Integration by Substitution

L

lamaclass

Junior Member
Joined
Oct 18, 2009
Messages
69
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

u=1−x2\displaystyle u=1-x^2

Then dudx=−2x, so −52du=5xdx\displaystyle \frac{du}{dx}=-2x,\ so\ -\frac{5}{2}du=5xdx

Now integrate −52u13du\displaystyle -\frac{5}{2}u^{\frac{1}{3}}du

For 2.... use the substitution

u=1−x2 so du=−2xdx and continue as for 1.\displaystyle u=1-x^2\ so\ du=-2xdx\ and\ continue\ as\ for\ 1.
 
For 3....

(1+1t)3=(t+1t)3=(t+1)3t3\displaystyle (1+\frac{1}{t})^3=(\frac{t+1}{t})^3=\frac{(t+1)^3}{t^3}

Then, (t+1)3=t3+3t2+3t+1, giving (t+1)3t3t2=t3+3t2+3t+1t5\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}

Now divide t5 into all terms of the numerator and integrate term by term.\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

−∫(1+u)3du=−14(1+u)4+C\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.

∫x(1−x2)2dx\displaystyle \int\frac{x}{(1-x^{2})^{2}}dx

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

Make the subs and we get:

∫sin(t)cos5(t)dt\displaystyle \int\frac{sin(t)}{cos^{5}(t)}dt

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

−∫1u5du\displaystyle -\int\frac{1}{u^{5}}du
 
They all lend themselves to trigsubs!

1.

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

Right−angled triangle, perpendicular sides 1−x2 and x,\displaystyle Right-angled\ triangle,\ perpendicular\ sides\ \sqrt{1-x^2}\ and\ x,

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

Sinθ or Cosα=x, choose Sinθ=x.\displaystyle Sin\theta\ or\ Cos\alpha=x,\ choose\ Sin\theta=x.

dxdθ=Cosθ, dx=Cosθdθ\displaystyle \frac{dx}{d\theta}=Cos\theta,\ dx=Cos\theta d\theta

1−x2=1−Sin2θ=Cos2θ\displaystyle 1-x^2=1-Sin^{2}\theta=Cos^{2}\theta

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

Substitute Cosθ=u, dudθ=−Sinθ, −du=Sinθdθ\displaystyle Substitute\ Cos\theta=u,\ \frac{du}{d\theta}=-Sin\theta,\ -du=Sin\theta d\theta

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

=−158(1−x2)83+C=−158(1−x2)1283+C=−158(1−x2)43+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,

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

u=1−x2, dudx=−2x, −52du=5xdx\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.

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

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

Sin2θ=11+1t, 1Sin2θ=1+1t=Sec2θ\displaystyle Sin^2 \theta=\frac{1}{1+\frac{1}{t}},\ \frac{1}{Sin^{2}\theta}=1+\frac{1}{t}=Sec^2 \theta

ddtSec2θ=−1t2\displaystyle \frac{d}{dt}Sec^2 \theta=-\frac{1}{t^2}

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

=−14(1+1t)4+C\displaystyle =-\frac{1}{4}(1+\frac{1}{t})^4+C


Using basic substitutions,

u=1+1t, dudt=−1t2\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 . . .

1)    ∫5x(1−x2)13dx\displaystyle 1)\;\;\int 5x\left(1-x^2\right)^{\frac{1}{3}}dx
Click to expand...

We have:   5∫(1−x2)12(x dx)\displaystyle \text{We have: }\;5\int(1-x^2)^{\frac{1}{2}}(x\,dx)

Let:  u = 1−x2⇒du = −2x dx⇒x dx = −12du\displaystyle \text{Let: }\:u \:=\:1-x^2 \quad\Rightarrow\quad du \:=\:-2x\,dx \quad\Rightarrow\quad x\,dx \:=\:-\tfrac{1}{2}du

Substitute:  5∫u12(−12du)  =  −52∫u12du  =  −53 u32+C\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

Back-substitute:  −53(1−x2)32+C\displaystyle \text{Back-substitute: }\:-\tfrac{5}{3}(1-x^2)^{\frac{3}{2}} + C



We have:  ∫(1−x2)−3(x dx)\displaystyle \text{We have: }\:\int(1-x^2)^{-3}(x\,dx)

Let:  u = 1−x2⇒du = −2x dx⇒x dx = −12 du\displaystyle \text{Let: }\:u \:=\:1-x^2 \quad\Rightarrow\quad du \:=\:-2x\,dx \quad\Rightarrow\quad x\,dx \:=\:-\tfrac{1}{2}\,du

Substitute:  ∫u−3(−12 du) = −12∫u−3du = 14 u−2+C\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

Back-substitute:  14(1−x2)−2+C\displaystyle \text{Back-substitute: }\:\tfrac{1}{4}(1-x^2)^{-2} + C


3)    ∫(1+1t)3(1t2)dt\displaystyle 3)\;\;\int \left(1+\frac{1}{t}\right)^3 \left(\frac{1}{t^2}\right) dt
Click to expand...

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

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

Substitute:  ∫u3(−du)  =  −∫u3du  =  −14u4+C\displaystyle \text{Substitute: }\:\int u^3(-du) \;=\;-\int u^3du \;=\;-\tfrac{1}{4}u^4 + C

back-substitute:  −14(1+1t)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.
 
You must log in or register to reply here.
Top