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Integration

Integration by Trigonometric Substitution 2

We assume that you are familiar with the material in integration by trigonometric substitution 1 and geometric representation of trigonometric functions.

Various methods are available for evaluating the integrals of rational functions (ratios of two polynomials). It has been discovered that a fractional expression involving trig functions can always be transformed into a rational function.

Let u = tan half x. We can represent this in the following diagram:

right angle triangle with angle half x, opposite side u, adjacent side 1  and hypotenuse square root of (1 plus u squared)

We deduce cos (half x) = 1 over  square root of (1 plus u squared) and sin (half x) = u over  square root of (1 plus u squared). Hence,

cos x = (1 - u squared) over (1 + u squared)

sin x = 2u over (1 + u squared)

Since all other trig functions can be written in terms of sin x and cos x all fractional expressions of trig functions can always be transformed into a rational function.

Example

Consider integral of reciprocal of (1 + cos x) dx.

Let u = tan half x then du = half (1 + u squared) dx and dx = 2 over (1 + u squared) du.

The integral simplifies to :

tan (half x) + c

Exercise

Try all 5 of these.

indefinite integral of dx =

Working:

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