integrate without complex factorisation

• January 3rd 2011, 01:18 AM
integrate without complex factorisation
integrate
1/[1+sin(x)+sin^2 (x)]
• January 3rd 2011, 02:21 AM
Chris L T521
Quote:

Originally Posted by ayushdadhwal
integrate
1/[1+sin(x)+sin^2 (x)]

Observe that $\sin^2 x+\sin x+1 = \left(\sin x+\tfrac{1}{2}\right)^2+\tfrac{3}{4}$.

Also observe that $\displaystyle\int\frac{\,dx}{\left(\sin x+\tfrac{1}{2}\right)^2+\tfrac{3}{4}} \sim \int\frac{\,du}{u^2+a^2}$

Can you figure out what substitution is required to complete the integration?
• January 3rd 2011, 05:24 AM
chisigma
Quote:

Originally Posted by ayushdadhwal
integrate
1/[1+sin(x)+sin^2 (x)]

If You have an integral containing only circular functions the substitution $t= \tan \frac{x}{2}$ ever conducts to an integral of a rational function. In particular...

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

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

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

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

In the case You have proposed the trasformation is...

$\displaystyle \int \frac{dx}{1+ \sin x + \sin^{2} x} \implies \int \frac{1+t^{2}}{1+2 t+ 6 t^{2} + 2 t^{3} + t^{4}}\ dt$

Kind regards

$\chi$ $\sigma$
• January 3rd 2011, 01:27 PM
mr fantastic
Quote:

Originally Posted by Chris L T521
Observe that $\sin^2 x+\sin x+1 = \left(\sin x+\tfrac{1}{2}\right)^2+\tfrac{3}{4}$.

Also observe that $\displaystyle\int\frac{\,dx}{\left(\sin x+\tfrac{1}{2}\right)^2+\tfrac{3}{4}} \sim \int\frac{\,du}{u^2+a^2}$

Can you figure out what substitution is required to complete the integration?

Nice idea but it won't fly. $dx \neq du$ ....
• January 3rd 2011, 02:35 PM
Plato
• January 5th 2011, 05:24 AM
simplependulum
Sub. $\displaystyle \tan(x/2) = \frac{1-t}{1+t}$ or

$\displaystyle x = \frac{\pi}{2} - 2\tan^{-1}(t)$

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

At the same time

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

We have , the integral

$\displaystyle = - 2 \int \frac{1+t^2}{ (1-t^2)^2 + (1-t^2)(1+t^2) + (1+t^2)^2 }~dt{$

$\displaystyle = -2 \int \frac{t^2+1}{t^4 + 3 }~dt$

$\displaystyle = -(1 + \frac{1}{\sqrt{3} } ) \int \frac{t^2 + \sqrt{3}}{t^4 + 3 }~dt - ( 1 - \frac{1}{\sqrt{3}} ) \int \frac{t^2 - \sqrt{3}}{t^4 + 3 }~dt$

Consider

$\displaystyle \int \frac{t^2 + \sqrt{3}}{t^4 + 3 }~dt = \int \frac{ 1 + \sqrt{3}/t^2 }{ ( t - \sqrt{3}/t )^2+ 2 \sqrt{3} } ~dt$

Sub. $\displaystyle t - \sqrt{3}/t = u$

$\displaystyle (1 + \sqrt{3}/t^2 )dt = du$

The integral

$\displaystyle = \int \frac{du}{u^2 + 2\sqrt{3} } = \frac{1}{\sqrt[4]{12}}\tan^{-1}( \frac{u}{\sqrt[4]{12}}) + C$

How about the second integral ?

$\displaystyle \int \frac{t^2 - \sqrt{3}}{t^4 + 3 }~dt$
• January 6th 2011, 04:03 AM
TheCoffeeMachine
Quote:

Originally Posted by simplependulum
How about the second integral ?

$\displaystyle \int \frac{t^2 - \sqrt{3}}{t^4 + 3 }~dt$

\begin{aligned} & \displaystyle t^4+3 = (t^2)^2+(\sqrt{3})^2 = (t^2+\sqrt{3})^2-2\sqrt{3}t^2 = (t^2-\sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3})(t^2+\sqrt{2}\sq rt{\sqrt{3}}t+\sqrt{3}). \\& \therefore ~ I = \int\frac{t^2-\sqrt{3}}{t^4+3}\;{dt} = \int\frac{t^2-\sqrt{3}}{ (t^2-\sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3})(t^2+\sqrt{2}\sq rt{\sqrt{3}}t+\sqrt{3})}\;{dt}.\end{aligned}

$\displaystyle u := \frac{t^2-\sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3}}{t^2+\sqrt{2}\sq rt{\sqrt{3}}t+\sqrt{3}} \Rightarrow \frac{du}{dt} = \frac{2\sqrt{2}\sqrt{\sqrt{3}}\left(t^2-\sqrt{3}\right)}{(t^2+\sqrt{2}\sqrt{\sqrt{3}}t+\sq rt{3})^2} \Rightarrow dt = \frac{(t^2+\sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3})^2}{2 \sqrt{2}\sqrt{\sqrt{3}}\left(t^2-\sqrt{3}\right)}\;{du}.$

\begin{aligned} \therefore ~ I & = \int\frac{(t^2-\sqrt{3})(t^2+\sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3})^2 }{2\sqrt{2}\sqrt{\sqrt{3}}(t^2+\sqrt{2}\sqrt{\sqrt {3}}t+\sqrt{3})(t^2-\sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3})\left(t^2-\sqrt{3}\right)}\;{du} \\& = \frac{1}{2\sqrt{2}\sqrt{\sqrt{3}}}\int\frac{(t^2+\ sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3})}{(t^2-\sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3})}\;{du} = \frac{1}{2\sqrt{2}\sqrt{\sqrt{3}}}\int\frac{1}{u}\ ;{du} \\& = \frac{1}{2\sqrt{2}\sqrt{\sqrt{3}}}\ln{u}+k = \boxed{\frac{1}{2\sqrt{2}\sqrt{\sqrt{3}}}\ln{\frac {(t^2-\sqrt{2}\sqrt{\sqrt{3}}t+\sqrt{3})}{(t^2+\sqrt{2}\ sqrt{\sqrt{3}}t+\sqrt{3})}}+k}.\end{aligned}

I received an offer from one of my university choices while typing this up! (Sun)
• January 6th 2011, 06:28 AM
Krizalid
Quote:

Originally Posted by simplependulum
How about the second integral ?

$\displaystyle \int \frac{t^2 - \sqrt{3}}{t^4 + 3 }~dt$

i'd write it as in the same fashion,

$\dfrac{1-\frac{\sqrt{3}}{{{t}^{2}}}}{{{\left( t+\frac{\sqrt{3}}{t} \right)}^{2}}-2\sqrt{3}},$

and substitution $t+\dfrac{\sqrt{3}}{t}=\sqrt{2}\sqrt[4]{3}u$ will solve the problem.