solving integrals

[Guest]

Hi!
I can't solve such integrals.

f[tex]\frac{dx}{\sqrt{2-x^{2}}}[/tex]

f[tex]\frac{dx}{2x^{2}-6}[/tex]

Please help

[HallsofIvy]

Those are fairly standard integrals. You should be able to find them in any Calculus text.

The first comes from the fact that [tex]sin^2(t)+ cos^2(t)= 1[/tex] so that [tex]cos^2(t)= 1- sin^2(t)[/tex] and [tex]cos(t)= \sqrt{1- sin^2(t)}[/tex].

Compare that to [tex]\sqrt{2- x^2}= \sqrt{2}\sqrt{1- \left(\frac{x}{\sqrt{2}}\right)^2}[/tex].
Use the substitution [tex]\frac{x}{\sqrt{2}}= sin(t)[/tex]. Then [tex]\frac{1}{\sqrt{2}}dx= cos(t)dt[/tex] so [tex]dx= \sqrt{2}cos(t)dt[/tex] and [tex]\sqrt{1- \left(\frac{x}{\sqrt{2}}\right)^2}= \sqrt{1- sin^2(t)}= cos(t)[/tex]. The integral becomes [tex]2\int cos(t) dt[/tex].

For the second one use [tex]a^2- b^2= (a- b)(a+ b)[/tex] with [tex]a= \sqrt{2}x[/tex] and [tex]b= 6[/tex] to write the integral as [tex]\int \frac{dx}{(\sqrt{2}x- \sqrt{6})(\sqrt{2}x+ \sqrt{6})}[/tex].

Now use "partial fraction": Find A and B so that [tex]\frac{1}{(\sqrt{2}x- \sqrt{6})(\sqrt{2}x+ \sqrt{6})}=[/tex][tex]\frac{A}{\sqrt{2}x- \sqrt{6}}+ \frac{B}{\sqrt{2}x+ \sqrt{6}}[/tex].

[Guest]

From the formula: [tex]\int \frac{dx}{\sqrt{a^2-x^2}}=Arcsin \frac{x}{a}+C[/tex]
So we have, [tex]\int \frac{dx}{\sqrt{2-x^2}}=Arcsin \frac{x}{\sqrt{2}}+C[/tex]

For [tex]\int \frac{dx}{2x^{2}-6}[/tex], you can factor out 2, so you have,

[tex]\frac{1}{2} \int \frac{dx}{x^{2}-3} = \frac{1}{2} \int \frac{dx}{(x - \sqrt{3})(x+\sqrt{3})}[/tex]

Then use partial fractions:

[tex]= \frac{1}{2} \int \frac{dx}{(2 \sqrt{3})(x+\sqrt{3})} - \frac{1}{2} \int \frac{dx}{(2 \sqrt{3})(x - \sqrt{3})}[/tex]

[tex]=\frac{1}{4 \sqrt{3}} ( \ln \mid {x+ \sqrt{3}} \mid - \ln \mid {x - \sqrt{3}} \mid ) +C[/tex]

[tex]=\frac{1}{4 \sqrt{3}} \ln \mid \frac { {x+ \sqrt{3} } } { {x - \sqrt{3} } } \mid +C[/tex]

depending if its ok that there is a radical in the denominator..