1. ## anti deriviative

the question is :
anti deriviative of : 4/(1+4t^2)
How do I figure out questions like this? I really cant see this as product or chain rule so kind of stumped.

2. Originally Posted by frankinaround
the question is :
anti deriviative of : 4/(1+4t^2)
How do I figure out questions like this? I really cant see this as product or chain rule so kind of stumped.
antiderivative will be an arctan function

3. Originally Posted by frankinaround
the question is :
anti deriviative of : 4/(1+4t^2)
How do I figure out questions like this? I really cant see this as product or chain rule so kind of stumped.
Product and chain rules are used for differentiation NOT anti-differentiation.

To do this particular anti-differentiation you make the substitution $\displaystyle 2t = \tan(\theta)$.

Alternatively you can substitute $\displaystyle u = 2t$ and recognise a standard form.

4. but how do you know when its integration by parts or trig substitution?

5. With practice it becomes clearer. You can either learn the trig forms or try integration by parts and when that fails badly it's a trig sub

6. for this problem my integral became : 2t + 1/(8cos^theyta) * theyta
theyta = arctan(2t)
t=1/2 (because its an area problem from t= 0 to t = 1/2)

The thing is im not allowed to use a calculator, and even if I did it would not help. the possible answers are:
-pie
3/2 * pie
1/2 * pie
pie
0

How do I get the right answer? How do I calculate the arctan without a calculator? ect?

7. $\displaystyle \displaystyle \int_0^{\frac{1}{2}} \frac{4}{1+4t^2} \, dt$

$\displaystyle u = 2t$ ... $\displaystyle du = 2 \, dt$

$\displaystyle \displaystyle 2\int_0^{\frac{1}{2}} \frac{2}{1+(2t)^2} \, dt$

$\displaystyle \left[2\arctan(2t)\right]_0^{\frac{1}{2}}$

$\displaystyle 2\arctan(1) - 2\arctan(0)$ ... note the following:

(1) these are unit circle values. you should already be familiar with them.
(2) the arctangent function has range $\displaystyle -\dfrac{\pi}{2} < y < \dfrac{\pi}{2}$ ... you should be familiar with this also.

$\displaystyle 2 \cdot \dfrac{\pi}{4} - 2(0) = \dfrac{\pi}{2}$

8. Originally Posted by frankinaround
for this problem my integral became : 2t + 1/(8cos^theyta) * theyta
theyta = arctan(2t)
t=1/2 (because its an area problem from t= 0 to t = 1/2)

The thing is im not allowed to use a calculator, and even if I did it would not help. the possible answers are:
-pie
3/2 * pie
1/2 * pie
pie
0

How do I get the right answer? How do I calculate the arctan without a calculator? ect?
If you have arctan(1) then ask yourself the question for what values of theta is tan(theta) = 1?

By the way, it's "theta" not "theyta" and "pi" not "pie."

-Dan

9. so if your using U substitution how does it become u = 2t? are you saying u^2 = 4t^2 and then u=2t that way? Also im familier with trig substitution and u substitution. this problem is making me nuts. look at what Im getting with trig substitution for example :

In = integral.

In(4/(1+4t^2))
4t^2=tan^2theta
2t=tan(Theta)
dt = (sec(theta))/2 + (tan(theta))/4 dtheta == 1/(2 cos(theta)) + sin(theta)/(4 cos(theta)) dtheta

in(4/1+tan^2Theta) dt = in(4cos^2Theta) dt =
in(4cos^2Theta) ( 1/(2 cos(theta)) + sin(theta)/(4 cos(theta)) ) dtheta ==
in(2costheta + cos(theta)sin(theta) dTheta)
= 2 sin theta + 1/2 sin^2 theta.

evaluate at 1/2 and 0.
theta = arctan 2t. arctan 1 = 45* or pie/4
2sin(45) = 1/2 sin^2(45) doesnt equal any answer on test. Test answers all have pie in them.

Someone in this forum said use U substitution, but yeah how do you get 2t = u. Another person said answer will be an arctan function. Can anyone explain that one to me? Yeah this problem is just beating me up. badly.

right now to solve integrals I have these methods :
Trig substitution
u substitution
factoring
and sometimes its the equation of a circle.
so what am i missing? arctan substitution? what?

