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:
3/2 * pie
1/2 * pie
How do I get the right answer? How do I calculate the arctan without a calculator? ect?
... note the following:
(1) these are unit circle values. you should already be familiar with them.
(2) the arctangent function has range ... you should be familiar with this also.
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.
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 :
and sometimes its the equation of a circle.
so what am i missing? arctan substitution? what?
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 or 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 as well.