Integrating a Definite Integral with Trigonometric Functions

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Homework Statement


∫dt/(t^2 +2tcos a + 1)
(Limits of the integral are from 0 to 1)
(0<a<π)

Homework Equations

The Attempt at a Solution


Put t=sin a
dt=cosa da
∫dt/(t^2 +2tcos a + 1) = ∫cos a da/(sin^2 a + sin 2a + 1) [ limits of integration changed to 0 to π/2]
= ((cosec a)/2) ∫sin 2a da/(sin^2 a + sin 2a + 1)

I couldn't figure out what to do next... Please help!
 
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utkarsh009 said:

Homework Statement


∫dt/(t^2 +2tcos a + 1)
(Limits of the integral are from 0 to 1)
(0<a<π)

Homework Equations

The Attempt at a Solution


Put t=sin a
dt=cosa da
∫dt/(t^2 +2tcos a + 1) = ∫cos a da/(sin^2 a + sin 2a + 1) [ limits of integration changed to 0 to π/2]
= ((cosec a)/2) ∫sin 2a da/(sin^2 a + sin 2a + 1)
I don't know what you did in this last step, but you can't bring csc(a) out as if it were a constant.
utkarsh009 said:
I couldn't figure out what to do next... Please help!
 
Oh yes... By mistake i typed the integration symbol after it. Cosec a should be inside the integral. So, how should i proceed??
 
I would try a different substitution -- let t = cos(a).That should get you an integral that's easier to do.
 
Oh yah... Thanks ... I solved it... But have a look at this file... I couldn't get an answer for this...
New Doc 1_1.jpg
 
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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