Question

compute integrals
im confused

Answer 1

\[\int\limits_{-1}^{3}\frac{ t^3 }{ 1+t^2 }\]

2) \[\int\limits_{}^{}Sec^3xtan^3xdx\]

Answer 2

How about for the first one that you try it with a long hand division first? and then maybe apply some partial fraction decomposition?

Answer 3

yea, i am really confused! i fist find u and du..

Answer 4

integral (v) du = u*v - integral(u)dv

Answer 5

let me try:

\[\frac{ t^3 }{ t^2+1 }=t-\frac{ t }{ t^2+1 }\]

Answer 6

the right hand side is an easy integral, it's just an u substitution for u=t^2+1

Answer 7

i got that too for u. but how about du ? :/

Answer 8

du=2tdt

Answer 9

mmm would it be t?

Answer 10

well solve the above for dt and plug it back into the integral for the quotient. But basically it will turn out to be a logarithm. So the answer should be a logarithmic property.

Answer 11

could you show me steps ?

Answer 12

yes one second.

Answer 13

great, thank you

Answer 14

\[\int\limits \frac{ t }{ t^2+1 }dt\]

let u= t^2+1 such that:
du=2tdt

So you can solve the above for dt, or you can also solve it for t directly because t is in the numerator.
\[t=\frac{ du }{ 2dt }\]
Substitute that back in the above:

\[\int\limits \frac{ \frac{ du }{ 2dt } }{ u }dt=\frac{ 1 }{ 2 } \int\limits \frac{ 1 }{ u }du\]

Answer 15

why its t=du/2dt?

Answer 16

algebra.

\[u=t^2+1\]
so differentiate that with respect to t, that's just a regular differentiation of a polynom of second order. So use the Leibniz Notation you end up at:
\[\frac{ du }{ dt }=2t\]
Now you have two options:
1) solve for dt and plug it back into the integral
2) solve for t (because t appears in the numerator) and plug it back into the integral.

If you decide for 2, then by algebra:
\[t= \frac{ du }{ 2dt }\]
division by 2

Answer 17

I see. wwhat is the next step? do i need to do antiderivtive?

Answer 18

yes, that step is imperative if you apply aan integral, especially when it's a definite integral like in your case. You want to apply the fundamental theorem of single variable calculus:

\[\int\limits_a^b f(x)dx=F(b)-F(a)\]
Where F(x) is the antiderivative of f(x)

Answer 19

I see

Answer 20

\[\int\limits_{?}^{?} du/2dt/u=1/2\int\limits_{?}^{?}1/(t^2+1)?\]

Answer 21

Do you know the anti derivative of 1/x?

Answer 22

Yes! In|x|?

Answer 23

exactly, So you should be able to do the integral above:

\[\frac{ 1 }{ 2 } \int\limits \frac{ 1 }{ u }du\]

Answer 24

\[\int\limits_{}^{} \ln|u|du \]1/2

Answer 25

1/2 in front

Answer 26

exactly, the one half as a constant factor infront of it. So you're almost done with taht exercise, remember that you need to integrate t also because, above I demonstrated that:
\[\frac{ t^3 }{ t^2+1 }=t-\frac{ t }{ t^2+1 }\] so what you computed waht the integral on the very right end side of this equation, you need to integrate the entire right hand side, but the integral of t with respect to t is very easy.

Answer 27

is this u?

Answer 28

excuse me, I don't understand your question, what is supposed to be u?

Answer 29

u=t^2+1? insert this into
1/2