# Thread: Integral. Substituting makes it more complex

1. ## Integral. Substituting makes it more complex

$\int x \arccos{\sqrt{x}}dx$
$\arccos{\sqrt{x}}=u$
$-\frac{1}{2\sqrt{x-x^2}}dx=du$
$\int udv=uv-\int vdu$
then
$\frac{\arccos{\sqrt{x}}x^2}{2}+\frac{1}{4}\int\fra c{{x^2}}{{\sqrt{x-x^2}}}dx$

2. Originally Posted by totalnewbie
$\int x \arccos{\sqrt{x}}dx$
$\arccos{\sqrt{x}}=u$
$-\frac{1}{2\sqrt{x-x^2}}dx=du$
$\int udv=uv-\int vdu$
then
$\frac{\arccos{\sqrt{x}}x^2}{2}+\frac{1}{4}\int\fra c{{x^2}}{{\sqrt{x-x^2}}}dx$
I am note sure what you are doing.
Given the integral,
$\int x\cos^{-1} \sqrt{x} dx$
You can do it by parts,
$u=x$ and $v'=\cos^{-1} \sqrt{x}$
Thus,
$u'=1$ and $v=\int \cos^{-1} \sqrt{x} dx$

But the question is,
What is,
$\int \cos^{-1} \sqrt{x} dx$
You can write,
$2\int \sqrt{x} \cos^{-1} \sqrt{x} \cdot \frac{1}{2\sqrt{x}} dx$
Let,
$u=\sqrt{x}$ then, $u'=\frac{1}{2\sqrt{x}}$
Thus,
$2\int u\cos^{-1} u du$
You can let,
$a=u$ and $b'=\cos^{-1}u$
Thus,
$a'=1$ and $b=\int \cos^{-1} u du$
Then, we need to find what,
$\int \cos^{-1}udu$ is?
To do that we can write,
$\int (1)\cos^{-1} udu$
Let,
$c'=1$ and $d=\cos^{-1} u$
Then,
$c=u$ and $d'=\frac{1}{\sqrt{1-x^2}}$.
Thus, by parts,
$u\cos^{-1} u-\int \frac{u}{\sqrt{1-u^2}} du$
But then what is?
$\int \frac{u}{\sqrt{1-u^2}} du$
You can write,
$-\frac{1}{2}\int \frac{-2u}{\sqrt{1-u^2}} du$
Let, $t=1-u^2$ then $t'=-2u$.
Thus,
$-\frac{1}{2} \int \frac{1}{\sqrt{t}}dt$
This is certainly doable!

The details, and the actual integral I leave to you do complete. (Maybe, there is an easier way, but this is one of the longest I seen).

3. Hello, totalnewbie!

Make a substitution first . . . then do it by parts.

$\int x\cdot\arccos(\sqrt{x})\,dx$

Let $\arccos(\sqrt{x}) \:=\:y\quad\Rightarrow\quad \sqrt{x} \:=\:\cos y\quad\Rightarrow\quad x \:=\:$ $\cos^2\!y\quad\Rightarrow\quad dx \:=\:-2\cos y\!\cdot\!\sin y\!\cdot\!dy$

Substitute: . $\int\left(\cos^2\!y\right)(y)(-2\cos y\!\cdot\!\sin y\!\cdot\!dy) \;=\;-2\int y\cdot\cos^3\!y\!\cdot\!\sin y\!\cdot\!dy$

By parts: . $\begin{array}{cc} u \\ du\end{array}\begin{array}{cc}= \\ = \end{array}\begin{array}{cc} y \\ dy\end{array}\quad \begin{array}{cc}dv\\ v\end{array} \begin{array}{cc}=\\=\end{array} \begin{array}{cc}\cos^3\!y\!\cdot\!\sin y\!\cdot\!dy \\ -\frac{1}{4}\cos^4\!y\end{array}$

And we have: . $-\frac{1}{4}y\cos^4\!y \;+ \;\frac{1}{4}\!\int\cos^4\!y\,dy$

Can you finish it now?