trig substitution: integral((1)/(x^2+3)^(3/2))dx

[saiyanmx89]

integral((1)/(x^2+3)^(3/2))dx
x = sqrt(3)tan(theta)
dx = sqrt(3)sec^2(theta)

= integral((1)/(3sec^2(theta))^(3/2)) * sqrt(3)sec^2(theta)d(theta)
= (1/3)integral(cos^2(theta)d(theta)
= (1/3)integral(1+cos2(theta))d(theta)
= (1/3)((theta) + (1/2)sin(2(theta)) + C
= (1/3)(theta) + (1/6)sin(theta)cos(theta) + C

I don't know know where to go from here. What do I plug in for sin(theta) and cos(theta)????

 

[galactus]

Re: trig substitution help

\(\displaystyle \int \frac{1}{(x^{2}+3)^{\frac{3}{2}}}dx\)


Solve your original sub for theta and sub back in to get it back in terms of x.

That is, sub back in \(\displaystyle {\theta}=tan^{-1}(\frac{x}{\sqrt{3}})\)

It should whittle down nicely. You have the correct sub, but I think you may have erred somewhere.

\(\displaystyle x=\sqrt{3}tan(t), \;\ dx=\sqrt{3}sec^{2}(t)dt\)

Then we get:

\(\displaystyle \frac{1}{3}\int cos(t)dt\)

Then, resub as I showed in the beginning.