# Indefinite Integral

• Jun 12th 2009, 11:42 PM
simplependulum
Indefinite Integral
a)

solve $\int \frac{\sin{2x}}{ \sin^4{x} + \cos^4{x} } ~dx$

by substituting

i) $u = \tan^2{x}$

ii) $u = \cos{2x}$

b)

Can we deduce that $\tan^{-1}({\cos{2x}}) = - \tan^{-1}({\tan^2{x}}) ~~ ?$
• Jun 13th 2009, 12:04 AM
matheagle
I would let $u=\sin x$ or $u=\cos x$.

If $u=\sin x$, then $du=\cos x dx$

and our integral becomes

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

$=\int {2u du\over 2u^4 -2u^2 +1}$.

Next, let $w=u^2$ giving us $\int {dw\over 2w^2 -2w +1}$

$=\int {dw\over 2(w^2 -w) +1} =\int {dw\over 2(w-1/2)^2 +1/2}$.

From this you can get the arctangent function.
• Jun 13th 2009, 12:27 AM
Moo
Hello,

1. Substituting $u=\tan^2(x)$ is indeed more thorough.

$du=2 \cdot\frac{1}{\cos^2(x)} \cdot \frac{\sin(x)}{\cos(x)} ~dx$

Thus the integral is :

$\int\frac{2\sin(x)\cos(x)}{\sin^4(x)+\cos^4(x)} \cdot\frac{\cos^3(x)}{2\sin(x)} ~du=\int\frac{\cos^4(x)}{\sin^4(x)+\cos^4(x)} ~du=\int\frac{du}{1+u^2}$

2. Substitute $u=\cos(2x)$

$du=-2\sin(2x) ~dx$

Thus the integral is :

$\int\frac{\sin(2x)}{\sin^4(x)+\cos^4(x)}\cdot\frac {1}{-2\sin(2x)} ~du=-\frac 12\int\frac{du}{\sin^4(x)+\cos^4(x)}$

Now you have to transform $\sin^4(x)+\cos^4(x)$ with respect to u.

$\cos(2x)=\begin{cases} 1-2\sin^2(x) \\ 2\cos^2(x)-1 \end{cases} \Rightarrow \begin{cases} \sin^4(x)=\left(\frac{1-\cos(2x)}{2}\right)^2=\frac 14 (1-u)^2 \\ \cos^4(x)=\left(\frac{\cos(2x)+1}{2}\right)^2=\fra c 14 (1+u)^2 \end{cases}$

Thus $\sin^4(x)+\cos^4(x)=\frac 14 (2u^2+2)=\frac 12(u^2+1)$

And the integral is :
$-\int\frac{du}{u^2+1}$

And from all this stuff, you can answer question b) (which is not possible with baldeagle's method).
• Jun 13th 2009, 06:29 AM
Jester
One thing to add - be careful with your constants of integration. Clearly

$
\tan^{-1}({\cos{2x}}) = - \tan^{-1}({\tan^2{x}})$

can't be right - it's not true for $x = 0$.