Integration by Substitution

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lamaclass

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I'm having a hard time trying to figure out how to solve these problems:

1. INT [(5x (1-x[sup:3nr42ds0]2[/sup:3nr42ds0])[sup:3nr42ds0]1/3[/sup:3nr42ds0] dx)]

2. INT [(X/(1-x[sup:3nr42ds0]2[/sup:3nr42ds0])[sup:3nr42ds0]3[/sup:3nr42ds0] dx)]

3. INT {(1+1/t)[sup:3nr42ds0]3[/sup:3nr42ds0] (1/t[sup:3nr42ds0]2[/sup:3nr42ds0]) dt)]

Any help is greatly appreciated!
 
For 1.... use the substitution

\(\displaystyle u=1-x^2\)

Then \(\displaystyle \frac{du}{dx}=-2x,\ so\ -\frac{5}{2}du=5xdx\)

Now integrate \(\displaystyle -\frac{5}{2}u^{\frac{1}{3}}du\)

For 2.... use the substitution

\(\displaystyle u=1-x^2\ so\ du=-2xdx\ and\ continue\ as\ for\ 1.\)
 
For 3....

\(\displaystyle (1+\frac{1}{t})^3=(\frac{t+1}{t})^3=\frac{(t+1)^3}{t^3}\)

\(\displaystyle Then,\ (t+1)^3=t^3+3t^2+3t+1,\ giving\ \frac{(t+1)^3}{t^3t^2}=\frac{t^3+3t^2+3t+1}{t^5}\)

\(\displaystyle Now\ divide\ t^5\ into\ all\ terms\ of\ the\ numerator\ and\ integrate\ term\ by\ term.\)
 
lamaclass said:
I'm having a hard time trying to figure out how to solve these problems:



3. INT {(1+1/t)[sup:3tx2qvty]3[/sup:3tx2qvty] (1/t[sup:3tx2qvty]2[/sup:3tx2qvty]) dt)]

u = 1/t

du = -1/t[sup:3tx2qvty]2[/sup:3tx2qvty] dt

\(\displaystyle -\int (1+u)^3du = -\frac{1}{4}(1+u)^4 + C\)

Any help is greatly appreciated!
Click to expand...
 
#2. An alternative method could be trig sub.

\(\displaystyle \int\frac{x}{(1-x^{2})^{2}}dx\)

Let \(\displaystyle x=sin(t), \;\ dx=cos(t)dt\)

Make the subs and we get:

\(\displaystyle \int\frac{sin(t)}{cos^{5}(t)}dt\)

Now, let \(\displaystyle u=cos(t), \;\ du=-sin(t)dt\)

\(\displaystyle -\int\frac{1}{u^{5}}du\)
 
They all lend themselves to trigsubs!

1.

\(\displaystyle \int{5x(1-x^2)^{\frac{1}{3}}dx\)

\(\displaystyle Right-angled\ triangle,\ perpendicular\ sides\ \sqrt{1-x^2}\ and\ x,\)

\(\displaystyle hypotenuse=1,\ \theta\ either\ of\ the\ two\ acute\ angles.\)

\(\displaystyle Sin\theta\ or\ Cos\alpha=x,\ choose\ Sin\theta=x.\)

\(\displaystyle \frac{dx}{d\theta}=Cos\theta,\ dx=Cos\theta d\theta\)

\(\displaystyle 1-x^2=1-Sin^{2}\theta=Cos^{2}\theta\)

\(\displaystyle 5\int[Cos^{2}\theta]^{\frac{1}{3}}Cos\theta Sin\theta d\theta=5\int{Cos^{\frac{5}{3}}\theta Sin\theta d\theta\)

\(\displaystyle Substitute\ Cos\theta=u,\ \frac{du}{d\theta}=-Sin\theta,\ -du=Sin\theta d\theta\)

\(\displaystyle -5\int{u^{\frac{5}{3}}du=-5\frac{u^{\frac{8}{3}}}{\frac{8}{3}}+C=-5\frac{Cos^{\frac{8}{3}}\theta}{\frac{8}{3}}+C\)

\(\displaystyle =-\frac{15}{8}(\sqrt{1-x^2})^{\frac{8}{3}}+C=-\frac{15}{8}(1-x^2)^{\frac{1}{2}\frac{8}{3}}+C=-\frac{15}{8}(1-x^2)^{\frac{4}{3}}+C\)


Using basic substitutions,

\(\displaystyle \int{5x(1-x^2)^{\frac{1}{3}}dx\)

\(\displaystyle u=1-x^2,\ \frac{du}{dx}=-2x,\ -\frac{5}{2}du=5xdx\)

\(\displaystyle -\frac{5}{2}\int{u^{\frac{1}{3}}du=-\frac{5}{2}\frac{u^{\frac{4}{3}}}{\frac{4}{3}}+C=-\frac{15}{8}(1-x^2)^{\frac{4}{3}}+C\)


3.

