# trig-substitution!

• Mar 26th 2013, 03:34 AM
lawochekel
trig-substitution!
i have this $\displaystyle \intk\sqrt{\frac{1-z}{z}}dz$

i set $\displaystyle z=sin^2\theta$
differentiating z w.r.t $\displaystyle \theta$, i have $\displaystyle dz=2sin\theta cos\theta d\theta$
substituting

$\displaystyle 2k\int\sqrt{\frac{1-sin^2\theta}{sin^2\theta}}sin\theta cos\theta d\theta$

i recall that $\displaystyle cos^2\theta=1-sin^2\theta$

i tend to be confuse from here onward, pls help!
• Mar 26th 2013, 04:09 AM
smatik
Re: trig-substitution!
you'll get cos^2thetha when you'll put 1-sin^2theta=cos^2theta. then you can expand cos^2theta as (cos2theta+1)/2...then integrate cos2theta/2 and 1/2 separately...
• Mar 29th 2013, 04:01 PM
x3bnm
Re: trig-substitution!
$\displaystyle \text{You know that } \int \sqrt{\frac{1-z}{z}}dz = 2\int\sqrt{\frac{1-\sin^{2}(\theta)}{\sin^{2}(\theta)}}\sin(\theta) \cos(\theta)\,\, d\theta$

$\displaystyle \text{ by letting } z = \sin^{2}{(\theta)} \text{ and so } dz = 2\sin{(\theta)}\cos{(\theta)}\,\,d\theta$

Now:

\displaystyle \begin{align*}2\int\sqrt{\frac{1-\sin^{2}(\theta)}{\sin^{2}(\theta)}}\sin(\theta) \cos(\theta)\,\, d\theta =& 2 \int \frac{\sqrt{(\cos^{2}{(\theta)})}}{\sqrt{(\sin^{2} {(\theta)})}} \sin(\theta) \cos(\theta)\,\, d\theta..........[\text{ because } 1 - \sin^{2}(\theta) = \cos^{2}(\theta)] \\ =& 2\int\frac{\cos(\theta)}{\sin(\theta)}\sin(\theta) \cos(\theta)\,\, d\theta \\ =& 2\int\cos^{2}(\theta)\,\, d\theta .....[\text{by using simple algebra}] \\ =& 2\int\frac{\cos(2\theta) + 1}{2}\,\, d\theta ......[\text{because }\cos(2\theta) = 2 \cos^{2}(\theta) - 1] \\ =& 2\left( \int \frac{\cos(2\theta)}{2}d\theta + \int \frac{1}{2}\,\,d\theta\right) \\ =& 2\left(\frac{\sin(2\theta)}{4} + \frac{\theta}{2}\right) + C \\ =& \frac{ \sin(2\theta)}{2} + \theta + C \\ =& \frac{2\sin(\theta)\cos(\theta)}{2} + \theta + C ......[\text{because } \sin(2\theta) = 2\sin(\theta)\cos(\theta)] \\ =& \sin(\theta) \cos(\theta) + \theta + C \end{align*}

By plugging into $\displaystyle \sin(\theta) = \sqrt{z}$, $\displaystyle \cos(\theta) = \sqrt{1 - z}$ and $\displaystyle \theta = \sin^{-1}(\sqrt{z})$ we get:

$\displaystyle \int k\sqrt{\frac{1-z}{z}}dz = \sqrt{z} \sqrt{1 - z} + \sin^{-1}(\sqrt{z}) + C$

And that's it. That's the answer. If you differentiate the new found answer with CAS(like maple) you'll get what you started with which is the given integral in the question. Hope this helps.