Problems with integration substitution

[lippy]

The first problem I cannot figure out is;

Evaluate the following integral using the substitution 4u = 3x

of integral dx/(16+9x^2)

4u=3x
du = 3/4dx
dx = 4/3du

so integral [(4/3)du] / (16+12u^2) = [4/3 du] /4(3u^2+4) = 1/3u^2+4 du =
[ -u^-3 - something for the 4 ] = [ -3/4x^-3 ]

I know this is way off...

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Also

Convert the following integral into an integral involving Ø (i think that is theta?) using the substitution 3x = 4sinØ. Simplify the integrand

Integral from 2*sqrt(2)/3 to 2/3 x^3dx/sqrt(16-9x^2)

Uhm, I don't even know where to start on this?

Ø = arcsin 3/4x
dØ = dx / (3/4) sqrt (1-9/16x^2) <- chain rule w/ u = 3/4x
dx = 3/4 sqrt (1-9/16x^2)

from there, no clue? can anyone help?

 

[lippy]

Can i do this?

dØ = dx / (1-9/16x^2) * 16/16 = dx [16/16-9x^x] = 4dx / sqrt(16-9x^2)

dx = 1/4 dØ sqrt(16-9x^2)

so the integral becomes

1/4 (4/3 sinØ)^2 = 4/9 sin^2Ø

?!?!

then, what does simplifying the integrand mean? should i plug the original x value back in, or what? i'm probably waaaaaaaaaaay off

 

[galactus]

For #1, let \(\displaystyle \L\\x=\frac{4}{3}tan(u); dx=\frac{4}{3}sec^{u}du\)

That gives you: \(\displaystyle \L\\\frac{\frac{4}{3}sec^{2}(u)du}{16+9(\frac{4}{3}tan(u))^{2}}\)

=\(\displaystyle \L\\\frac{\frac{4}{3}sec^{2}(u)du}{16+9(\frac{16}{9}tan^{2}(u))}\)

Now, can you finish?. Remember \(\displaystyle 1+tan^{2}(u)=sec^{2}(u)\)

It whittles down to something very easy. Don't forget to re-sub

 

[skeeter]

\(\displaystyle \frac{1}{16+9x^2}\,=\)

\(\displaystyle \frac{1}{16(1 + \frac{9x^2}{16})}\,=\)

\(\displaystyle \frac{1}{16}\,\frac{1}{1+(\frac{3x}{4})^2}\,=\)

\(\displaystyle \frac{1}{12}\,\frac{\frac{3}{4}}{1+(\frac{3x}{4})^2\)

integrate the last expression ...

\(\displaystyle \frac{1}{12} arctan(\frac{3x}{4}) + C\)

 

[galactus]

See there?. There's more than one way to skin an integral. Try it both ways.