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Let $
$$ I_n = intcos^n{x} dx, ;text{ for } n=0,1,2,3, ldots$

Prove the reduction formula
$$I_n = frac{1}{n} cos^{n-1}(x) sin(x) + frac{n-1}{n} I_{n-2}, ; n geq 2.$$

How do I approach this question? Is there anything I should look out for?

You can conventionally use Integration by parts.

Here it another method:

$$dfrac{d(cos^mxsin x)}{dx}=cos^{m+1}x-mcos^{m-1}x(1-cos^2x)$$

Integrate both sides wrt $x$ to find $$cos^mxsin x+K=(m+1)I_{m+1}-mI_{m-1}$$

We use integration by parts;

$$I_{n}=int cos^{n}(x)dx=int cos^{n-1}(x)cos(x)dx=intfrac{d}{dx}left(sin(x)cos^{n-1}(x)right)+(n-1)sin^{2}(x)cos^{n-2}(x)dx$$

Then recalling $sin^{2}(x)+cos^{2}(x)=1$, we have;

$$I_{n}=sin(x)cos^{n-1}(x)+(n-1)I_{n-2}-(n-1)I_{n}$$

Then solving for $I_{n}$ gives the result.

$$I_n=intcos^n(x)dx=intcos^{n-2}(x).cos^2(x)dx=intcos^{n-2}(x)left[1-sin^2(x)right]dx$$$$
=I_{n-2}-intcos^{n-2}(x)sin^2(x)dx$$$$
=I_{n-2}-frac{cos^{n-1}(x)sin(x)}{n-1}+intfrac{cos^n(x)}{n-1}dx$$
so:
$$I_n=I_{n-2}-frac{cos^{n-1}(x)sin(x)}{n-1}+frac{I_n}{n-1}$$
$$I_n=frac{n-1}{n-2}I_{n-2}-frac{cos^{n-1}(x)sin(x)}{n-2}$$

A bit more detailed, but essentially the same answer as some already given

$$I_n=intcos^nt dt$$
$$I_n=intcos^{n-1}t cos t dt$$
Integration by parts:
$$dv=cos t dtRightarrow v=sin t\u=cos^{n-1}tRightarrow du=-(n-1)cos^{n-2}t sin t dt$$
$$I_n=uv-int vdu$$
$$I_n=cos^{n-1}t sin t-intsin t (-(n-1)cos^{n-2}t sin t)dt$$
$$I_n=cos^{n-1}t sin t+(n-1)intcos^{n-2}t sin^2t dt$$
$$I_n=cos^{n-1}t sin t+(n-1)intcos^{n-2}t (1-cos^2t)dt$$
$$I_n=cos^{n-1}t sin t+(n-1)intcos^{n-2}t dt-(n-1)intcos^nt dt$$
$$I_n=cos^{n-1}t sin t+(n-1)I_{n-2}-(n-1)I_n$$
$$I_n+(n-1)I_n=cos^{n-1}t sin t+(n-1)I_{n-2}$$
$$nI_n=cos^{n-1}t sin t+(n-1)I_{n-2}$$
$$I_n=frac{cos^{n-1}t sin t}n+frac{n-1}nI_{n-2}$$
QED

Nice explanation. What if you change to $$int_{0}^{frac{pi}{2}}cos^{n}xmathrm{d}x$$? What is then change?