# Thread: integration substitution

1. ## integration substitution

Hi , after a substitution with x=3 cos u the answer in the text gives the following...see image

I can't see how it tidies up to u cos u
thank you

2. ## Re: integration substitution

Could you please post all of the original question?

3. ## Re: integration substitution

the question is.. by considering the substitution x=3cos u ,evaluate

4. ## Re: integration substitution

Originally Posted by minicooper58
the question is.. by considering the substitution x=3cos u ,evaluate
Let \displaystyle \begin{align*} x = 3\cos{u} \implies dx = -3\sin{u}\,du \end{align*}, and noting that when \displaystyle \begin{align*} x = 0, u = \frac{\pi}{2} \end{align*} and when \displaystyle \begin{align*} x = \frac{3}{2}, u = \frac{\pi}{3} \end{align*}, the integral becomes

\displaystyle \begin{align*} \int_0^{\frac{3}{2}}{\frac{x\arccos{\left(\frac{x} {3}\right)}}{\sqrt{9 - x^2}}\,dx} &= \int_{\frac{\pi}{2}}^{\frac{\pi}{3}}{\frac{3\cos{u }\arccos{\left(\frac{3\cos{u}}{3}\right)}}{\sqrt{9 - \left(3\cos{u}\right)^2}}\cdot -3\sin{u}\,du} \\ &= -9\int_{\frac{\pi}{2}}^{\frac{\pi}{3}}{\frac{u\cos{ u}\sin{u}}{\sqrt{9 - 9\cos^2{u}}}\,du} \\ &= 9\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{u\cos{ u}\sin{u}}{\sqrt{9\sin^2{u}}}\,du} \\ &= 9\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{\frac{u\cos{ u}\sin{u}}{3\sin{u}}\,du} \\ &= 3\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}{u\cos{u}\,du } \end{align*}

which you can now solve using Integration by Parts

5. ## Re: integration substitution

The terms on the top are multiplied together so 3sinu cancels out.

6. ## Re: integration substitution

Originally Posted by minicooper58
the question is.. by considering the substitution x=3cos u ,evaluate
After substitution integral becomes :

$I=-3\cdot \int\limits_{\frac{\pi}{2}}^{\frac{\pi}{3}} u \cdot \cos u\,du=3\cdot \int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} u \cdot \cos u\,du$

which can be solved using integration by parts...