Step-by-Step Guide to Evaluating a Tricky Definite Integral: sinx + 2cosx + 3

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1. Evaluate exactly (in terms of \pi) the definite integral \int^{\pi/2}_{-\-\pi/2} \frac{dx}{sinx + 2cosx +3}



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How do i do this? Step by step instructions if possible.


The Attempt at a Solution


I've tried to manipulate the integral but still don't get anything. I also set the denominator as u. but then i cannot substitute du.

Help??
 
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Substitute trig. objects with parametric formulae:
sinx=2t/(1+t^2), cosx=(1-t^2)/(1+t^2), dx=2dt/(1+t^2)
and your trig. integrale becomes the rational integral
2*Integ[-1,1](1/(t^2+2*t+5)dt = pi/4
 
That's clever. Took me a moment to figure out why dx= 2dt/(1+t^2).
 
awesome... thanks a lot guys.

We didn't spend that much time on that identity so it totally slipped my mind. Thanks for the reminder!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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