- #1
kallazans
- 5
- 0
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?
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?
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