Substitution and Integration of Rational Functions

[rheighton]

I have three similar problems, and figured out the substitutions for them, but am not sure what processes to use to solve them:

1) (from x=0 to x=4) ∫ (1-root(x)) / (1+root(x))
letting u=root(x), dx=2udu
giving: (from x=0 to x=2) ∫ [(1-u) / (1+u)]2udu

2) ∫ dx / (root(x) + cubert(x))
letting u=x^(1/6), x=u^6, dx=6u^5 du
giving 6∫ (u^3) / (u+1) du

3) (from x=(pi/3) to x=(pi/2)) ∫ (sin(theta)d(theta)) / (cos^2(theta)+cos(theta)-2)
letting u=cos(theta), du=-sin(theta)
giving (interval?) ∫ -du / (sin^2(theta)+sin(theta)-2)

I have no idea what steps to take now, or if my work is even correct so far... do I use partial fractions, with polynomial division for 2)? And I have absolutely no clue what to do with 3)... if somebody would be gracious enough to help me, it would be awesome!

 

[daon]

1. if u = sqrt(x), then dx = 2udu

so,

\(\displaystyle \L
\\ \int\limits_0^4 \frac{1+\sqrt{x}}{1-\sqrt{x}}dx = 2\int\limits_0^4 \frac{1+u}{1-u}du \\= 2\int\limits_0^4 \frac{1}{1-u}du + 2\int\limits_0^4 \frac{u}{1-u}du =...\)

 

[soroban]

Hello, rheighton!

\(\displaystyle 1)\;\L\int^{\;\;\;4}_0\frac{1\,-\,\sqrt{x}}{1\,+\,\sqrt{x}}\,dx\)

Let \(\displaystyle u\:=\:\sqrt{x}\;\;\Rightarrow\;\;dx\:=\:2u\,du\)

giving: \(\displaystyle \L\,\int^{\;\;\;2}_0\frac{1\,-\,u}{1\,+\,u}\)\(\displaystyle (2u\,du)\;\;\) . . . yes!

You're doing great!

We have: \(\displaystyle \L\,2\int\frac{u\,-\,u^2}{1\,+\,u}\,du\)

Use long division: \(\displaystyle \L\,2\int\left(2\,-\,u\,-\,\frac{2}{1\,+\,u}\right)\,du\)

\(\displaystyle 2)\;\L\int\frac{dx}{\sqrt{x}\,+\,\sqrt[3]{x}}\)

Let \(\displaystyle u\,=\,x^{\frac{1}{6}\;\;\Rightarrow\;\;x\,=\,u^6\;\;\Rightarrow\;\;dx\,=6u^5\,du\)

giving: \(\displaystyle \L\,6\int \frac{u^3}{u\,+\,1}\,du\;\) . . . good!

Long division: \(\displaystyle \L\,6\int\left(u^2\,-\,u\,+\,1\,-\,\frac{1}{u\,+\,1}\right)\,du\)

\(\displaystyle 3)\;\L\int^{\;\;\;\frac{\pi}{2}}_{\frac{\pi}{3}} \frac{\sin\theta\,d\theta}{\cos^2\theta \,+\,\cos\theta \,-\,2}\)

Let \(\displaystyle u\,=\,\cos\theta\;\;\Rightarrow\;\;du\,=\,-\sin\theta\,d\theta\)

Note that the denominator factors: \(\displaystyle \L\;\int\frac{\sin\theta\,d\theta}{(\cos\theta\,-\,1)(\cos\theta\,+\,2)}\)

Giving: \(\displaystyle \L\,\int\frac{-du}{(u\,-\,1)(u\,-\,2)\;\;\) . . . now apply Partial Fractions.

 

[wjm11]

Note: In problem 3, the second term in the denominator, (u-2), should be (u+2), I believe.