\(\displaystyle \int{(1+\frac{1}{t})^{3}(\frac{1}{t^2})dt\)

\(\displaystyle Right-angled\ triangle,\ \theta=acute\ angle,\ \sqrt{1+\frac{1}{t}}=hypotenuse,\ \sqrt{\frac{1}{t}}=adjacent,\ 1=opposite.\)

\(\displaystyle Sin^2 \theta=\frac{1}{1+\frac{1}{t}},\ \frac{1}{Sin^{2}\theta}=1+\frac{1}{t}=Sec^2 \theta\)

\(\displaystyle \frac{d}{dt}Sec^2 \theta=-\frac{1}{t^2}\)

\(\displaystyle -\int{(Sec^2 \theta)^3 dSec^2 \theta=-\int{u^{3}du=-\frac{u^4}{4}+C=-\frac{Sec^8 \theta}{4}+C\)

\(\displaystyle =-\frac{1}{4}(1+\frac{1}{t})^4+C\)


Using basic substitutions,

\(\displaystyle u=1+\frac{1}{t},\ \frac{du}{dt}=-\frac{1}{t^2}\)

\(\displaystyle -\int{u^{3}du=-\frac{u^4}{4}+C=-\frac{1}{4}(1+\frac{1}{t})^4+C\)
 
Hello,l lamaclass!

These are all "simple" substitutions . . .

\(\displaystyle 1)\;\;\int 5x\left(1-x^2\right)^{\frac{1}{3}}dx\)
Click to expand...

\(\displaystyle \text{We have: }\;5\int(1-x^2)^{\frac{1}{2}}(x\,dx)\)

\(\displaystyle \text{Let: }\:u \:=\:1-x^2 \quad\Rightarrow\quad du \:=\:-2x\,dx \quad\Rightarrow\quad x\,dx \:=\:-\tfrac{1}{2}du\)

\(\displaystyle \text{Substitute: }\:5\int u^{\frac{1}{2}}\left(-\tfrac{1}{2}du\right) \;=\;-\tfrac{5}{2}\int u^{\frac{1}{2}}du \;=\;-\tfrac{5}{3}\,u^{\frac{3}{2}} + C\)

\(\displaystyle \text{Back-substitute: }\:-\tfrac{5}{3}(1-x^2)^{\frac{3}{2}} + C\)


\(\displaystyle 2)\;\; \int \frac{x}{(1-x^2)^3}\,dx\)
Click to expand...

\(\displaystyle \text{We have: }\:\int(1-x^2)^{-3}(x\,dx)\)

\(\displaystyle \text{Let: }\:u \:=\:1-x^2 \quad\Rightarrow\quad du \:=\:-2x\,dx \quad\Rightarrow\quad x\,dx \:=\:-\tfrac{1}{2}\,du\)

\(\displaystyle \text{Substitute: }\:\int u^{-3}\left(-\tfrac{1}{2}\,du\right) \:=\:-\tfrac{1}{2}\int u^{-3}du \:=\:\tfrac{1}{4}\,u^{-2} + C\)

\(\displaystyle \text{Back-substitute: }\:\tfrac{1}{4}(1-x^2)^{-2} + C\)


\(\displaystyle 3)\;\;\int \left(1+\frac{1}{t}\right)^3 \left(\frac{1}{t^2}\right) dt\)
Click to expand...

\(\displaystyle \text{We have: }\:\int \left(1+ t^{-1}\right)^3\left(t^{-2}dt\right)\)

\(\displaystyle \text{Let: }\:u \:=\:1 + t^{-1} \quad\Rightarrow\quad du \:=\:-t^{-2}\,dt \quad\Rightarrow\quad t^{-2}dt \:=\:-du\)

\(\displaystyle \text{Substitute: }\:\int u^3(-du) \;=\;-\int u^3du \;=\;-\tfrac{1}{4}u^4 + C\)

\(\displaystyle \text{back-substitute: }\:-\tfrac{1}{4}\left(1 + \frac{1}{t}\right)^4 + C\)

 
Plenty to practice with.

As you can see, there are many ways to skin a crocodile.
 
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