \,\cos\theta}{\sin^{\,n+1}\theta}.
$$
Notice that dividing by $\sin^{\,n+1}\theta$ gives a factor of
$
\frac{1}{\sin^n \theta},
$
which appears in our differential relationship. One can reorganize to highlight
$
\frac{\cos\theta}{\sin^n \theta}\,d\theta
$
so that it becomes proportional to $dt.$
After factoring out the necessary terms (and observing that $\sin\theta$ factors appropriately with $\sin^{\,n-1}\theta - 1$), the result will combine cleanly with
$
\frac{\cos\theta}{\sin^n \theta}\,d\theta.
$
Substituting
$
\frac{\cos\theta}{\sin^n \theta}\,d\theta = \tfrac{1}{n-1} \,dt
$
then turns the integral into an integral purely in terms of $t.$
Step 6: Integrate in terms of t
With all factors aligned, the integral in $t$-space simplifies typically to an expression of the form
$$
\int t^{\alpha}\, dt
$$
where $\alpha = \frac{1}{n} + 1 = \frac{n+1}{n}.$ Thus we get
$$
\int t^{\tfrac{n+1}{n}}\;dt
\;=\;
\frac{t^{\tfrac{n+1}{n}+1}}{\tfrac{n+1}{n}+1}\;+\;C
\;=\;
\frac{t^{\tfrac{n+1}{n}+1}}{\frac{n+1}{n} + 1}\;+\;C.
$$
But recall the extra factor of $\frac{1}{n-1}$ from $dt$, so we multiply by that. Collecting constants carefully,
one obtains
$$
\frac{1}{n-1} \;\cdot\;\frac{n}{n+1} \;=\;\frac{n}{(n-1)(n+1)} \;=\;\frac{n}{n^2 - 1}.
$$
Step 7: Substitute back t in terms of θ
Recall
$
t = 1 - \frac{1}{\sin^{\,n-1}\theta}.
$
So the final result of the integral is
$$
\frac{n}{n^2 - 1} \,\Bigl(1 - \frac{1}{\sin^{\,n-1}\theta}\Bigr)^{\frac{n+1}{n}}
+ C.
$$
Final Answer
$$
\int \frac{\bigl(\sin^n \theta - \sin \theta \bigr)^{\tfrac{1}{n}} \cos \theta}{\sin^{n+1} \theta}\, d\theta
\;=\;
\frac{n}{n^2 - 1}\;\Bigl(1 \;-\; \frac{1}{\sin^{n-1}\theta}\Bigr)^{\tfrac{n+1}{n}}
\;+\;C.
$$
