# Techniques of Integration

# Just Knowing

Sometimes the integrand is a derivative that is known. There is nothing wrong with taking advantage of this situation. The answer can be verified by taking the derivative if there is any doubt. For example:

{$$ \Large \int \cos\theta d\theta $$}

It is known {$ \cos\theta $} is the derivative of {$ \sin\theta $} and therefore the antiderivative of {$\cos\theta$} is {$\sin\theta$}.

{$$ \Large \int \cos\theta d\theta = \sin\theta + C $$}

*Just knowing* means recognizing that an integrand is exactly a known derivative. When the integrand is very similar to a known derivative, apparently differing only by a constant factor or something nearly as simple, *advanced guessing* may be appropriate.

# Advanced Guessing

Sometimes the integrand is very like a known derivative.

The technique of *advanced guessing* may work when the integrand is nearly or substantially the same as a known derivative. For example:

{$$ \Large \int \sin\theta d\theta $$}

Of course the difference between knowing and guessing is often how many derivatives are known. In this case, the derivative of {$ \cos\theta $} is taken to be known to be {$ -\sin\theta $}. It might be necessary to guess that {$ \cos\theta $} is closely related to the wanted antiderivative.

From

{$$ \Large {d \over {dx}}\cos\theta = -\sin\theta $$}

the guess is that that the antiderivative of {$ \sin\theta $} might be {$ - \cos\theta $}. While this example is more obvious fact than guess, careless errors can be avoided by taking the derivative of the guess:

{$$ \Large {d \over {dx}}-\cos\theta = -(-1)\sin\theta = \sin\theta $$}

Therefore:

{$$ \Large \int \sin\theta d\theta = -\cos\theta + C $$}

# Manipulation

All that is meant by manipulation is changing the form of the integrand to one that is recognized or otherwise much easier to integrate. In many cases this means using algebra or trigonometry on the integrand.

# Substitution

# Partial Fractions

# Integration by Parts

{$$ \Large \int uv' dx = uv - \int u'v dx + C $$} {$$ \Large \int_a^b uv' dx = uv\vert_a^b - \int_a^b u'v dx $$}

This is easy to see from:

{$$ \large \eqalign{ (uv)' &= u'v + uv' \\ uv' &= (uv)' - u'v \\ \int uv' dx &= uv - \int u'v dx } $$}

*Sources:*

- Love, Clyde E., Earl David Rainville
*Differential and Integral Calculus*(Macmillian, 1916) Google Books - Lecture 30: Integration by parts, reduction formulae Instructor: David Jerison

*Recommended:*

** Categories:** IntegralCalculusTopics

This is a student's notebook. I am not responsible if you copy it for homework, and it turns out to be wrong.

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