Integrating Trigonometric Functions with Substitution

[darioe]

Homework Statement

Integrate at interval [0,T] (T and k are given real numbers) the

2. Relevant equation

$$_{0}^{T}\int \frac{sin(p)}{\sqrt{k+p}}\ dp$$

The Attempt at a Solution

$$\ Using\ substitution\ u\ =\ tan(p/2),\ results\ as\ :\ p\ =\ 2*arctan(u)\ \ ;\ \ dp\ =\ \frac{2}{1+u^2}\ du\ ;\$$

$$sin(p)\ =\ \frac{2*u}{1+u^2} ;\ cos(p)\ =\ \frac{1-u^2}{1+u^2} ;\$$

$$_{0}^{T}\int \frac{sin(p)}{\sqrt{k+p}}\ dp \ \ =\ _{0}^{2*arctan(T)}\int \frac{2*u*2}{(1+u^2)\ *\ \sqrt{k+2*arctan(u)}\ *\ (1+u^2)}\ du$$

$$\ ¿\ Could\ someone\ get\ a\ better\ result\ ?$$

(maybe with the substitution u = 2* sin(p) )


...

 

[Count Iblis]

See e.g. 16 http://mathworld.wolfram.com/FresnelIntegrals.html"

 

[darioe]

Should be:

$$_{0}^{T}\int \frac{sin(p)}{\sqrt{k+p}}\ dp \ \ =\ _{0}^{tan(T/2)}\int \frac{2*u*2}{(1+u^2)\ *\ \sqrt{k+2*arctan(u)}\ *\ (1+u^2)}\ du$$

but it looks like I could have to know about Fresnel Integrals. Thank you for the help.


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