Trig substitution

[JJ007]

\(\displaystyle \int \frac {\ \sqrt{a^2-x^2}}{x}dx\)

\(\displaystyle x=a sin \theta\)
\(\displaystyle \sqrt{a^2-x^2}=\sqrt{a^2-a^2sin^2 \theta}=\)

Not sure how to do the rest.

 

[Deleted member 4993]

\(\displaystyle \int \frac {\ \sqrt{a^2-x^2}}{x}dx\)

\(\displaystyle x=a sin \theta\)....................\(\displaystyle dx = a cos\theta d\theta\)

\(\displaystyle \sqrt{a^2-x^2} = \sqrt{a^2-a^2sin^2 \theta} = a\cdot \sqrt{cos^2 \theta}\)

Not sure how to do the rest.

 

[JJ007]

\(\displaystyle \int \frac {\ \sqrt{a^2-x^2}}{x}dx\)

\(\displaystyle x=a sin \theta\)....................\(\displaystyle dx = a cos\theta d\theta\)

\(\displaystyle \sqrt{a^2-x^2} = \sqrt{a^2-a^2sin^2 \theta} = a\cdot \sqrt{cos^2 \theta}\)

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

 

[Deleted member 4993]

\(\displaystyle \int \frac {\ \sqrt{a^2-x^2}}{x}dx\)

\(\displaystyle x=a sin \theta\)....................\(\displaystyle dx = a cos\theta d\theta\)

\(\displaystyle \sqrt{a^2-x^2} = \sqrt{a^2-a^2sin^2 \theta} = a\cdot \sqrt{cos^2 \theta}\)

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

\(\displaystyle cos^2\theta = 1 - sin^2\theta\)

I don't think you are ready to do these types of problems.

 

[JJ007]

Which part were you referring to?

\(\displaystyle \int \frac {\ a cos \theta * a cos\theta}{a sin\theta}d\theta= a \int (1 - sin^2\theta)d \theta= a(cos\theta-\theta)+C ?\)

 

[Deleted member 4993]

Which part were you referring to?

\(\displaystyle \int \frac {\ a cos \theta * a cos\theta}{a sin\theta}d\theta= a \int (\frac{1}{sin\theta} - sin\theta)d \theta= a \left [\int (csc \theta d\theta - \int sin\theta d\theta \right ]\)

 

[BigGlenntheHeavy]

\(\displaystyle \int \frac{[a^{2}-x^{2}]^{1/2}}{x}dx \ = \ aln\bigg|\frac{|x|}{|[a^{2}-x^{2}]^{1/2}+a|}\bigg|+[a^{2}-x^{2}]^{1/2}+C\)

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