10. note the following basic rule for taking the derivative of $\displaystyle y = \arctan{u}$ where $\displaystyle u$ is a function of $\displaystyle x$ ...

$\displaystyle \dfrac{d}{dx} \arctan{u} = \dfrac{u'}{1+u^2}$

now, what is the antiderivative of $\displaystyle \dfrac{u'}{1+u^2}$ ?

11. Originally Posted by frankinaround
so if your using U substitution how does it become u = 2t? are you saying u^2 = 4t^2 and then u=2t that way? Also im familier with trig substitution and u substitution. this problem is making me nuts. look at what Im getting with trig substitution for example :

In = integral.

In(4/(1+4t^2))
4t^2=tan^2theta
2t=tan(Theta)
dt = (sec(theta))/2 + (tan(theta))/4 dtheta == 1/(2 cos(theta)) + sin(theta)/(4 cos(theta)) dtheta

in(4/1+tan^2Theta) dt = in(4cos^2Theta) dt =
in(4cos^2Theta) ( 1/(2 cos(theta)) + sin(theta)/(4 cos(theta)) ) dtheta ==
in(2costheta + cos(theta)sin(theta) dTheta)
= 2 sin theta + 1/2 sin^2 theta.

evaluate at 1/2 and 0.
theta = arctan 2t. arctan 1 = 45* or pie/4
2sin(45) = 1/2 sin^2(45) doesnt equal any answer on test. Test answers all have pie in them.

Someone in this forum said use U substitution, but yeah how do you get 2t = u. Another person said answer will be an arctan function. Can anyone explain that one to me? Yeah this problem is just beating me up. badly.

right now to solve integrals I have these methods :
Trig substitution
u substitution
factoring
and sometimes its the equation of a circle.
so what am i missing? arctan substitution? what?
If you want to use trigonometric substitution, then as was suggested, you let $\displaystyle \displaystyle t = \frac{1}{2}\tan{\theta}$ (or $\displaystyle \displaystyle 2t = \tan{\theta}$) so that $\displaystyle \displaystyle dt = \frac{1}{2}\sec^2{\theta}\,d\theta$.

Then your integral $\displaystyle \displaystyle \int{\frac{4\,dt}{1 + 4t^2}}$ becomes

$\displaystyle \displaystyle \int{\frac{4\cdot \frac{1}{2}\sec^2{\theta}\,d\theta}{1 + 4\left(\frac{1}{2}\tan{\theta}\right)^2}}$

$\displaystyle \displaystyle = \int{\frac{2\sec^2{\theta}\,d\theta}{1 + 4\cdot \frac{1}{4}\tan^2{\theta}}}$

$\displaystyle \displaystyle = \int{\frac{2\sec^2{\theta}\,d\theta}{1 + \tan^2{\theta}}}$

$\displaystyle \displaystyle = \int{\frac{2\sec^2{\theta}\,d\theta}{\sec^2{\theta }}}$

$\displaystyle \displaystyle = \int{2\,d\theta}$

$\displaystyle \displaystyle = 2\theta + C$.

Now remembering that $\displaystyle \displaystyle 2t = \tan{\theta}$, this means $\displaystyle \displaystyle \theta = \arctan{\left(2t\right)}$.

So $\displaystyle \displaystyle 2\theta + C = 2\arctan{\left(2t\right)} + C$.

12. but how do you make the substitution t for 1/2 tan theta?

from what I understand :

sqrt(1-x^2) can be a trig substitution of a triangle with a hypotenuse of 1 and a side x and another side sqrt(1-x^2)

So what is the triangle with 1/2tan theta = t? can you help me see the triangle this is all coming from?

13. I don't use a triangle AT ALL. The point is to turn the denominator into either $\displaystyle \displaystyle 1 - \sin^2{\theta}$ or $\displaystyle \displaystyle 1 - \tan^2{\theta}$ so that you can simplify everything using a trigonometric identitity. You can substitute ANY function you like that will make this happen, as long as you substitute its derivative for $\displaystyle \displaystyle dx$ as well.