Trig substitution

JJ007

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Nov 7, 2009
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27
 a2x2xdx\displaystyle \int \frac {\ \sqrt{a^2-x^2}}{x}dx

x=asinθ\displaystyle x=a sin \theta
a2x2=a2a2sin2θ=\displaystyle \sqrt{a^2-x^2}=\sqrt{a^2-a^2sin^2 \theta}=

Not sure how to do the rest.
 
JJ007 said:
 a2x2xdx\displaystyle \int \frac {\ \sqrt{a^2-x^2}}{x}dx

x=asinθ\displaystyle x=a sin \theta....................dx=acosθdθ\displaystyle dx = a cos\theta d\theta

a2x2=a2a2sin2θ=acos2θ\displaystyle \sqrt{a^2-x^2} = \sqrt{a^2-a^2sin^2 \theta} = a\cdot \sqrt{cos^2 \theta}

Not sure how to do the rest.
 
Subhotosh Khan said:
JJ007 said:
 a2x2xdx\displaystyle \int \frac {\ \sqrt{a^2-x^2}}{x}dx

x=asinθ\displaystyle x=a sin \theta....................dx=acosθdθ\displaystyle dx = a cos\theta d\theta

a2x2=a2a2sin2θ=acos2θ\displaystyle \sqrt{a^2-x^2} = \sqrt{a^2-a^2sin^2 \theta} = a\cdot \sqrt{cos^2 \theta}

 acosθacosθasinθdθ?\displaystyle \int \frac {\ a cos \theta * a cos\theta}{a sin\theta}d\theta?
 
JJ007 said:
Subhotosh Khan said:
JJ007 said:
 a2x2xdx\displaystyle \int \frac {\ \sqrt{a^2-x^2}}{x}dx

x=asinθ\displaystyle x=a sin \theta....................dx=acosθdθ\displaystyle dx = a cos\theta d\theta

a2x2=a2a2sin2θ=acos2θ\displaystyle \sqrt{a^2-x^2} = \sqrt{a^2-a^2sin^2 \theta} = a\cdot \sqrt{cos^2 \theta}

 acosθacosθasinθdθ?\displaystyle \int \frac {\ a cos \theta * a cos\theta}{a sin\theta}d\theta?

cos2θ=1sin2θ\displaystyle cos^2\theta = 1 - sin^2\theta

I don't think you are ready to do these types of problems.
 
Which part were you referring to?

 acosθacosθasinθdθ=a(1sin2θ)dθ=a(cosθθ)+C?\displaystyle \int \frac {\ a cos \theta * a cos\theta}{a sin\theta}d\theta= a \int (1 - sin^2\theta)d \theta= a(cos\theta-\theta)+C ?
 
JJ007 said:
Which part were you referring to?

 acosθacosθasinθdθ=a(1sinθsinθ)dθ=a[(cscθdθsinθdθ]\displaystyle \int \frac {\ a cos \theta * a cos\theta}{a sin\theta}d\theta= a \int (\frac{1}{sin\theta} - sin\theta)d \theta= a \left [\int (csc \theta d\theta - \int sin\theta d\theta \right ]
 
[a2x2]1/2xdx = alnx[a2x2]1/2+a+[a2x2]1/2+C\displaystyle \int \frac{[a^{2}-x^{2}]^{1/2}}{x}dx \ = \ aln\bigg|\frac{|x|}{|[a^{2}-x^{2}]^{1/2}+a|}\bigg|+[a^{2}-x^{2}]^{1/2}+C

Here is the solution. Have fun if you decide to do the grunt work in deriving it.\displaystyle Here \ is \ the \ solution. \ Have \ fun \ if \ you \ decide \ to \ do \ the \ grunt \ work \ in \ deriving \ it.
 
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