Three Weird Questions - Integration. (1 Viewer)

[Mathematician]

Integrate (x^2+1)^1/2
Integrate (4-x^2)^1/2
Integrate x^2(1+x^2)^1/2

You have to do these by parts.

U cant use substitution and for the first q ,the book tells u a hint.
Hint: [(x^2+1)^1/2]dx= [(x^2+1)^1/2 * 1 ]dx

u cant use x=tanu.

Also for the first one, u get to a stage when u have to integrate
x^2(1+x^2)^1/2 - And obviously ur not allowed to use substition - since it would have been easier to use substition from the beginning - which ur not allowed to , since u have to do it by parts.

Haha i write crap.

 

[McLake]

Um, are these for people to work out, or do you wan't help?

 

[Lazarus]

Originally posted by Mathematician
Integrate (4-x^2)^1/2

Can you solve this one by declaring that it's equal to the area of a semi-circle with radius 2? (i.e. 2pi)

Repeatedly IBP is too time-consuming for me to solve the rest of them for fun.

 

[Mathematician]

....

Ah sorry i wasnt clear, i wanted help lol.


U CANT USE TRIG SUBSTITUTION AT ALL.

Lazurus u could do that if it was a definite integral.

 

[Lazarus]

Oh yeah. OK.

 

[McLake]

F is the intergral symbol

Integrate (x^2+1)^1/2

Hint: [(x^2+1)^1/2]dx= [(x^2+1)^1/2 * 1 ]dx



u = (x^2+1)^1/2
du = x(x^2+1)^-1/2

v = x
dv = 1

so x(x^2+1)^1/2 - Fx^2(x^2+1)^-1/2
= xI - F[x^2*I/(x^2 + 1)]
= xI - F[I - I/(x^2 + 1)]

...

 

[bobo123]

mmm
wana hint for the first one?

when your doing the second integration from IBP
you gotta manipulate it a bit
eg: x^2 = x^2 +1 - 1

hope it helps