If someone wouldn't mind helping me with these two integrals, I would really appreciate it, thanks!
$\displaystyle \int^{\sqrt2}_1\frac{dx}{x^2 \sqrt{2-x^2}}$
$\displaystyle \int^{\frac{1}{2}}_0 x \arctan({2x}) dx$
If someone wouldn't mind helping me with these two integrals, I would really appreciate it, thanks!
$\displaystyle \int^{\sqrt2}_1\frac{dx}{x^2 \sqrt{2-x^2}}$
$\displaystyle \int^{\frac{1}{2}}_0 x \arctan({2x}) dx$
YUCK!!!!!
Ok, First rearrange your integral so you can use a u-substitution:
$\displaystyle \int^{\sqrt2}_1\frac{dx}{x^2 \sqrt{2-x^2}}$
$\displaystyle \int^{\sqrt2}_1 \frac{dx}{2 \cdot \frac{x^2}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} - \frac{x}{\sqrt{2}}}$
$\displaystyle u = \frac{x}{\sqrt{2}}$
$\displaystyle
du = \frac{1}{\sqrt{2}} dx$
So your integral becomes:
\$\displaystyle \int \frac{du}{2 u^2 \sqrt{1-u^2}}$
Factor out your constants & then use the $\displaystyle sin^{-1}$ substitution for $\displaystyle \sqrt{1-u^2}$
This should give you a fairly easy integral. Just remember to plug back in all of your substitutions or change your limits accordingly. Hope that helps!
Wow, these are probably the ugliest integrals I've seen all day!
Ok for the second one:
$\displaystyle \int^{\frac{1}{2}}_0 x \arctan({2x}) dx$
First do a u substitution with u = 2x
Factor out your constants.
Then you are going to have to do integration by parts:
$\displaystyle u = tan^{-1}$, $\displaystyle dv = u du$
Now, after you do parts, in the part still under the integral, you will probably have to do long division to get the exponent on top to be less than the one on the bottom.
You should get something in the form of $\displaystyle \int 1 + \frac{1}{u^2+1}$ *I didn't do this long division out, but I think that is probably right for your new integral after long division* This will allow you to break up the integral into two separate integrals.
Integrate and then replace your u-substitution values.
Sorry I didn't have to time to work this out for you. Hope this helps!!
I'm not quite sure how the two expressions are the same...
But an easy solution to the first one is to use the substitution: $\displaystyle x = \frac{1}{t} \ \Rightarrow \ \ dx = -\frac{1}{t^2} \ dt$
So our integral becomes: $\displaystyle \int \frac{dx}{x^2\sqrt{2-x^2}} = \int \frac{-\frac{1}{t^2} \ dt}{\frac{1}{t^2} \sqrt{2 - \frac{1}{t^2}}} = - \int \frac{dt}{\sqrt{2 - \frac{1}{t^2}}}$
Simplify the expression a little bit more and you'll get a pretty easy integral.
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Second one, use integration by parts: $\displaystyle \begin{array}{lcl} u = \arctan (2x) & \Rightarrow & du = \frac{2}{1+(2x)^2} \ dx \\ dv = x & \Rightarrow & v = \frac{1}{2}x^2 \end{array}$
So: $\displaystyle \int x \arctan({2x}) dx = \frac{1}{2}x^2\arctan (2x) - \frac{1}{2} \int \frac{x^2}{1+(2x)^2} \ dx$
For the first use the Trig substitution $\displaystyle x=\sqrt{2}\sin(\theta) \implies dx=\sqrt{2} \cos(\theta)d\theta$
using this we can change the limits of integration
$\displaystyle 1=\sqrt{2}\sin(\theta) \iff \theta=\frac{\pi}{4}$ and
$\displaystyle \sqrt{2}=\sqrt{2}\sin(\theta) \iff \theta =\frac{\pi}{2}$
Now the integral becomes
$\displaystyle \int_{\pi/4}^{\pi/2}\frac{\sqrt{2}\cos(\theta}{2\sin^2(\theta)\sqrt{ 2-2\sin^{2}(\theta)}}d\theta=\int_{\pi/4}^{\pi/2}\csc^2(\theta)d\theta=-\cot(\theta) \bigg|_{\pi/4}^{\pi/2}$