Changes

Given an integral  

Given an integral  

:<big><math>\int\frac{1}{\sqrt{9-x^2}}dx</math></big>

:<math>\int\frac{1}{\sqrt{9-x^2}}dx</math>

By looking at the radical we can determine that it represents the base of a right triangle by understanding the [[Pythagorean theorem]].

By looking at the radical we can determine that it represents the base of a right triangle by understanding the [[Pythagorean theorem]].

 

:<math>\sqrt{9-x^2}=3+x</math>

:<big><math>\sqrt{9-x^2}=3+x</math></big> where 3 is the hypotenuse and x is the height of the triangle.

where 3 is the hypotenuse and ''x'' is the height of the triangle.

This allows us to rewrite the expression to :<big><math>sin\theta=\frac{x}{3}</math></big>. This allows us to substitute x with :<big><math>3sin\theta</math></big>.   

This allows us to rewrite the expression to <math display="inline">\sin\theta=\frac{x}{3}</math>. This allows us to substitute ''x'' with <math display="inline">3\sin\theta</math>.   

Now to do the substitution

Now to do the substitution

:<big><math>9-x^2=9-(3sin\theta )^2</math></big>

:<math>9-x^2=9-(3\sin\theta )^2</math>

:<big><math>9-(3sin\theta )^2=9-9sin^2\theta </math></big>

:<math>9-(3\sin\theta )^2=9-9\sin^2\theta </math>

:<big><math>9-9sin^2\theta=9(1- sin^2\theta) </math></big> 

:<math>9-9\sin^2\theta=9(1- \sin^2\theta) </math>

And by use of trigonometric identities we know that  

And by use of trigonometric identities we know that  

:<big><math>1- sin^2\theta=cos^2\theta </math></big>

:<math>1- \sin^2\theta=\cos^2\theta </math>

:<big><math>9(1- sin^2\theta)=9cos^2\theta </math></big>  

:<math>9(1- \sin^2\theta)=9\cos^2\theta </math>

Therefore,

Therefore,

:<big><math>\sqrt{9-x^2}=3cos\theta </math></big>

:<math>\sqrt{9-x^2}=3\cos\theta </math>

We are not done yet, we must also take the derivative of :<big><math>3sin\theta</math></big>

We are not done yet, we must also take the derivative of <math display="inline">3\sin\theta</math>

:<big><math>\frac{dx}{d\theta}3sin\theta=3cos\theta </math></big>  

:<math>\frac{dx}{d\theta} 3\sin\theta=3\cos\theta </math>

By partial derivatives we move the :<big><math>{d\theta}</math></big> over.

By partial derivatives we move the <math display="inline">d\theta</math> over.

:<big><math>dx=3cos\theta d\theta </math></big>

:<math>dx=3\cos\theta d\theta </math>

Now we are ready to rewrite our integral.

Now we are ready to rewrite our integral.

:<big><math>\int\frac{1}{\sqrt{9-x^2}}dx=\int\frac{3cos\theta}{3cos\theta}d\theta=\int d\theta=\theta+c</math></big>

:<math>\int\frac{1}{\sqrt{9-x^2}}dx=\int\frac{3\cos\theta}{3\cos\theta}d\theta=\int d\theta=\theta+c</math>

 

:<big><math>\int\frac{1}{\sqrt{9-x^2}}dx=sin^{-1}\left(\frac{x}{3}\right)+c </math></big>

:<math>\theta=\sin^{-1}\left(\frac{x}{3}\right)+c</math>

:<math>\int\frac{1}{\sqrt{9-x^2}}dx=\sin^{-1}\left(\frac{x}{3}\right)+c </math>

== Other methods of integration ==

== Other methods of integration ==