inverse trig integration

[paulxzt]

the integral of arcsinx dx from 0 to 1/2

do i use integration by parts? u = sinx what do i use for dV?

also..

2) integral of ( 1 / t^3 * sqrt[t^2-1])) dt from sqrt(2) to 2
u = t^2 -1 , t = sqrt(u+1)

du/dt = 2t
du = 2tdt

rewritten: integral of (1 / (sqrt(u+1))^3 (sqrt(u))]
am i doing this right?

 

[galactus]

Yes, parts is a good way.

Let \(\displaystyle \L\\u=sin^{-1}(x), \;\ dv=dx, \;\ v=x, \;\ du=\frac{1}{\sqrt{1-x^{2}}}dx\)

This leads to:

\(\displaystyle \L\\xsin^{-1}(x)-\int\frac{x}{\sqrt{1-x^{2}}}dx\)

Can you finish?.


For the second one, trig sub isn't bad.

\(\displaystyle \L\\\int\frac{1}{t^{3}\sqrt{t^{2}-1}}dt\)

Let \(\displaystyle \L\\t=sec{\theta}, \;\ dt=sec{\theta}tan{\theta}d{\theta}\)

Make the subs and it whittles down to:

\(\displaystyle \L\\\int{cos^{2}{\theta}d{\theta}\)

 

[paulxzt]

Yes, parts is a good way.

Let \(\displaystyle \L\\u=sin^{-1}(x), \;\ dv=dx, \;\ v=x, \;\ du=\frac{1}{\sqrt{1-x^{2}}}dx\)

This leads to:

\(\displaystyle \L\\xsin^{-1}(x)-\int\frac{x}{\sqrt{1-x^{2}}}dx\)

Can you finish?.


For the second one, trig sub isn't bad.

\(\displaystyle \L\\\int\frac{1}{t^{3}\sqrt{t^{2}-1}}dt\)

Let \(\displaystyle \L\\t=sec{\theta}, \;\ dt=sec{\theta}tan{\theta}d{\theta}\)

Make the subs and it whittles down to:

\(\displaystyle \L\\\int{cos^{2}{\theta}d{\theta}\)

for the 2nd one, you can just randomly let t = sec (-) ?

 

[galactus]

No, not randomly. Are you familiar with trig substitution?.

After we make the subs, we get \(\displaystyle \sqrt{sec^{2}{\theta}-1}\).

Which simplifies to \(\displaystyle tan{\theta}\)

 

[paulxzt]

No, not randomly. Are you familiar with trig substitution?.

After we make the subs, we get \(\displaystyle \sqrt{sec^{2}{\theta}-1}\).

Which simplifies to \(\displaystyle tan{\theta}\)

I don't get where you can get any sort of trig functions with the original that i wrote.. could you explain where you can bring in the trig substitutions when there's only the variable t in the orig. problem

 

[galactus]

You don't see how?. Make the substitutions I gave you and then you'll see how.

Do you have a calc book?. If so, go to the trigonometric substitution section.

If you have an integral of the following forms, you can make the given subs:

\(\displaystyle \L\\\sqrt{a^{2}-x^{2}}, \;\ use \;\ x=a\cdot{sin{\theta}}\)

\(\displaystyle \L\\\sqrt{a^{2}+x^{2}}, \;\ use \;\ x=a\cdot{tan{\theta}}\)

\(\displaystyle \L\\\sqrt{x^{2}-a^{2}}, \;\ use \;\ x=a\cdot{sec{\theta}}\)


The latter is your form.