Evaluate this integral (1 Viewer)

[qqmore]

Is there a quick way to evaluate this:

Integrate from 0 to pi/2 [ 1 / ( (cos x)^2 + 2(sin x)^2 ) ]

otherwise the t formula will take 2 pages to write it all.....

 

[tommykins]

I'm guessing change the cos²x to 1-sin²x you get

int. 1/1-s²+2s² dx = 1/1+s² dx

that takes the form of an inverse tan

hence evaluate [atan(sinx)]pi/2->0

 

[lolokay]

divide it out by cos²x
so you get
sec²/(1+2tan²)
= atan(rt2 tanx)/rt2

(I looked at the answer on the integrator. you should probably do that when you want to find an easier way to do something)

 

[miche11e]

i woulda done it tommykins way...

 

[Yamiyo]

Tommykins' way only works if it's cosx/(cos^2(x)+2sin^2(x))
i.e. (du/dx)/(1+u^2)