# How to integrate without trigonometric substitution

1. Dec 13, 2003

### kallazans

S(x²(1-x²)¹/²)dx
without use x=sen(T)?

Sx^2(1-x^2)^1/2dx

x=sen(t)
dx=cos(t)dt (-pi/2<=t<=pi/2)

S(x^2(1-x^2)^1/2)dx=S(sen^2(t)cos^2(t))dt
S(sen^2(t)cos^2(t)^)dt=S((1-cos^2(2t))/4)dt
=S1/4dt - S((co^2(2t))/4)dt= t + c1 - S((1-cos(4t))/8)dt
=t/4 + c1 - S1/8dt + S(cos(4t)/8)dt= t/4 + c1 - t/8 + sen(4t)/32 + c2
=t/8 + sen(4t)/32 + C
but:
sen(4t)=2sen(2t)cos(2t)=2(2sentcost)(1-2sen^2(t))
=4sentcost-8sen^3(t)cost

and t=arcsen(x)

logo
Sx^2(1-x^2)^1/2dx= 1/8arcsen(x)+ 1/8x(1-x^2)^1/2 - 1/4x^3(1-x^2)^1/2 + C

but show that D(1/8arcsen(x)+ 1/8x(1-x^2)^1/2 - 1/4x^3(1-x^2)^1/2 + C)=x^2(1-x^2)^1/2 is another history!is a very hard work? someone agree with me?

Last edited: Dec 13, 2003
2. Dec 13, 2003

### Hurkyl

Staff Emeritus
Why without substitution?

Anyways, what approaches have you tried so far? Have you tried any substitutions? What were the results? Have you tried integration by parts? Have you tried something else?

3. Dec 16, 2003

### Sting

You could use tabular integration (which is a form of integration by parts).

The only problem in that method would be integrating (1 - x2)1/2dx