# Trig And calculus

Need help in integration and trig
How do you integrate
1. 1/(1-x^5)
2. 1/(1+x^4)

and the trig question.
Show that

(a^2 - b^2)/c^2 = sin(A-B)/sin(A+B)

## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
Every polynomial, such as 1- x5 and 1+ x4 can be factored, using real numbers, into a product of linear or quadratic terms.
To factor 1- x5, find all complex roots to x5= 1. One is, of course 1 the others are complex conjugates which can be paired to give two quadratic factors. Then use "partial fractions".
Same for 1/(1+x4).

In the trig question, are we to assum that a, b, and c are lengths of sides opposite angles A, B, C? In a right triangle or general triangle?

general triangle!
a,b,c are length's
A,B,C are opposite angles

Sorry but i made a mistake in the first integration question
its suppose to be 1/sqrt(1-x^5)

When i used this integration in mathematica 5 : i got something like hypergeometric2f1....

Can u help me out!

HallsofIvy
Science Advisor
Homework Helper
The 5 "fifth roots of unity" lie on a circle, in the complex plane of radius 1, equally spaced around the circle. The angle between them is 360/5= 72 degrees so they are;
1, cos(72)+ i sin(72), cos(144)+ i sin(144), cos(216)+ i sin(216), cos(288)+ i sin(288).

Since cos(72)= cos(288), sin(72)= -sin(288), cos(144)= cos(216), and sin(144)= sin(216), these are in pairs of complex conjugates (as they have to be in order to satisfy and equation with real coefficients.

The solutions to x5= 1 are: 1, cos(72)+ i sin(72), cos(72)- i sin(72), cos(144)+ i sin(144), cos(144)- i sin(144) and so
1- x= -(x-1)(x- cos(72)+ i sin(72))(x- 72- i sin(72))(x- cos(144)+ isin(144))(x- cos(144)- i sin(144))= -(x-1)((x-cos(72))2+ sin2(72))((x-cos(144)2+ sin2(144))

= -(x-1)(x2- 2cos(72)+ 1)(x2-2cos(144)+ 1).

Once you have that factorization you can expand 1/(1- x5) in partial fractions.