Let $ $$ I_n = \int\cos^n{x} \ dx, \;\text{ for } n=0,1,2,3, \ldots$
Prove the reduction formula $$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^mx\sin x)}{dx}=\cos^{m+1}x-m\cos^{m-1}x(1-\cos^2x)$$
Integrate both sides wrt $x$ to find $$\cos^mx\sin 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=\int\frac{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=\int\cos^n(x)dx=\int\cos^{n-2}(x).\cos^2(x)dx=\int\cos^{n-2}(x)\left[1-\sin^2(x)\right]dx$$$$ =I_{n-2}-\int\cos^{n-2}(x)\sin^2(x)dx$$$$ =I_{n-2}-\frac{\cos^{n-1}(x)\sin(x)}{n-1}+\int\frac{\cos^n(x)}{n-1}dx$$ 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
$$I_n=\int\cos^nt\ dt$$ $$I_n=\int\cos^{n-1}t\ \cos t\ dt$$ Integration by parts: $$dv=\cos t\ dt\Rightarrow v=\sin t\\u=\cos^{n-1}t\Rightarrow du=-(n-1)\cos^{n-2}t\ \sin t\ dt$$ $$I_n=uv-\int vdu$$ $$I_n=\cos^{n-1}t\ \sin t-\int\sin t\ (-(n-1)\cos^{n-2}t\ \sin t)dt$$ $$I_n=\cos^{n-1}t\ \sin t+(n-1)\int\cos^{n-2}t\ \sin^2t\ dt$$ $$I_n=\cos^{n-1}t\ \sin t+(n-1)\int\cos^{n-2}t\ (1-\cos^2t)dt$$ $$I_n=\cos^{n-1}t\ \sin t+(n-1)\int\cos^{n-2}t\ dt-(n-1)\int\cos^nt\ dt$$ $$I_n=\cos^{n-1}t\ \sin t+(n-1)I_{n-2}-(n-1)I_n$$ $$I_n+(n-1)I_n=\cos^{n-1}t\ \sin t+(n-1)I_{n-2}$$ $$nI_n=\cos^{n-1}t\ \sin t+(n-1)I_{n-2}$$ $$I_n=\frac{\cos^{n-1}t\ \sin t}n+\frac{n-1}nI_{n-2}$$ QED
Nice explanation. What if you change to $$\int_{0}^{\frac{\pi}{2}}\cos^{n}x\mathrm{d}x$$? What is then change?
