1. ## [SOLVED] trig subs

$

\int\sqrt{25-t^2}\,dt
$

a = 5 u = t so....

t = 5 sin $\theta$

sqrt(25-(5 sin $\theta$)^2 )

sqrt(25 cos^2 $\theta$)

25 cos $\theta$

25 sin $\theta$

25 sin(t/5) is what I get after substituting in for t. I know this isn't right but could someone explain why?

2. Originally Posted by ur5pointos2slo
$

\int\sqrt{25-t^2}\,dt
$

a = 5 u = t so....

t = 5 sin $\theta$

sqrt(25-(5 sin $\theta$)^2 )

sqrt(25 cos^2 $\theta$)

25 cos $\theta$

25 sin $\theta$

25 sin(t/5) is what I get after substituting in for t. I know this isn't right but could someone explain why?
$\sqrt{25} \ne 25$

3. Originally Posted by Jhevon
$\sqrt{25} \ne 25$

AHHHHHH ok so lets keep working

5 cos $\theta$

so

5 sin $\theta$

I still get

5 sin(t/5)

which is definately nowhere near the answer in the back of my text.

they have....
25/2 sin^-1(t/5) + (t * sqrt(25-t^2)/2 + C

4. Originally Posted by ur5pointos2slo
AHHHHHH ok so lets keep working

5 cos $\theta$

so

5 sin $\theta$

I still get

5 sin(t/5)

which is definately nowhere near the answer in the back of my text.

they have....
25/2 sin^-1(t/5) + (t * sqrt(25-t^2)/2 + C
again, you back substituted wrong. $\theta \ne \frac t5$. it is $\sin \theta = \frac t5$.

secondly, where is your $d \theta$ ?!

lets start over....

$\int \sqrt{25 - t^2}~dt$

let $t = 5 \sin \theta$

$\Rightarrow \color{red} dt = 5 \cos \theta ~d \theta$

so our integral becomes:

$\int \sqrt{25 - 25 \sin^2 \theta} \cdot {\color{red}5 \cos \theta} ~d \theta = 25 \int \cos^2 \theta ~d \theta$

now finish up

5. scratch that I think you edited.

Ok I think I got lost in that last statement because shouldnt this

sqrt(25 cos^2 theta) 5 cos theta dtheta

be

5 cos theta 5 cos theta dtheta

=

25 int(cos^2 theta) dtheta?

6. Originally Posted by Jhevon
again, you back substituted wrong. $\theta \ne \frac t5$. it is $\sin \theta = \frac t5$.

secondly, where is your $d \theta$ ?!

lets start over....

$\int \sqrt{25 - t^2}~dt$

let $t = 5 \sin \theta$

$\Rightarrow \color{red} dt = 5 \cos \theta ~d \theta$

so our integral becomes:

$\int \sqrt{25 - 25 \sin^2 \theta} \cdot {\color{red}5 \cos \theta} ~d \theta = 25 \int \cos^2 \theta ~d \theta$

now finish up
ok

25/2 sin^-1(t/5) (1 + cos(2theta) ) / 2

trig integrals always give me a hard time. Did I substitute in correctly for this? If so what would I do with that 1 + cos(2theta) / 2

7. Originally Posted by ur5pointos2slo
ok

25/2 sin^-1(t/5) (1 + cos(2theta) ) / 2

trig integrals always give me a hard time. Did I substitute in correctly for this? If so what would I do with that 1 + cos(2theta) / 2
no you didn't. you didn't even integrate correctly

$25 \int \cos^2 \theta ~d \theta = 25 \int \frac {1 + \cos 2 \theta}2~d \theta = \frac {25}2 \int (1 + \cos 2 \theta) ~ d \theta$

now what?

8. Originally Posted by Jhevon
no you didn't. you didn't even integrate correctly

$25 \int \cos^2 \theta ~d \theta = 25 \int \frac {1 + \cos 2 \theta}2~d \theta = \frac {25}2 \int (1 + \cos 2 \theta) ~ d \theta$

now what?
hmmm I really am not sure. I think what is confusing me is the fact that you have 1 + cos 2 theta. so you have to break the integrals up? and wouldnt this be a u-sub now? with
u = 2theta du = 2 so put 2 in 1/2 out the integral?

but then I get 25/4 sin 2theta

9. Originally Posted by ur5pointos2slo
hmmm I really am not sure. I think what is confusing me is the fact that you have cos 2 theta. wouldnt this be a u-sub now? with
u = 2theta du = 2 so put 2 in 1/2 out the integral?

but then I get 25/4 sin 2theta
yes...u-sub is the hard way. you should know that $\int \cos k \theta ~d \theta = \frac 1k \sin k \theta + C$, for $k \ne 0$ a constant.

thus you get $\frac {25}2 \left( \theta + \frac 12 \sin 2 \theta \right) + C = \frac {25}2 \theta + \frac {25}4 \sin 2 \theta + C = \frac {25}2 \theta + \frac {25}2 \sin \theta \cos \theta + C$

now use your trig sub to back substitute to get your answer in terms of $t$.

10. Originally Posted by Jhevon
yes...u-sub is the hard way. you should know that $\int \cos k \theta ~d \theta = \frac 1k \sin k \theta + C$, for $k \ne 0$ a constant.

thus you get $\frac {25}2 \left( \theta + \frac 12 \sin 2 \theta \right) + C = \frac {25}2 \theta + \frac {25}4 \sin 2 \theta + C = \frac {25}2 \theta + \frac {25}2 \sin \theta \cos \theta + C$

now use your trig sub to back substitute to get your answer in terms of $t$.
ok another probably obvious question. When you have

sin 2 theta how would you go about using substitutions when you dont know what theta is? we know that theta = sin^-1(t/5) but how would that even help here?

11. Originally Posted by ur5pointos2slo
ok another probably obvious question. When you have

sin 2 theta how would you go about using substitutions when you dont know what theta is? we know that theta = sin^-1(t/5) but how would that even help here?
did you even read the chapter on trig subs? were you paying attention in class?

since we know $\sin \theta = \frac t5 = \frac {\text{opposite}}{\text{hypotenuse}}$, we have the following triangle. use it to find $\cos \theta = \frac {\text{adjacent}}{\text{hypotenuse}}$, and then you can change your answer to be in terms of $t$. We find the other side by using Pythagoras' theorem

12. Originally Posted by Jhevon
yes...u-sub is the hard way. you should know that $\int \cos k \theta ~d \theta = \frac 1k \sin k \theta + C$, for $k \ne 0$ a constant.

thus you get $\frac {25}2 \left( \theta + \frac 12 \sin 2 \theta \right) + C = \frac {25}2 \theta + \frac {25}4 \sin 2 \theta + C = \frac {25}2 \theta + \frac {25}2 \sin \theta \cos \theta + C$

now use your trig sub to back substitute to get your answer in terms of $t$.
Thanks for all your help!!! I have solved my problem.

13. Originally Posted by Jhevon
did you even read the chapter on trig subs? were you paying attention in class?

since we know $\sin \theta = \frac t5 = \frac {\text{opposite}}{\text{hypotenuse}}$, we have the following triangle. use it to find $\cos \theta = \frac {\text{adjacent}}{\text{hypotenuse}}$, and then you can change your answer to be in terms of $t$
I did read the chapter and have a lot of notes. The trig substitutions section just expects you to know your trig identities which I do not know all of them by heart. I know all the basic identities. The sin2theta I had to look up. Thats what I was confused about.