Intergration (1 Viewer)

[Shady01]

Ok this is piss easy question I've used t= tanx.. t= cotx x= 1 +t^2 I just cannot get it help please .

The intergral of dt/(1+t^2) between x= cotx and x= tanx
or u can go to Fitzparticks 33(b) question 26
cheers

 

[Riviet]

Integral of dt/(1+t²) is tan^-1t

cot(x)=tan(pi/2 - x)

Remember that tan[tan^-1A] = A

Sub. in your values and you should get the answer at the back of book (pi/2 - x). Hope that helps.

 

[Shady01]

ahhhhhhhhhhhh cheers Riv makes sense now
I knew it was Tan-1t but i kept getting
tan^(-1)cotx- tan^(-1)tanx
which gives
Tan^(-1)cotx-x never thought of 90-x cheers thanx fer that champ
Nick

 

[lala2]

Since this is an integration thread...I'll hook on!

How do you find the following integrals?

1. x/(x^2 + x +1)
2. x^3/(x^2 +1)
3. x^2/(x^2 + 1)

 

[bboyelement]

Since this is an integration thread...I'll hook on!

How do you find the following integrals?

1. x/(x^2 + x +1)
2. x^3/(x^2 +1)
3. x^2/(x^2 + 1)

for 2 and 3 you just divide the top by the bottom and integrate and for the first one ... let u = x+1/2 ... therefore you will get S (u-1/2)(/u^2+3/4) and you split it up ... and integrate ... im not at home so i cant do all the working out.. i hope that helps

 

[Riviet]

Since this is an integration thread...I'll hook on!

How do you find the following integrals?

1. x/(x^2 + x +1)
2. x^3/(x^2 +1)
3. x^2/(x^2 + 1)

1. x/(x²+x+1) = 1/2. (2x+1-1)/(x²+x+1)
= 1/2. (2x+1)/(x²+x+1) - 1/2. 1/[(x + 1/2)²+3/4]
*cough* log tan inverse *cough*

2. Use long division or manipulation:
x³/(x²+1)=[x(x²+1)-x]/(x²+1)=x - x/(x²+1)
= x - 1/2. 2x/(x²+1)

3. You could use long division, but manipulation is much easier:
x²/(x²+1)=(x²+1-1)/(x²+1)
=1 - 1/(x²+1)

Have fun integrating.

 

[lala2]

Thank you Riviet and bboyelement!