Trigonometric substitutions

[flakine]

\(\displaystyle \L\\\int_{\2sqrt{2}}^{4}\frac{1}{t^{3}sqrt{t^{2}-4}\)

Please help

 

[tkhunny]

Trig? Maybe later. Try rewriting it, first. Note that everything is nice and smooth and continuous and positive and finite on the interval you have been given.

\(\displaystyle \L\,\int_{2\sqrt{2}}^{4}{\frac{t}{t^{4}\sqrt{t^{2}-4}}}\,dt\)

Perhaps, after a simpler substitution, maybe $\displaystyle u^{2} = t^{2} - 4$, it will all be more plain?

 

[galactus]

TKH's method is more efficient, but if you must use trig sub, try

\(\displaystyle \L\\t=2sec({\theta})\;\ and\;\ dt=2sec({\theta})tan({\theta})d{\theta}\)

 

[flakine]

Ok, what next? 1/2du=tdt

$\displaystyle \int\frac{1} {t^4U}du$

 

[soroban]

Hello, flakine!

\(\displaystyle \L \int_{_{2sqrt{2}}}^{\;\;\;^4}\frac{dt}{t^{3}sqrt{t^2-4}}\)

Let $\displaystyle t\,=\,2\sec\theta\;\;\Rightarrow\;\;dt\,=\,2\sec\theta\tan\theta\,d\theta$

Substitute: \(\displaystyle \L\,\int\frac{2\sec\theta\tan\theta\,d\theta}{(2\sec\theta)^3(2\tan\theta)} \;=\;\frac{1}{8}\int\frac{d\theta}{\sec^2\theta}\;=\;\frac{1}{8}\int\)\(\displaystyle \cos^2\theta\,d\theta \;= \;\frac{1}{16}\L\int\)$\displaystyle \left(1\,+\,\cos2\theta\right)\,d\theta$

Can you finish it now?

 

[galactus]

Ok, what next? 1/2du=tdt

$\displaystyle \int\frac{1} {t^4U}du$

It MUST be ENTIRELY in terms of 'u'.

 

[flakine]

Still need help, How do you convert it in respect to t.

 

[tkhunny]

Really strong hint.

\(\displaystyle \L\,u^{2} = t^{2} - 4\)

\(\displaystyle \L\,u^{2} + 4 = t^{2}\)

\(\displaystyle \L\,(u^{2} + 4)^{2} = (t^{2})^{2} = t^{4}\)

You can get to the trig substitution in a moment. This will just get you there.

 

[tkhunny]

Ok, what next? 1/2du=tdt

$\displaystyle \int\frac{1} {t^4U}du$

Whiile we're at it, you may wish to look at this piece again. It doesn't look quite right.

Maybe, 2*u*du = 2*t*dt