Integration which possibly involves partial fractions.

[kirakun]

Homework Statement

Well this is part of an integration process, namely:

\int \frac {sin^2x}{4+3cos^2x}dx

Homework Equations

My attempt involved using a u-substitution, namely t = tan x

The Attempt at a Solution

Using t = tan x, sin^2 x = \frac {t^2}{1+t^2} and cos^2 x = \frac {1}{1+t^2}

we have \int \frac {sin^2x}{4+3cos^2x}dx

which becomes

1. \int \frac {t^2}{(1+t^2)(4+ \frac {3}{1+t^2})} \times \frac {1}{1+t^2}

2. \int \frac {t^2}{(1+t^2)( \frac {4+4t^2+3}{1+t^2})} \times \frac {1}{1+t^2}

and finally yields

\int \frac {t^2}{(7+4t^2)(1+t^2)}dt

So i am basically stuck at this point and cannot proceed further.
Maybe i should have used another substitution or I started it wrongly.
I cannot seem to get to the solution though i suspect arctan might be present in the answer.
Please do guide me.
Thanks again.

 

[SteamKing]

It's not clear why you chose t = tan x as your substitution.

It's also not clear how you obtained the integral in t after substituting. You clearly wound up with a more complicated integral using the t substitution. You should show all the steps you took to make sure you haven't made any mistakes.

 

[kirakun]

It's not clear why you chose t = tan x as your substitution.

It's also not clear how you obtained the integral in t after substituting. You clearly wound up with a more complicated integral using the t substitution. You should show all the steps you took to make sure you haven't made any mistakes.

What method do u think would be best to tackle this one? well i chose tan x because with the denominator containing terms squared, i thought i would end up with an arctan case, but the t^2 in the numerator after simplifying makes it harder to achieve the form if not impossible.

Also when t = tan x, sin^2 x = \frac {t^2}{1+t^2} and cos^2 x = \frac {1}{1+t^2}

Upon substituting including the dx i ended up simplifying it like i showed.

I am still trying some other methods...

 

[haruspex]

\int \frac {t^2}{(7+4t^2)(1+t^2)}dt

Partial fractions, as you suggested. Did you try that?

 

[kirakun]

Partial fractions, as you suggested. Did you try that?

It will be of the form

\frac {At+B}{1+t^2} + \frac {Ct+D}{7+4t^2} right?

Consequently

For the numerator,

t^2 = (At+B)(7+4t^2) + (Ct+D)(1+t^2)

(4A+C)t^3 + (4B+D) t^2 + (7A+C)t + (7B+D) = t^2

Comparing coefficients of t^2 and comparing constants we get the 2 equations respectively;

4B + D = 1 and 7B+D = 0

which gives B = - 1/3 and D = 7/3
A and C being both 0

Thus the partial fraction is

\frac {7}{3(7+4t^2)} - \frac {1}{3(1+t^2)}

But on expanding it does not give t^2 in the numerator, am I doing something wrong?

 

[haruspex]

It will be of the form

\frac {At+B}{1+t^2} + \frac {Ct+D}{7+4t^2} right?
I'll be trying this.

Since your integrand fraction only has t² terms, no odd powers, the partial fraction expansion must have A = C = 0. (Imagine substituting u = t² first.)

 

[kirakun]

Since your integrand fraction only has t² terms, no odd powers, the partial fraction expansion must have A = C = 0. (Imagine substituting u = t² first.)

Alright. But it should give the same answer as in my edited post

 

[haruspex]

\frac {7}{3(7+4t^2)} - \frac {1}{3(1+t^2)}

But on expanding it does not give t^2 in the numerator, am I doing something wrong?

Check that again. It does for me.

 

[kirakun]

Check that again. It does for me.

Oh god yes, i forgot the 9 in the denominator. Well that partial fraction makes the integrand now easier.
I should complete the denominators to the square and use arctan integral?

 

[haruspex]

Oh god yes, i forgot the 9 in the denominator. Well that partial fraction makes the integrand now easier.
I should complete the denominators to the square and use arctan integral?

Use arctan, yes. No need to complete any squares.

 

[kirakun]

Use arctan, yes. No need to complete any squares.

Thank you!