Yes, it's called trigonometric substitution. Try r = tan u.Mathpower said:The Attempt at a Solution
∴t = (1+r2)-1
I then tried substituting u = 1+r^2 but that didn't work!
Is there a trick with this?
No no, put the two attempts together:Mathpower said:NO idea! I tried r= tan u
1/(1+(tan u)^2)...
Too me that just seems even more worse!
Then (LET: < be integral sign/ integrand)
[itex]
= <1/(1+u) dr
= <(1/(1+u)) (secu)^2 du
= <((secu)^2/(1+u)) du [/itex]
Stuck...
If you meanMathpower said:oh wait is the answer u because 1+tan^2 u = sec^2 u
then <1
Do you meanMathpower said:=u
=tan(r)^-1 + C
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Integrating a polynomial is the process of finding its antiderivative or the original function from which the polynomial was derived. This can be used to calculate the area under a curve or solve certain types of equations.
Integrating a complicated polynomial involves using a combination of algebraic techniques, such as substitution and integration by parts, along with knowledge of basic integration rules and properties. It may require multiple steps and careful manipulation to arrive at the final answer.
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