1. ## integration

$\int_0^{1/5} x\sin^{-1}(5 x) dx =$

i did the integration by parts first:

$u = arcsin(5x), dv = x,$
$du = \frac{5}{\sqrt{1-25x^2}} dx, v = \frac{1}{2}x^2,$

$uv-\int{vdu} = arcsin(5x)\frac{1}{2}x^2-\int_0^{1/5}\frac{1}{2}x^2(\frac{5}{\sqrt{1-25x^2}})dx$

need help from this point

2. First, set $\mu=5x$, then the integration by parts process won't be so tedious.

3. After this, a trig. substitution would work with the remaining integral.

4. ## Re:

RE:

I haven't worked much with trig sub. At this time I would probably go for a table/manual...

5. Hello, viet!

$\int_0^{\frac{1}{5}} x\sin^{-1}(5x)\,dx$

i did the integration by parts first:

$\begin{array}{ccccccc}u &= & \arcsin(5x) & \quad & dv & = & x \\
du &= &\frac{5}{\sqrt{1-25x^2}} dx& \quad & v & = & \frac{1}{2}x^2
\end{array}$

$uv-\int{v\,du} \;= \;\arcsin(5x)\frac{1}{2}x^2 \,-\,\int_0^{\frac{1}{5}}\frac{1}{2}x^2\left(\frac{5} {\sqrt{1-25x^2}}\right)dx$ . Good!

We have: . $\frac{1}{2}x^2\arcsin(5x) - \frac{5}{2}\int\frac{x^2}{\sqrt{1 - 25x^2}}\,dx$

. . Let: $5x = \sin\theta\quad\Rightarrow\quad x = \frac{1}{5}\sin\theta\quad\Rightarrow\quad dx = \frac{1}{5}\cos\theta\,d\theta$

. . The integral becomes: . $\int\frac{\frac{1}{25}\sin^2\!\theta}{\cos\theta}\ left(\frac{1}{5}\cos\theta\,d\theta\right) \;=\;\frac{1}{125}\int\sin^2\!\theta\,d\theta$

. . . . $=\;\frac{1}{250}\int (1 - \cos2\theta)d\theta\;= \;\frac{1}{250}\left(\theta - \frac{1}{2}\sin2\theta\right) \;=\;\frac{1}{250}(\theta - \sin\theta\cos\theta)$

. . Back-substitute: . $\frac{1}{250}\left[\arcsin(5x) - 5x\sqrt{1 - 25x^2}\right]$

We have: . $\frac{1}{2}x^2\arcsin(5x) - \frac{5}{2}\left[\frac{1}{250}\left(\arcsin(5x) - 5x\sqrt{1-25x^2}\right)\right]$

. . $= \;\frac{1}{2}x^2\arcsin(5x) - \frac{1}{250}\arcsin(5x) + \frac{1}{20}x\sqrt{1-25x^2}$

. . $=\;\frac{1}{500}\left[250x^2\arcsin(5x) - 2\arcsin(5x) + 25x\sqrt{1-25x^2}\right]^{\frac{1}{5}}_0$

Now evaluate . . .
. . $\left(\text{I got: }\frac{\pi}{125}\right)$

6. $\int {x\arcsin 5x~dx}$

Let's set $\theta=5x\implies{d\theta=5~dx}$

The integral becomes to

$\frac{1}
{25}\int {\theta \arcsin \theta~d\theta }$

Now let $u=\arcsin\theta\implies{du=\frac1{\sqrt{1-\theta^2}}~d\theta}$ and $dv=\theta~d\theta\implies{v=\frac12\theta^2}$, this yields

$\int {x\arcsin 5x~dx} = \frac{1}
{2}x^2 \arcsin 5x - \frac{1}
{{100}}\arcsin 5x + \frac{1}
{{20}}x\sqrt {1 - 25x^2 } + c$

That's my contribution