trigonometric substitution

[nickname]

My question about the following problem is: how does it go from ?(1/?4-4sin^2thetha) * 2cos theta dtheta

to ?(1/8costhetha)(8costheda dtheta)


?1/?(4-x^2) dx
x=2sin theta
dx= 2costheta dtheta

= ?1/?(4-4sin^2theta) * 2costheta dtheta

= ?(1/?(4) ?(1-sin^2theta)) * 2costheta dtheta

= ?(1/8cos theta * 8 costheta dtheta = = ? dtheta= theta + c = arc sin x/a + c ------> thetha= arcsin x/a +c = arcsin x/2 + c


Thank you for your help!

 

[BigGlenntheHeavy]

$\displaystyle \int\frac{dx}{\sqrt{4-x^{2}}}, \ Let \ x \ = \ 2sin(\theta), \ then \ dx \ = \ 2cos(\theta)d\theta$

\(\displaystyle Hence, \ \int\frac{2cos(\theta)d\theta}{\sqrt{4-4sin^{2}(\theta)}} \ = \ \int\frac{2cos(\theta)d\theta}{2\sqrt(1-sin^{2}(\theta)}}\)

$\displaystyle = \ \int\frac{2cos(\theta)d\theta}{2cos(\theta)} \ = \ \int d\theta \ = \ \theta+C, \ 2sin(\theta) \ = \ x, \ sin(\theta) \ = \ \frac{x}{2}, \ \theta \ = \ arcsin(x/2)$

$\displaystyle Ergo, \ \int\frac{dx}{\sqrt{4-x^{2}}} \ = \ arcsin(x/2) \ + \ C.$

$\displaystyle Check: \ D_x[arcsin(x/2)+C] \ = \ \frac{1}{\sqrt{4-x^{2}}}$

 

[nickname]

Thank you so much for your help! i understand it better now.

Thanks again!

 

[Aladdin]

$\displaystyle An \ easier \ way \ for \ \int\frac{dx}{\sqrt{4-x^{2}}}$

$\displaystyle \int\frac{dx}{2\sqrt{1-\frac{x^{2}}{4}}}$ . . . $\displaystyle Let \ u=\frac{x}{2} ---. du=\frac{dx}{2}$

$\displaystyle \int\frac{2du}{2\sqrt{1-u^2}}$ ----- $\displaystyle Arcsin(u) +C$ which equals : $\displaystyle Arcsin(x/2) +C$

:wink: