Thread: Somemore integration help on trigo needed

1. Somemore integration help on trigo needed

1.
$\displaystyle \int \frac {x\arcsin (2x)}{\sqrt{1-4x^2}} dx$
i tried using substitution by letting u= $\displaystyle \arcsin (2x)$but got stucked.

2.
Evaluate $\displaystyle \int \sqrt{x} \arctan\sqrt{x} dx$
totally no idea on how to start this.

2. Re: Somemore integration help on trigo needed

Hey tempq1.

Try using a substitution u = 2x instead of arcsin(2x) and then see what the form of the integral is (and after that consider integrating by parts to get a cancellation).

3. Re: Somemore integration help on trigo needed

Originally Posted by chiro
Hey tempq1.

Try using a substitution u = 2x instead of arcsin(2x) and then see what the form of the integral is (and after that consider integrating by parts to get a cancellation).
sry, but i don't get you. How would letting u=2x solve the equation?

4. Re: Somemore integration help on trigo needed

It certainly wouldn't hurt to substitute u=2x (I'll use w=2x):

$\displaystyle \int \frac {x\arcsin (2x)}{\sqrt{1-4x^2}}\ dx = \frac{1}{4} \int \frac {w\arcsin (w)}{\sqrt{1-w^2}}\ dw$

Now try integrating by parts using $\displaystyle u = \arcsin(w)$ and $\displaystyle dv = \frac {w}{\sqrt{1-w^2}}\ dw$.

- Hollywood

5. Re: Somemore integration help on trigo needed

Originally Posted by hollywood
It certainly wouldn't hurt to substitute u=2x (I'll use w=2x):

$\displaystyle \int \frac {x\arcsin (2x)}{\sqrt{1-4x^2}}\ dx = \frac{1}{4} \int \frac {w\arcsin (w)}{\sqrt{1-w^2}}\ dw$

Now try integrating by parts using $\displaystyle u = \arcsin(w)$ and $\displaystyle dv = \frac {w}{\sqrt{1-w^2}}\ dw$.

- Hollywood
thanks. Anyone knows Q2?

6. Re: Somemore integration help on trigo needed

Substitute $\displaystyle w=sqrt(x)$ to get rid of the radicals, then integrate by parts with $\displaystyle u=arctan(w)$ and $\displaystyle dv=w^2\ dw$.

- Hollywood

7. Re: Somemore integration help on trigo needed

Originally Posted by tempq1
thanks. Anyone knows Q2?
We know that:

$\displaystyle \int u \, dv=uv-\int v \, du.\!$

So for $\displaystyle \int \sqrt{x} \tan^{-1}(\sqrt{x})\;\;dx$

$\displaystyle \text{Let } u = \tan^{-1}(\sqrt{x}) \therefore du = \frac{1}{2\sqrt{x}(1+x)}\;\; dx$

$\displaystyle \text{Let } dv = \sqrt{x} \therefore v = \frac{2x^{\frac{3}{2}}}{3}$

So now:

\displaystyle \begin{align*}\int \sqrt{x} \tan^{-1}(\sqrt{x})\;\;dx =& \frac{2x^{\frac{3}{2}}}{3}\tan^{-1}(\sqrt{x}) - \int \left(\frac{2x^{\frac{3}{2}}}{3}\right)\left(\frac {1}{2\sqrt{x}(1+x)}\right) dx\\ =& \frac{2x^{\frac{3}{2}}}{3} \tan^{-1}(\sqrt{x}) -\int \frac{x}{3(1+x)} dx \end{align*}

For:
$\displaystyle \int \frac{x}{3(1+x)} dx$

Let $\displaystyle u = 1 + x \text{ so } x = u - 1 \text{ and } du = dx \text{ and } x = u -1$

\displaystyle \begin{align*}-\int \frac{x}{3(1+x)} dx =& -\int \frac{(u-1)}{3u} du\\ =& -\int (\frac{1}{3} -\frac{1}{3u}) du\\ =& -\frac{u}{3} + \frac{\ln{(u)}}{3} + C \\ =& -\frac{1+x}{3} +\frac{\ln{(1+x)}}{3} + C\end{align*}

$\displaystyle \therefore \int (\sqrt{x} \tan^{-1}(\sqrt{x}))\;\; dx = \frac{2x^{\frac{3}{2}}}{3} \tan^{-1}(\sqrt{x}) + \frac{\ln{(1+x)}}{3} - \frac{1+x}{3} + C$

Check:

http://www.wolframalpha.com/input/?i=differentiation+2*x^(3%2F2)*tan^(-1)(x^(1%2F2))%2F3+%2B+ln(1%2Bx)%2F3-(1%2Bx)%2F3+%2B+C