1. ## integration question help

ok. so the book says to prove the reduction formula for:
integral of cos^n x dx
the says this equals 1/n cos^(n-1) x * sin x + ((n-1)/n) * integral of cos^n-2 x dx

i tried to do this by breaking cos^n x into cos^n-1 * cos x
i let u equal cos^n-1 x and dv = cos x dx
for du i got -(n-1)(cos^n-2 x)(sin x) dx
and for v i got sin x
so the answer i got was cos^n-1 * sin x + (n-1) * the integral of cos^n-2 x * sin x dx

what did i do wrong

2. Originally Posted by pcgamer03
ok. so the book says to prove the reduction formula for:
integral of cos^n x dx
the says this equals 1/n cos^(n-1) x * sin x + ((n-1)/n) * integral of cos^n-2 x dx

i tried to do this by breaking cos^n x into cos^n-1 * cos x
i let u equal cos^n-1 x and dv = cos x dx
for du i got -(n-1)(cos^n-2 x)(sin x) dx
and for v i got sin x
so the answer i got was cos^n-1 * sin x + (n-1) * the integral of cos^n-2 x * sin x dx

what did i do wrong
you started out well. see here and match up with your work to see where you went wrong

3. Originally Posted by pcgamer03
ok. so the book says to prove the reduction formula for:
integral of cos^n x dx
the says this equals 1/n cos^(n-1) x * sin x + ((n-1)/n) * integral of cos^n-2 x dx
Let $I_n = \int \cos^n x dx$.

Then if $n\geq 2$ we can write, $\int \cos^{n-1} x \cos^1 x dx$

Let, $s'=\cos x \mbox{ and }t=\cos^{n-1} x$
Then, $s = -\sin x \mbox{ and }t ' = -(n-1)\cos^{n-2} \sin x$

Apply integration by parts, $-\sin x \cos^{n-1} x + (n-1)\int \sin^2 x \cos^{n-2} x dx$

Now, $\int \sin^2 x \cos^{n-2} xdx = \int (1-\cos^2 x)\cos^{n-2} dx = I_{n-2} - I_n$

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

Solve, $I_n = -\frac{1}{n}\sin x \cos^{n-1} x + \frac{(n-1)}nI_{n-2}$

There should not be a negative. But I am too lazy to check where is the mistake.