Question

Integration question (inside)

Answer 1

\[
\int\frac{dx}{2+\tan x}
\]
I'm guessing \(u=\tan x\), which gets me to:
\[
\int\frac{du}{(u^2+1)(u+2)}=\int\frac{du}{u^3+2u^2+u+2}=\int\frac{du}{u(u+1)^2+2}
\]
But I don't know where to go from there.

Answer 2

change tanx in terms of sinx and cosx and
the q becomes cosx/(2cosx-sinx)
now express cosx==A(2cosx -sinx)+G(-2sinx+cosx)
and then derive ur answer which may be of the form Ax+Bln|2cosx-sinx| +c

Answer 3

why don't you make u=tan(x)+2 to get rid of the +2?

Answer 4

Well, then dx will become more complex, so the net result will be equally complex.

Answer 5

We'll have instead:
\[
\int\frac{du}{u((u+2)^2+1)}
\]

Answer 6

The other solution I'd suggest is a Weierstrass substitution.

Answer 7

That was mentioned in chat, but he doesn't introduce that method until question 12 of this chapter, and this question is 5(v)

Answer 8

If this follows the pattern of the previous four sub-questions in this question, there will be some type of algebraic manipulation to simplify the integral.

Answer 9

What about partial fractions, because the second line in your first reply can be broken up.

Answer 10

Yep, that is probably it, I'm just struggling with partial fractions.

Answer 11

There is a decently sized section in the text of this chapter on rational functions and partial fraction decomposition, but it is going straight over my head.

Answer 12

\[\frac{1}{(u^2+1)(u+2)}=\frac{Au+B}{u^2+1}+\frac{C}{u+2}\\
1=(Au+B)(u+2)+C(u^2+1)\\
1=Au^2+u(2A+B)+2B+Cu^2+C\\
1=u^2(A+C)+u(2A+B)+(2B+C)\]
Solve the system:
\[A+C=0\\
2A+B=0\\
2B+C=1\]
to get A, B, C.

Answer 13

Solution to that system I'm getting A=-1/5, B=2/5, C=1/5

Answer 14

So does that mean the integral can be written as:
\[
\int\frac{\frac{-1}{5}u+\frac{2}{5}}{u^2+1}du+\int\frac{\frac{1}{5}}{u+2}du\text{ ?}
\]

Answer 15

Yep, and that's more easily evaluated.

Answer 16

Huh. Okay. Do you know of any good resources for learning about this partial fractions magic?

Answer 17

http://tutorial.math.lamar.edu/Classes/CalcII/PartialFractions.aspx

Answer 18

Thanks! :)

Answer 19

For the final result of the integral I am getting:
\[
\frac{-1}{10}\log(u^2+1)+\frac{2}{5}\arctan u + \frac{1}{5}\log(u+2)\\
\frac{-1}{10}\log(\sec^2x)+\frac{2}{5}x + \frac{1}{5}\log(\tan x+2)\\
\]
which looks pretty reasonable.