1. I'll have to play with this one ... even powers of sine and cosine are a pain-in-the-...
2.
3. how is it negative ?
from 1 to sqrt(3), the function y = arctan(1/x) is greater than 0.
I get positive .468
Alright. I already put a lot of thought into some of these but I cannot get the right answer out or am stuck in the middle.
1) The integral, from 0 to pi/4, of ((sinx)^4)((cosx)^2)dx
My thought process has gotten to (1/16)integral(1-cos4x)dx -integral((1/8)((cosx)^2)((sin2x)^2)dx
What do i do from here? Or is there a different way to go at it?
2) Integral, from pi/4 to pi/2 of (cotx)^2 for this one, I got all the way to solving the integral, which i Think is -.5(cotx)^2 -ln(abs value of)sinx ...
I am stuck here, since pi/2 cant go into cotangent... what do I do?
3) The integral of 1 to radical 3 of arctan(1/x)dx. I solved this several times, and I got the integral to come out as xarctan(1/x) +.5ln(x^2 +1)
However, this is wrong, because i know that the answer SHOULD be -.468. How do i get -.468?
THANK YOU! for your helP!
Hello, 3deltat!
How can the answer be negative? . . . The graph is above the x-axis.
I got the integral to come out as: . . . . . Right!
However, this is wrong, because i know that the answer SHOULD be -0.468 . ??Ah, I see skeeter already beat me to it!Code:| | ..* | .*:::::| | *:::::::::| | * |:::::::::| |* |:::::::::| | |:::::::::| - * - + - - - - + - | 1 √3
We have: .
. .
. .
. .
. .
. .
When both are even, you need to apply these identities: and :
Now split up the integral:
Let's focus on this integral:
This is the same as saying
Evaluating, we get
Now let's evaluate
Break off a factor of and apply the identity
Thus, the integral becomes
Now let
We can change the limits of integration as well.
The integral can now be written as
Finally, our total solution is
Does this make sense?
--Chris