1. ## trig substitution

Hi there any help showing the below integral equals the function +C?

integral= sqrt[(a^2-x^2]dx

function=(a^2/2)(theta+sin(theta)cos(theta))+c

using Substitution x=(a)sin(theta).

tried using

integral dx/sqrt[(a^2-x^2]=theta + C but can't seem to crunch it!

used dx=acos(theta).

any help appreciated.

2. ## Re: trig substitution

Originally Posted by Neverquit
Hi there any help showing the below integral equals the function +C?

integral= sqrt[(a^2-x^2]dx

function=(a^2/2)(theta+sin(theta)cos(theta))+c

using Substitution x=(a)sin(theta).

tried using

integral dx/sqrt[(a^2-x^2]=theta + C but can't seem to crunch it!

used dx=acos(theta).

any help appreciated.
\displaystyle \begin{align*} \int{ \frac{ dx }{ \sqrt{ a^2 - x^2 } } } \end{align*}

Make the substitution \displaystyle \begin{align*} x = a\sin{\theta} \implies dx = a\cos{\theta}\,d\theta \end{align*} and the integral comes

\displaystyle \begin{align*} \int{\frac{dx}{\sqrt{a^2 - x^2}}} &= \int{\frac{a\cos{\theta}\,d\theta}{\sqrt{a^2 - \left(a\sin{\theta}\right)^2}}} \\ &= \int{ \frac{ a \cos{\theta} \,d\theta }{ \sqrt{ a^2 - a^2\sin^2{\theta} } } } \\ &= \int{ \frac{a\cos{\theta}}{\sqrt{a^2\left(1 - \sin^2{\theta}\right)}} } \\ &= \int{\frac{a\cos{\theta}\,d\theta}{\sqrt{a^2\cos^2 {\theta}}}} \\ &= \int{\frac{a\cos{\theta}\,d\theta}{\pm a\cos{\theta}}} \\ &= \int{\pm 1\,d\theta} \\ &= \pm \theta + C \\ &= \pm \arcsin{\left(\frac{x}{a}\right)} + C \end{align*}