1. ## Interesting Integral

Compute
$\displaystyle \int_0^{\pi/2}\frac{\sin x}{9+16\sin(2x)}\,dx$

2. ## Re: Interesting Integral

Use the substitutions...

$\displaystyle t= \tan \frac{x}{2}$

$\displaystyle dx = 2\ \frac{dt}{1+t^{2}}$

$\displaystyle \sin x = \frac{2 t}{1+t^{2}}$

$\displaystyle \cos x = \frac{1-t^{2}}{1+t^{2}}$

... and take into account the identity...

$\displaystyle \sin 2x = 2\ \sin x\ \cos x$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. ## Re: Interesting Integral

I am not sure with that. The rational function will be very ugly ...

4. ## Re: Interesting Integral

Originally Posted by watchmath
I am not sure with that. The rational function will be very ugly ...
Not so bad, though, from a Calculus standpoint. Notice that the denominator is a biquadratic: $\displaystyle at^4 + bt^2 + c$. Use a substitution u = t^2 and the form will be much more recognizable.

-Dan

5. ## Re: Interesting Integral

I got
$\displaystyle \frac{4t}{9t^4-64t^3+18t^2+64t+9}$
and I would say that is ugly .

6. ## Re: Interesting Integral

Originally Posted by watchmath
I got
$\displaystyle \frac{4t}{9t^4-64t^3+18t^2+64t+9}$
and I would say that is ugly .
The bottom factorises to $\displaystyle \displaystyle (t^2 - 8t + 9)(9t^2 + 8t + 1)$, so the next step is partial fractions.

7. ## Re: Interesting Integral

Originally Posted by Prove It
The bottom factorises to $\displaystyle \displaystyle (t^2 - 8t + 9)(9t^2 + 8t + 1)$, so the next step is partial fractions.
How did you obtain this factorisation, if you don't mind me asking? I'm not seeing it; neither can I find a clever/painless way of solving the integral!

8. ## Re: Interesting Integral

Originally Posted by TheCoffeeMachine
How did you obtain this factorisation, if you don't mind me asking? I'm not seeing it; neither can I find a clever/painless way of solving the integral!
I used Wolfram Alpha, lol...

9. ## Re: Interesting Integral

Originally Posted by Prove It
I used Wolfram Alpha, lol...
lol!!