1. ## Integration!!

Show that

2. Hello,

Write : $\sin^n x=\sin^{n-2} x \cdot \sin^2 x=\sin^{n-2}x \cdot (1-\cos^2 x)=\sin^{n-2}x-\cos^2x \cdot \sin^{n-2}x$

Integrate by parts the latter term

3. ## Thank you(Further Qn)

Hey Moo could you give me a further tip on integrating especially
sin^(n-2)x plz!!!!

4. Originally Posted by matty888
Hey Moo could you give me a further tip on integrating especially
sin^(n-2)x plz!!!!
Actually, you can't integrate it ^^
I'll give it a try :

As I told you, $\int \sin^n x dx=\int \sin^{n-2}x dx -\int \cos^2 x \sin^{n-2} x dx$

~~~~~~~~~~~~
Let's consider $\int \cos^2 x \sin^{n-2} x dx$

$\int \cos^2 x \sin^{n-2} x dx =\int \cos x (\cos x \cdot \sin^{n-2} x) dx \ \leftarrow \text{IBP}$
$v'(x)=\cos x \cdot \sin^{n-2} x \longrightarrow v(x)=\frac{1}{n-1} \cdot \sin^{n-1} x$
$u(x)=\cos x \longrightarrow u'(x)=-\sin x$

Therefore :
\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}

~~~~~~~~~~~~

Getting back to the original integral :

$\int \sin^n x dx=\int \sin^{n-2}x dx -\frac{1}{n-1} \cdot \sin^{n-1} x \cdot \cos x-\frac{1}{n-1} \int \sin^n x dx$