# integration

• Mar 7th 2011, 10:42 PM
Ilsa
integration
∫Sin2x/Sin^4x+cos^4x
can u pls help me
• Mar 7th 2011, 11:36 PM
Prove It
This is almost undreadable... is it $\displaystyle \displaystyle \int{\frac{\sin{2x}}{\sin^4{x} + \cos^4{x}}\,dx}$ or $\displaystyle \displaystyle \int{\frac{\sin{2x}}{\sin^4{x}} + \cos^4{x}\,dx}$?
• Mar 8th 2011, 02:08 AM
HallsofIvy
Undreadable? We shouldn't dread it then? I will assume that you mean
$\displaystyle \int \frac{sin(2x)}{sin^4(x)+ cos^4(x)}dx$
since the other would be almost trivial.

Use trig identities:
$\displaystyle sin^2(x)= \frac{1}{2}(1- cos(2x))$ and $\displaystyle cos^2(x)= \frac{1}{2}(1+ cos(2x))$

$\displaystyle sin^4(x)= \left(\frac{1}{2}(1- cos(2x))\right)^2= \frac{1}{4}(1- 2cos(2x)+ cos^2(2x)$
$\displaystyle cos^4(x)= \left(\frac{1}{2}(1+ cos(2x))\right)^2= \frac{1}{4}(1+ 2cos(2x)+ cos^2(2x)$
$\displaystyle sin^4(x)+ cos^4(x)= \frac{1}{2}(1+ cos^2(2x))$
• Mar 8th 2011, 03:01 AM
Prove It
Quote:

Originally Posted by HallsofIvy
Undreadable? We shouldn't dread it then? I will assume that you mean
$\displaystyle \int \frac{sin(2x)}{sin^4(x)+ cos^4(x)}dx$
since the other would be almost trivial.

Use trig identities:
$\displaystyle sin^2(x)= \frac{1}{2}(1- cos(2x))$ and $\displaystyle cos^2(x)= \frac{1}{2}(1+ cos(2x))$

$\displaystyle sin^4(x)= \left(\frac{1}{2}(1- cos(2x))\right)^2= \frac{1}{4}(1- 2cos(2x)+ cos^2(2x)$
$\displaystyle cos^4(x)= \left(\frac{1}{2}(1+ cos(2x))\right)^2= \frac{1}{4}(1+ 2cos(2x)+ cos^2(2x)$
$\displaystyle sin^4(x)+ cos^4(x)= \frac{1}{2}(1+ cos^2(2x))$

Alternatively:

$\displaystyle \displaystyle \int{\frac{\sin{2x}}{\sin^4{x} + \cos^4{x}}\,dx} = \int{\frac{2\sin{x}\cos{x}}{\sin^4{x} + (1 - \sin^2{x})^2}}$

$\displaystyle \displaystyle = \int{\frac{2\sin{x}\cos{x}}{\sin^4{x} + 1 - 2\sin^2{x} + \sin^4{x}}\,dx}$

$\displaystyle \displaystyle = \int{\frac{2\sin{x}\cos{x}}{2\sin^4{x} - 2\sin^2{x} + 1}\,dx}$.

Now make the substitution $\displaystyle \displaystyle u = \sin^2{x} \implies du = 2\sin{x}\cos{x}\,dx$ and the integral becomes

$\displaystyle \displaystyle \int{\frac{1}{2u^2 - 2u + 1}\,du}$

$\displaystyle \displaystyle = \frac{1}{2}\int{\frac{1}{u^2 - u + \frac{1}{2}}\,du}$

$\displaystyle \displaystyle = \frac{1}{2}\int{\frac{1}{u^2 - u + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 + \frac{1}{2}}\,du}$

$\displaystyle \displaystyle = \frac{1}{2}\int{\frac{1}{\left(u - \frac{1}{2}\right)^2 + \frac{1}{4}}\,du}$

Now make the substitution $\displaystyle \displaystyle u - \frac{1}{2} = \frac{1}{2}\tan{\theta} \implies du = \frac{1}{2}\sec^2{\theta}\,d\theta$ and the integral becomes

$\displaystyle \displaystyle \frac{1}{2}\int{\frac{\frac{1}{2}\sec^2{\theta}}{\ left(\frac{1}{2}\tan{\theta}\right)^2 + \frac{1}{4}}\,d\theta}$

$\displaystyle \displaystyle = \frac{1}{4}\int{\frac{\sec^2{\theta}}{\frac{1}{4}\ tan^2{\theta} + \frac{1}{4}}\,d\theta}$

$\displaystyle \displaystyle = \int{\frac{\sec^2{\theta}}{\tan^2{\theta} + 1}\,d\theta}$

$\displaystyle \displaystyle = \int{\frac{\sec^2{\theta}}{\sec^2{\theta}}\,d\thet a}$

$\displaystyle \displaystyle = \int{1\,d\theta}$

$\displaystyle \displaystyle = \theta + C$

$\displaystyle \displaystyle = \arctan{(2u-1)} + C$

$\displaystyle \displaystyle = \arctan{(2\sin^2{x} - 1)} + C$.
• Mar 8th 2011, 07:59 AM
TheCoffeeMachine
It was calculated in here (through many different methods).
• Mar 9th 2011, 05:04 AM
Ilsa
Quote:

Originally Posted by Ilsa
∫Sin2x/Sin^4x+cos^4x
can u pls help me

thanku it was the first one by the way. thank u for ur help
• Mar 9th 2011, 05:07 AM
Ilsa
thanku mr fantastic it was the frst one by the way