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Trig integration

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Remember that integration is the inverse procedure to differentiation. So, if you can do trigonometric differentiation, you can do trig integration. (You can brush up on your trig differentiation in the calculus section of this site.)

You should know that {d \over {dx}}\left( {\sin x} \right)= \cos x

so it follows that\int {\cos xdx}= \sin x + C

And because {d \over {dx}}\left( {\cos x} \right)= - \sin x

so \int {\sin xdx}= - \cos x + C

Similarly, differentiating functions of (ax+b), by applying the chain rule, leads to integrals of such functions.

Because {d \over {dx}}\left\( {\sin \left( {ax + b} \right)} \right\)= \cos \left( {ax + b} \right)a

so \int {\cos \left( {ax + b} \right)dx}= {1 \over a}\sin \left( {ax + b} \right) + C

And since {d \over {dx}}\left\( {\cos \left( {ax + b} \right)} \right\)= - \sin \left( {ax + b} \right)a

so \int {\sin \left( {ax + b} \right)dx}= - {1 \over a}\cos \left( {ax + b} \right) + C

For example,

\int {\sin \left( {3x + 1} \right)dx}= - {1 \over 3}\cos \left( {3x + 1} \right) + C

Look at the following example

\eqalign{ && \int {\cos \left( {{1 \over 2}x} \right)dx} \cr && = \sin \left( {{1 \over 2}x} \right) \times {1 \over {{1 \over 2}}} + C \cr && = 2\sin \left( {{1 \over 2}x} \right) + C \cr }

You should know by heart the integrals of sin(ax + b) and cos(ax + b).

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