# Thread: Easy definite integral problem

1. ## Easy definite integral problem

So, i use substitution sinx=t, cosxdx=dt. When i put it in, i got ln(t)dt, now i know i have to use partial integration, but how do I set limits?

2. ## Re: Easy definite integral problem

Originally Posted by DonnieDarko

So, i use substitution sinx=t, cosxdx=dt. When i put it in, i got ln(t)dt, now i know i have to use partial integration, but how do I set limits?
$\displaystyle \int {\ln (t)dt} = t\ln (t) - t$

3. ## Re: Easy definite integral problem

Hello, DonnieDarko!

$\displaystyle \int^{\frac{2\pi}{3}}_{\frac{\pi}{6}} \cos x\ln(\sin x)\,dx$

$\displaystyle \text{So, I use substitution: }\:t \,=\, \sin x,\;dt\,=\,\cos x\,dx$

$\displaystyle \text{When I put it in, i got: }\:\int \ln t\,dt$

$\displaystyle \text{Now I know I have to use partial integration, but how do I set limits?}$

Plug the $\displaystyle x$-limits into your substitution.

You have: .$\displaystyle t \:=\:\sin x$

$\displaystyle \text{When }x = \tfrac{\pi}{6}\!:\;\;t \:=\:\sin\tfrac{\pi}{6} \:=\:\tfrac{1}{2}$

$\displaystyle \text{When }x = \tfrac{2\pi}{3}\!:\;\;t \:=\:\sin\tfrac{2\pi}{3} \:=\:\tfrac{\sqrt{3}}{2}$

$\displaystyle \displaystyle\text{The integral becomes: }\int^{\frac{\sqrt{3}}{2}}_{\frac{1}{2}}\!\! \ln t\,dt$

4. ## Re: Easy definite integral problem

Thank you very much Soroban, wasnt sure about it, so its always good to check

5. ## Re: Easy definite integral problem

I assume that by "partial integration", you mean what I would call "integration by parts":
$\displaystyle \int_a^b udv= \left[uv\right]_a^b- \int_a^b vdu$

Here, once you have made the substitution, t= sin(x), and have $\int ln(t)dt$, let u= ln(t), dv= dt. Then du= (1/t)dt and v= t so we have
$\displaystyle \left[t ln(t)\right]_{1/2}^{\sqrt{3}/2}- \int_{1/2}^{\sqrt{3}/2} dt= \left[tln(t)- t\right]_{1/2}^{\sqrt{3}/2}$
That's how Plato got his answer.