Trig Substitiution

B

Blitze105

New member
Joined
Aug 28, 2008
Messages
27
Hello,
I have a question that i am not too sure about.

The first one is the integral of:

x^2 / root(9 - x^2)
The answer is 1/2x rt(9-x^2) + arc sin(x/3) + 3 + c
From the formula i can gather the arcsin part, but i simply can't find an example or a helpful professor... on what to do with when the numerator is not 1.

One more integral i could use help with...

dt / t^2 (rt(1 + t^2))
the answer is: -rt(1 + t%2)/t + c

Any help with where to go on this would greatly help me. I have tried looking for examples in the book, however this book is completely useless. They do not show a single example of anything near this.
 
Blitze105 said:
Hello,
I have a question that i am not too sure about.

The first one is the integral of:

x^2 / root(9 - x^2)

Try

x =3⋅sin(θ)\displaystyle x \, = 3\cdot sin(\theta)

do the substitution (with paper and pencil-with-eraser) and tell us what did you get.

The answer is 1/2x rt(9-x^2) + arc sin(x/3) + 3 + c
From the formula i can gather the arcsin part, but i simply can't find an example or a helpful professor... on what to do with when the numerator is not 1.

One more integral i could use help with...

dt / t^2 (rt(1 + t^2))

Try

u = 1+t2)\displaystyle u \, = \, \sqrt{1+t^2)}


do the substitution (with paper and pencil-with-eraser) and tell us what did you get.

the answer is: -rt(1 + t%2)/t + c

Any help with where to go on this would greatly help me. I have tried looking for examples in the book, however this book is completely useless. They do not show a single example of anything near this.
Click to expand...
 
For the first problem...
I take x = 3sin(t) and dx = 3cos(t)

From here...
9sin^2(t)
3rt(1 - sin^2(t)

Multiply that by 3cos(t) and 9sin^2(t) is all that i am left with... and now i solve for (t) and plug that back in. If thats not close then i have no idea what i'm doing sadly. But if i am way off please let me know where and i'll keep trying
 
2nd one:\displaystyle 2nd \ one:

∫dtt2(1+t2), Let t = tan(x), then dt = sec2(x)dx.\displaystyle \int \frac{dt}{t^{2}\sqrt(1+t^{2})}, \ Let \ t \ = \ tan(x), \ then \ dt \ = \ sec^{2}(x)dx.

Hence, we get ∫sec2(x)dxtan2(x)(1+tan2(x)),\displaystyle Hence, \ we \ get \ \int \frac{sec^{2}(x)dx}{tan^{2}(x)\sqrt(1+tan^{2}(x))},

= ∫sec(x)dxtan2(x) = ∫cos(x)dxsin2(x).\displaystyle = \ \int \frac{sec(x)dx}{tan^{2}(x)} \ = \ \int \frac{cos(x)dx}{sin^{2}(x)}.

Another substitution: Let v = sin(x), dv = cos(x)dx.\displaystyle Another \ substitution: \ Let \ v \ = \ sin(x), \ dv \ = \ cos(x)dx.

Therefore, now we have; ∫dvv2 = −1v + C.\displaystyle Therefore, \ now \ we \ have; \ \int \frac{dv}{v^{2}} \ = \ \frac{-1}{v} \ + \ C.

= −1sin(x) + C, = −csc(x) + C, = −csc[arctan(t)] + C, x = arctan(t).\displaystyle = \ \frac{-1}{sin(x)} \ + \ C, \ = \ -csc(x) \ + \ C, \ = \ -csc[arctan(t)] \ + \ C, \ x \ = \ arctan(t).

Now let arctan(t) = θ, then tan(θ) = t, ergo −csc(θ) = −(t2+1)t.\displaystyle Now \ let \ arctan(t) \ = \ \theta, \ then \ tan(\theta) \ = \ t, \ ergo \ -csc(\theta ) \ = \ -\frac{\sqrt(t^{2}+1)}{t}.

Putting it all together, we have: ∫dtt2(1+t2) = −(t2+1)t+C.\displaystyle Putting \ it \ all \ together, \ we \ have: \ \int \frac{dt}{t^{2}\sqrt(1+t^{2})} \ = \ -\frac{\sqrt(t^{2}+1)}{t}+C.

