Changes

The rule for '''integration by parts''' is stated as follows:

The rule for '''integration by parts''' is stated as follows:

:<big><math>\int f(x) g'(x)\,dx = f(x) g(x) - \int f'(x) g(x)\,dx,</math></big>

:<big><math>\int f(x) g'(x)\,dx = f(x) g(x) - \int f'(x) g(x)\,dx,</math></big>

This rule is often useful when one function is a power of ''x'' and the other function  is a trigonometric function or ''e'' raised to a power of ''x''.

This rule is often useful when one function is a power of ''x'' and the other function  is a trigonometric function or ''e'' raised to a power of ''x''.

We can check this by taking the derivative of :<big><math>ln|x^2+3|</math></big>,

We can check this by taking the derivative of :<big><math>ln|x^2+3|</math></big>,

:<big><math>\frac{d}{dx}ln|x^2+3|=(\frac{1}{x^2+3})(2x)=\frac{2x}{x^2+3}</math></big>

:<big><math>\frac{d}{dx}ln|x^2+3|=\left (\frac{1}{x^2+3} \right)(2x)=\frac{2x}{x^2+3}</math></big>

==Trigonometric Substitution==

==Trigonometric Substitution==

From our trigonometric expression :<big><math>x=3sin\theta</math></big> we can see that  

From our trigonometric expression :<big><math>x=3sin\theta</math></big> we can see that  

:<big><math>\theta=sin^{-1}(\frac{x}{3})+c</math></big> giving us the final solution.  

:<big><math>\theta=sin^{-1}\left(\frac{x}{3}\right)+c</math></big> giving us the final solution.  

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

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

== Other methods of integration ==

== Other methods of integration ==