Trig Substitiution

B

Blitze105

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Aug 28, 2008
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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

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

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

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


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
 
\(\displaystyle 2nd \ one:\)

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

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

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

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

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

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

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

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

\(\displaystyle Check: \ D_t \ \bigg[-\frac{\sqrt(1+t^{2})}{t}+C\bigg] \ = \ \frac{1}{t^{2}\sqrt(1+t^{2})}.\)
 
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....
 
\(\displaystyle First \ one:\)

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

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

\(\displaystyle = \ \frac{27}{3} \int \frac{sin^{2}(\theta)cos(\theta)}{\sqrt(1-sin^{2}(\theta))}d\theta,\)

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

\(\displaystyle = \ 9\int\frac{1-cos(2\theta)}{2}d\theta \ = \ \frac{9}{2}\int(1-cos(2\theta))d\theta,\)

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

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

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

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

\(\displaystyle Check: \ D_x\bigg[\frac{9arcsin(x/3)-x\sqrt(9-x^{2})}{2}+C\bigg] \ = \ \frac{x^{2}}{\sqrt(9-x^{2})}.\)
 
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