u-substitution: int [x=0, 1] [dx/sqrt(4-x^2)], etc.

[math]

Using u-substitution,

1) integral from 0 to 1 of dx/sqrt(4-x^2)

2) integral of dx/sqrt(2x-x^2)

3) integral from 0 to 1/2 of (2-8x)/(1+4x^2) dx

4) integral from 0 to sqrt (2) / 2 of (2x)/(sqrt(1-x^4) dx

Whatever I let u equal for these problems, I cannot get the x's to cancel. I used my calculator and got the integrals to be different inverse trig functions. I don't know where to start and where the answers are coming from...any help would be appreciated! thanks!

 

[galactus]

Using u-substitution,

1) integral from 0 to 1 of dx/sqrt(4-x^2)

Let \(\displaystyle \L\\x=2sin({\theta}), \;\ dx=2cos({\theta})d{\theta}\)

[quote:17l2u1bk]2) integral of dx/sqrt(2x-x^2)

Let \(\displaystyle x=u+1, \;\ dx=du\)

The you'll have \(\displaystyle \L\\\frac{1}{\sqrt{1-u^{2}}}\)

Let \(\displaystyle u=sin({\theta}), \;\ du=cos({\theta})d{\theta}\)

3) integral from 0 to 1/2 of (2-8x)/(1+4x^2) dx

Partial fractions, \(\displaystyle \L\\\frac{2}{4x^{2}+1}-\frac{8x}{4x^{2}+1}\)

4) integral from 0 to sqrt (2) / 2 of (2x)/(sqrt(1-x^4) dx

Let \(\displaystyle \L\\u=x^{2}, \;\ du=2xdx\)

Then you will have \(\displaystyle \L\\\frac{1}{\sqrt{1-u^{2}}}du\)

Let \(\displaystyle \L\\u=sin({\theta}), \;\ du=cos({\theta})d{\theta}\)

Just some options to consider. Many ways to go about it.

Whatever I let u equal for these problems, I cannot get the x's to cancel. I used my calculator and got the integrals to be different inverse trig functions. I don't know where to start and where the answers are coming from...any help would be appreciated! thanks![/quote:17l2u1bk]

 

[soroban]

Hello, math!

Are you sure of the instructions?
. . These all require Inverse Trig formulas.

\(\displaystyle 1)\L\:\int^{\;\;\;1}_0\frac{dx}{\sqrt{4\,-\,x^2}\)

This has the \(\displaystyle \arcsin\) form.

Evaulate: \(\displaystyle \L\:\arcsin\left(\frac{x}{2}\right)\;\) from \(\displaystyle x=0\) to \(\displaystyle x=1\)

\(\displaystyle 2)\;\L\int \frac{dx}{\sqrt{2x\,-\,x^2}}\)

In the denominator, complete the square: \(\displaystyle \,2x\,-\,x^2\:=\:1\,-\,(x\,-\,1)^2\)

So we have: \(\displaystyle \L\:\int\frac{dx}{\sqrt{1\,-\,(x\,-\,1)^2}}\;=\;\arcsin(x\,-\,1)\,+\,C\)

\(\displaystyle 3)\;\L\int^{\;\;\;\frac{1}{2}}_0 \frac{2\,-\,8x}{1\,+\,4x^2}\,dx\)

Make two integrals . . .

. . \(\displaystyle 2\L\int\frac{dx}{1\,+\,4x^2} \:-\:\int\frac{8x}{1\,+\,4x^2}\,dx\)

The first integral is \(\displaystyle arctan\); the second is \(\displaystyle ln\)

\(\displaystyle 4)\;\L\int^{\;\;\;\frac{\sqrt{2}}{2}}_0 \frac{2x}{\sqrt{1\,-\,x^4}}\,dx\)

Let \(\displaystyle \,u\,=\,x^2\;\;\Rightarrow\;\;du\,=\,2x\,dx\)

The integral becomes: \(\displaystyle \L\:\int\frac{du}{\sqrt{1\,-\,u^2}}\) . . . arcsine!

 

[math]

thank you! we haven't done any inverse trig integrals yet. i know the derivatives, but how can i recognize when the integrals for each of the inverse trig functions?

How do you know what to make u equal to. In different types of examples we have set u equal to something whose derivative is a factor in the expression.

thanks again