Integration By Trig Substitution

[neshepard]

1. Homework Statement
∫√(4-x^2)/x dx


2. Homework Equations



3. The Attempt at a Solution
a^2=4 u^2=x^2 ⇒ u=asinθ
a=2 u=x
x=2sinθ sinθ
2cosθ=√(4-x^2)
dx=2cosθ dθ

∫√(4-4sin^2θ)/2sinθ 2cosθ dθ
∫2cos^2θ/2sinθ 2cosθ dθ
2∫cos^2θ/sinθ cosθ dθ

Now what? I have been working on this problem for the last 3 3 hours and I am at a stand still. Can somebody Please get me to the answer?

 

[rock.freak667]

2∫cos^2θ/sinθ cosθ dθ

Now use the formula cos²θ+sin²θ=1 and replace cos²θ and then try using t=sinθ as a substitution.

 

[Dick]

I don't think you want to use a trig substitution. Try u^2=4-x^2. Then use partial fractions on the u integral.

 

[neshepard]

To do partial fractions, don't I need to divide the numerator by the denominator because the exponent is larger in the numerator? If so, how? I've never tried to do synthetic division on a sqrt.

 

[Dick]

To do partial fractions, don't I need to divide the numerator by the denominator because the exponent is larger in the numerator? If so, how? I've never tried to do synthetic division on a sqrt.

Maybe you missed that it's u^2=4-x^2, not u=4-x^2. There's no sqrt left after the u substitution.

 

[neshepard]

What I have now is:
=2∫cos^2θ/sinθ dθ
=2∫1-sin^2θ/sinθ dθ
=2∫1-sinθ dθ
=2(cosθ + θ)
But I can't resub because there is no value for θ!

 

[Dick]

What I have now is:
=2∫cos^2θ/sinθ dθ
=2∫1-sin^2θ/sinθ dθ
=2∫1-sinθ dθ
=2(cosθ + θ)
But I can't resub because there is no value for θ!

You are also making algebra mistakes. (1-sin(t)^2)/sin(t) isn't 1-sin(t). And to get things back in terms of x, you want to put theta=arcsin(x/2).

 

[sachinism]

I don't think you want to use a trig substitution. Try u^2=4-x^2. Then use partial fractions on the u integral.

well that's right

whenever u have anything under the square root , try removing hte square root first