So we were taught in school that when integrating expressions of form $ \int f^{'}(x) f^n(x)$ dx

We get $f^{n+1} + C$

This way something like $\int sin(x)\sqrt{1-cos(x)}$ becomes $(1-cos(x))^{3/2} +C$

The Question now is that can't we integrate $ \sqrt{1-cos(x)}$ as such:

$\int \sqrt{1-cos(x)} = \frac {1}{sin(x)}\bigl(1-cos(x)\bigr) + C $

Is this Integral incorrect? Like I'm aware that this isn't defined for certain values of x, but integrating this way seems much simpler. Just like the reverse power rule i.e. increase the power by 1 and divide by the derivative of the inner function.

Can we do this??


Well, in general we have:


And that is not hard to prove using IBP.

So, when $\text{y}'\left(x\right)=\sin\left(x\right)$ we also have $\text{y}\left(x\right)=-\cos\left(x\right)$:


Or, when $\text{y}'\left(x\right)=\sqrt{1-\cos\left(x\right)}$ we also have $\text{y}\left(x\right)=-2\text{y}'\left(x\right)\cot\left(\frac{x}{2}\right)$:

$$\mathcal{I}_\text{n}\left(x\right)=\int\sqrt{1-\cos\left(x\right)}\cdot\left(-2\text{y}'\left(x\right)\cot\left(\frac{x}{2}\right)\right)^\text{n}\space\text{d}x=$$ $$\frac{\left(-2\text{y}'\left(x\right)\cot\left(\frac{x}{2}\right)\right)^{1+\text{n}}}{1+\text{n}}+\text{C}=\frac{\left(-2\sqrt{1-\cos\left(x\right)}\cot\left(\frac{x}{2}\right)\right)^{1+\text{n}}}{1+\text{n}}+\text{C}\tag3$$

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  • $\begingroup$ Yeah, so we can get y(x) by integrating y'(x) but, what I was wondering is do we need to have y'(x) with it ti integrate a function? Like can't we just integrate as usual and divide it by the derivate. So taking my example, If we write I= ∫$sqrt{1-cos(x)}$ we integrate it using "my way" which is raise power and divide by it's derivatives, which gets us I = $frac{1}{sin(x)}$$(1-cos(x))^{3/2} + C$ and if we were to differentiate it wouldn't we get the same result as the Integrand? Like does this make sense? $\endgroup$ – Shaheer ziya May 6 at 11:21
  • $\begingroup$ @Shaheerziya I do understand your question. $\endgroup$ – Jan Eerland May 6 at 11:22
  • $\begingroup$ I don't know what you're saaying. $\endgroup$ – Shaheer ziya Jul 22 at 13:08

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