Integrating a Definite Integral with Trigonometric Functions

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The discussion focuses on solving the definite integral ∫dt/(t² + 2t cos a + 1) with limits from 0 to 1, where 0 < a < π. The initial substitution t = sin a and subsequent transformation led to the integral being expressed as ∫cos a da/(sin² a + sin 2a + 1). A participant pointed out the incorrect handling of the cosec a term, which should remain within the integral. The conversation concluded with a suggestion to use the substitution t = cos(a) for simplification, which ultimately helped one participant solve the integral.

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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
 

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