computerages Posted December 14, 2006 Posted December 14, 2006 Hey everyone, I am trying to solve the following indefinite integral: [math]\int \frac{dx}{x\sqrt{4x^2-9}} [/math] And I have got 2Arcsecx+c... but I am afraid it could be wrong... can anyone please check it...
ajb Posted December 14, 2006 Posted December 14, 2006 I simply plugged it into mathematica and got [math]{\frac{-\arctan ({\frac{3} {{\sqrt{-9 + 4\,{x^2}}}}})}{ 3}}[/math] Maybe you can spot this by differentianting arctan.
Ragib Posted December 18, 2006 Posted December 18, 2006 [math]\int \frac {1}{x} \frac {1}{\sqrt {(2x)^2 - 3^2}} dx[/math] Integration by Parts: [math]\int u dv = uv - \int v du[/math] Choose u by LIATE, Logs, Inverse Trigs, Algebraic, Trigs, Exponentials. So in this, we only have algebraic...I would prefer to integrate the [math]\frac {1}{\sqrt {(2x)^2 - 3^2}}[/math]. So lets make [math]dv = \frac {1}{\sqrt {(2x)^2 - 3^2}} dx[/math] and [math]u= \frac {1}{x}[/math]. Now just use a lil trig sub for dv and your home free, Good Luck. O yea, Have you learned trig sub, maybe not, because if you had you would recognise that your integrals form is of sec substitution, which will always get your integral an arctan somewhere in it. Need help ask.
ssd Posted February 3, 2007 Posted February 3, 2007 Substitute, x= 1/z. Then, apart form the constants, the integral becomes of the form 1/sqrt(4/9-z^2). The final answer should be (1/3).arcCos{3/(2x)} +C [or, -(1/3).arcSin{3/(2x)} +C].
K!! Posted January 2, 2008 Posted January 2, 2008 Hey everyone, I am trying to solve the following indefinite integral:[math]\int \frac{dx}{x\sqrt{4x^2-9}} [/math] The faster way to solve this, it's by settin' [math]u^2=4x^2-9[/math]. The rest follows.
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