Integration

[tapout1829]

How would you integrate this?

. . . .Integral Ln (SQRT(x^2-1)) dx

And:

. . . .Integrate

. . . .Integral ((x^3) / (SQRT (4 +x^2)) dx

. . . .using trig subsitution with x = 2 tan(theta)

And:

. . . .Integrate the above, but

. . . .Use integration by parts with

. . . .u = x^2 and dv = ((x / SQRT(4 +x^2))

I'm having difficulties with both of problems :x Any help with be greatly appreciated.

 

[soroban]

Hello, tapout1829!

\(\displaystyle \L\;I\;=\;\int \ln\sqrt{x^2-1}\,dx\)

I would do it "by parts".

Note that: \(\displaystyle \,\ln\sqrt{x^2\,-\,1}\;=\;\ln(x^2\,-\,1)^{\frac{1}{2}} \;=\;\frac{1}{2}\cdot\ln(x^2\,-\,1)\)

Let \(\displaystyle u\;=\;\frac{1}{2}\ln(x^2\,-\,1)\) . . . . . . . . . . . . . \(\displaystyle dv\,=\,dx\)

Then: \(\displaystyle \,du\:=\:\frac{1}{2}\cdot\frac{2x}{x^2\,-\,1}\,dx\:=\:\frac{x\,dx}{x^2\,-\,1}\). . . . \(\displaystyle v\,=\,x\)

Then: \(\displaystyle \L\:I\;=\;\frac{1}{2}x\cdot\ln(x^2\,-\,1)\,-\,\int\underbrace{\frac{x^2}{x^2\,-\,1}}\,dx\)

. . . . . \(\displaystyle \L I\;=\;\frac{1}{2}x\cdot\ln(x^2\,-\,1)\,-\,\int\left(\overbrace{1\,+\,\frac{1}{x^2\,-\,1}\right)\,dx\;\;\) . . "division"

. . . . . \(\displaystyle \L I\;=\;\frac{1}{2}x\cdot\ln(x^2\,-\,1)\,-\,\int 1\,dx\,-\,\int\frac{dx}{x^2\,-\,1}\)

. . . . . \(\displaystyle \L I\;=\;\frac{1}{2}x\cdot\ln(x^2\,-\,1)\;-\;x\;-\;\frac{1}{2}\ln\left|\frac{x\,-\,1}{x\,+\,1}\right|\,+\,C\)
*

 

[ChaoticLlama]

I'll start you off on the trig substitution.

\(\displaystyle \L\int {\frac{{x^3 dx}}{{\sqrt {x^2 + 4} }}}\)

Whenever you recognize a trig substitution question, making an associated triangle is crucial. Your triangle is of course right-angled. Label one angel as \(\displaystyle \L\theta\), its adjacent side as 2, it's opposite side as x, and the hypotaneuse is therefore \(\displaystyle {\sqrt {x^2 + 4} }\). Look at why you are doing a tan trig substitution and the triangle you have constructed.

Now make your substitutions:

Let \(\displaystyle \L\ x=2tan(\theta)\)

Therefore...
\(\displaystyle \L\
x = 2\tan (\theta ) \\
x^2 = 4\tan ^2 (\theta ) \\
x^2 + 4 = 4\tan ^2 (\theta ) + 4 \\
x^2 + 4 = 4(\tan ^2 (\theta ) + 1) \\
x^2 + 4 = 4\sec ^2 (\theta ) \\
\sqrt {x^2 + 4} = 2\sec (\theta ) \\\)

Additionally...
\(\displaystyle \L\
x^3 = 8\tan ^3 (\theta ) \\
dx = 2\sec ^2 (\theta ) \\\)

And now your integral should look like:

\(\displaystyle \L\8\int {\tan ^3 (\theta )} \sec (\theta )d\theta\)

Note: After you have integrated, the triangle that you constructed will come into use again.