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.
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?
$\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.
$\displaystyle 2 \cdot \dfrac{\pi}{4} - 2(0) = \dfrac{\pi}{2}$
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?
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 $\displaystyle \dfrac{u'}{1+u^2}$ ?
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 $\displaystyle \displaystyle 2\theta + C = 2\arctan{\left(2t\right)} + C$.
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?
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.