integrate without complex factorisation

**ayushdadhwal** · January 3rd 2011, 01:18 AM

integrate
1/[1+sin(x)+sin^2 (x)]

**Chris L T521** · January 3rd 2011, 02:21 AM

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?

**chisigma** · January 3rd 2011, 05:24 AM

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

$\chi$ $\sigma$

**mr fantastic** · January 3rd 2011, 01:27 PM

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. $dx \neq du$ ....

**Plato** · January 3rd 2011, 02:35 PM

Take a look at this.

**simplependulum** · January 5th 2011, 05:24 AM

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 ?

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

**TheCoffeeMachine** · January 6th 2011, 04:03 AM

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!

**Krizalid** · January 6th 2011, 06:28 AM

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.

Thread: integrate without complex factorisation

LinkBack

Thread Tools

Display

integrate without complex factorisation

Similar Math Help Forum Discussions

estimate error of partial sum, terms too complex to integrate

complex numbers factorisation

More Complex Factorisation

Integrate using complex

Complex Factorisation

Search Tags