Integration by substiution?? Don't get??

**Reefer** · December 14th 2008, 06:51 PM

(secx)^2 sqrt((tanx)^3)dx

x^3/sqrt(x^2+25)

$(5x^2 + 6x - 12)/{{x^4}}$

Integral from 3 to sqrt(24) of x/sqrt(x^2-8)

Integral from 1/2 to 5/2 of 1/sqrt(6x+1)

I don't get any of this.. I have a test of this Tuesday and still have no idea how to do it..

**Mathstud28** · December 14th 2008, 07:11 PM

Originally Posted by Reefer

(secx)^2 sqrt((tanx)^3)dx

x^3/sqrt(x^2+25)

$(5x^2 + 6x - 12)/{{x^4}}$

Integral from 3 to sqrt(24) of x/sqrt(x^2-8)

Integral from 1/2 to 5/2 of 1/sqrt(6x+1)

I don't get any of this.. I have a test of this Tuesday and still have no idea how to do it..

The concept for most of these is all based on the Chain Rule: $\left(g(f(x))\right)'=g'(f(x))\cdot f'(x)$ . So now doesn't it make sense that $\int g'(f(x))\cdot f'(x)dx=g(f(x))$ ?

So lets take your first one

$\int \sec^2(x)\sqrt{\tan^3(x)}dx$ The key here is to identify your $f'$ . The first thing you should notice is that $\left(\tan(x)\right)'=\sec^2(x)$ . So now let us rewrite this as $\int\sec^2(x)\sqrt{\tan^3(x)}dx=\int\left(\tan(x)\ right)'\sqrt{\tan^3(x)}dx$ . The next step is to identify our $f$ since we have that our $f'=(\tan(x))'$ it makes sense that $f=\tan(x)$ . So now let us rewrite our integral as $\int\left(\tan(x)\right)'\cdot\sqrt{\left\{\tan(x) \right\}^3}dx$ . From here we see that $g'=\sqrt{x^3}$ so $g=\int\sqrt{x^3}=\frac{2x^{\frac{5}{2}}}{5}$ . So by the fact that $\int g'(f(x))\cdot f'(x)dx=g(f(x))$ we can see that $\int\sec^2(x)\sqrt{\tan^3(x)}dx=\frac{2\tan^{\frac {5}{2}}(x)}{5}+C$ . Now that was a lot of work wasn't it? There is a method that makes this much easier it is called u-substitution. It basically say that if I have $\int g'(f(x))\cdot f'(x)dx$ why not let $f(x)=u$ ...but I have only replaced half of my "symbols" in the integral. What about $f'(x)dx$ ? We take care of this by noting that if $u=f(x)\implies du=f'(x)dx$ . So after our substitution our integral becomes $\int g(u) du$ which usually makes this much easier. So in the previous example had we let $u=\tan(x)\implies du=\sec^2(x)dx$ our integral would have been transformed into $\int\sqrt{u^3}du$ which is pretty easy. Most of the ones you have here are of the form $\int g'(f(x))f'(x)dx$ ...see if you can find them and then see if you can do them. The ones that aren't should be apparent...ask back later for more help on those.

**Reefer** · December 14th 2008, 07:47 PM

Ok I got how to so the first one but...

I am still lost on every other one..

x^3/sqrt(x^2+25)

I have no idea.. I got my answer as x/sqrt(x^2+25) but the answer is something weird..

**Chop Suey** · December 15th 2008, 01:02 AM

Originally Posted by Reefer

Ok I got how to so the first one but...

I am still lost on every other one..

x^3/sqrt(x^2+25)

I have no idea.. I got my answer as x/sqrt(x^2+25) but the answer is something weird..

For $\int \frac{x^3}{\sqrt{x^2+25}}$ , try a sub such as $u^2 = x^2+25 \implies 2u~du=2x~dx$

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