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Actually, you can't integrate it ^^
I'll give it a try :
As I told you, $\displaystyle \int \sin^n x dx=\int \sin^{n-2}x dx -\int \cos^2 x \sin^{n-2} x dx$
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Let's consider $\displaystyle \int \cos^2 x \sin^{n-2} x dx$
$\displaystyle \int \cos^2 x \sin^{n-2} x dx =\int \cos x (\cos x \cdot \sin^{n-2} x) dx \ \leftarrow \text{IBP}$
$\displaystyle v'(x)=\cos x \cdot \sin^{n-2} x \longrightarrow v(x)=\frac{1}{n-1} \cdot \sin^{n-1} x$
$\displaystyle u(x)=\cos x \longrightarrow u'(x)=-\sin x$
Therefore :
$\displaystyle \begin{aligned} \int \cos x (\cos x \cdot \sin^{n-2} x) dx &= \frac{1}{n-1} \cdot \sin^{n-1} x \cdot \cos x - \frac{1}{n-1} \int (-\sin x) \cdot \sin^{n-1} x dx \\
&=\boxed{\frac{1}{n-1} \cdot \sin^{n-1} x \cdot \cos x+\frac{1}{n-1} \int \sin^n x dx} \end{aligned}$
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Getting back to the original integral :
$\displaystyle \int \sin^n x dx=\int \sin^{n-2}x dx -\frac{1}{n-1} \cdot \sin^{n-1} x \cdot \cos x-\frac{1}{n-1} \int \sin^n x dx$
Does it help you seeing through the problem ?