Check: Dt [−(1+t2)t+C] = 1t2(1+t2).\displaystyle Check: \ D_t \ \bigg[-\frac{\sqrt(1+t^{2})}{t}+C\bigg] \ = \ \frac{1}{t^{2}\sqrt(1+t^{2})}.
 
Blitze105 said:
For the first problem...
I take x = 3sin(t) and dx = 3cos(t)

From here...
9sin^2(t)
3rt(1 - sin^2(t)

Multiply that by 3cos(t) and 9sin^2(t) is all that i am left with... and now i solve for (t) and plug that back in. If thats not close then i have no idea what i'm doing sadly. But if i am way off please let me know where and i'll keep trying
Click to expand...

Looks good to me....
 
First one:\displaystyle First \ one:

∫x2(9−x2)dx, Let x = 3sin(θ), then dx = 3cos(θ)dθ.\displaystyle \int \frac{x^{2}}{\sqrt(9-x^{2})}dx, \ Let \ x \ = \ 3sin(\theta), \ then \ dx \ = \ 3cos(\theta)d\theta.

Ergo, we have: ∫[9sin2(θ)][3cos(θ)](9−9sin2(θ))dθ,\displaystyle Ergo, \ we \ have: \ \int \frac{[9sin^{2}(\theta)][3cos(\theta)]}{\sqrt(9-9sin^{2}(\theta))}d\theta,

= 273∫sin2(θ)cos(θ)(1−sin2(θ))dθ,\displaystyle = \ \frac{27}{3} \int \frac{sin^{2}(\theta)cos(\theta)}{\sqrt(1-sin^{2}(\theta))}d\theta,

= 9∫sin2(θ)cos(θ)cos(θ)dθ = 9∫sin2(θ)dθ,\displaystyle = \ 9\int \frac{sin^{2}(\theta)cos(\theta)}{cos(\theta)}d\theta \ = \ 9\int sin^{2}(\theta)d\theta,

= 9∫1−cos(2θ)2dθ = 92∫(1−cos(2θ))dθ,\displaystyle = \ 9\int\frac{1-cos(2\theta)}{2}d\theta \ = \ \frac{9}{2}\int(1-cos(2\theta))d\theta,

= 92[θ−sin(2θ)2]+C = 92[θ−sin(θ)cos(θ)]+C.\displaystyle = \ \frac{9}{2}\bigg[\theta-\frac{sin(2\theta)}{2}\bigg]+C \ = \ \frac{9}{2}[\theta-sin(\theta)cos(\theta)]+C.

Now x = 3sin(θ)   ⟹   sin(θ) = x3   ⟹   θ = arcsin(x3).\displaystyle Now \ x \ = \ 3sin(\theta) \ \implies \ sin(\theta) \ = \ \frac{x}{3} \ \implies \ \theta \ = \ arcsin(\frac{x}{3}).

Hence, we now have: 92[arcsin(x3)−(x3)((9−x2)3)]+C, cos(θ) = (9−x2)3.\displaystyle Hence, \ we \ now \ have: \ \frac{9}{2}\bigg[arcsin(\frac{x}{3})-(\frac{x}{3})(\frac{\sqrt(9-x^{2})}{3})\bigg]+C, \ cos(\theta) \ = \ \frac{\sqrt(9-x^{2})}{3}.

Therefore,putting it all together, we get: ∫x2(9−x2)dx = 9[arcsin(x/3)]−x(9−x2)2+C.\displaystyle Therefore, putting \ it \ all \ together, \ we \ get: \ \int \frac{x^{2}}{\sqrt(9-x^{2})}dx \ = \ \frac{9[arcsin(x/3)]-x\sqrt(9-x^{2})}{2}+C.QED\displaystyle QED

Check: Dx[9arcsin(x/3)−x(9−x2)2+C] = x2(9−x2).\displaystyle Check: \ D_x\bigg[\frac{9arcsin(x/3)-x\sqrt(9-x^{2})}{2}+C\bigg] \ = \ \frac{x^{2}}{\sqrt(9-x^{2})}.
 
You must log in or register to reply here.
Top