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I ran into a question with proving the reduction formula:

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

I then attempted to prove by differentiation with respect to $x$, but something strange happened (have I just violated the Fundamental Theorem of Calculus?)

After differentiation, the resulting expression is:

$\displaystyle\cos^n x=\bigg[\frac{1}{n}\bigg]\bigg[(n-1)\cos^{n-2}x(-\sin x)(\sin x)+\cos^{n-1}x\cos x+\frac{n-1}{n}\cos^{n-2}x\bigg]$

$\displaystyle=\bigg[\frac{n-1}{n}\cos^{n-2}x\bigg]\bigg[\cos^2x+\cos^n x\bigg]$

$\displaystyle \frac{n-1}{n}\bigg[\cos^nx+\cos^{2n-2}x\bigg]$

Edit: I corrected the derivative, but problem not solved.

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marked as duplicate by lab bhattacharjee, tetori, Daniel Rust, Eric Stucky, mathematics2x2life Jan 18 at 7:02

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You forgot to differentiate the sine in the product $\cos^{n-1} x\sin x$. $$\frac{d}{dx}\left(\cos^{n-1} x\sin x\right) = \cos^{n-1}x\cos x + (n-1) \cos^{n-2} x (-\sin x)\sin x = n \cos^n x - (n-1)\cos^{n-2} x.$$ –  Daniel Fischer Jan 17 at 18:45

2 Answers 2

Hint Write

$$\int\cos^n x dx=\int\cos x\cos^{n-1}x dx$$ and do an integration by parts using the relation $$\cos^2 x+\sin^2x =1$$

Edit

$$\int\cos^n x dx=\int\underbrace{\cos x}_{=f'(x)}\underbrace{\cos^{n-1}x }_{=g(x)}dx=\sin x\cos^{n-1}x+(n-1)\int\sin x\sin x\cos^{n-2}x dx$$ so $$\int\cos^n x dx=\int\cos x\cos^{n-1}x dx=\sin x\cos^{n-1}x+(n-1)\int(1-\cos^2 x)\cos^{n-2}x dx$$ so developp and you find $$n\int\cos^n x dx=\sin x\cos^{n-1}x+(n-1)\int\cos^{n-2}x dx$$ now divide by $n$ and you get your result.

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Hint:

$$\int \cos^nxdx=\int \cos^{n-1}xd(\sin x)=\cos^{n-1}x\sin x-(n-1)\int \cos^{n-2}x\sin^2xdx$$

$$=\cos^{n-1}x\sin x-(n-1)\int \cos^{n-2}x(1-\cos^2x)dx$$

$$=\cos^{n-1}x\sin x-(n-1)\int \cos^{n-2}dx-(n-1)\int \cos^nxdx$$

$$\to \int cos^nxdx=\frac{1}{n}\cos^{n-1}x\sin x-\frac{n-1}{n}\int \cos^{n-2}xdx \: \: \text{Q.E.D}$$

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I was unable to find the question. OK, I will delete it. –  user2213307 Jan 18 at 3:51

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