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## Integration by Trigonometric Substitution I

We assume that you are familiar with the material in integration by substitution 1 and integration by substitution 2 and inverse trigonometric functions. This page will use three notations interchangeably, that is, arcsin z, asin z and sin-1z all mean the inverse of sin z

When an integrand contains x2 + k2 but there is no way to obtain an x for replacing x dx by du, we may be able to use the trig identity,

1 + tan2x = sec2x or more generally k2 + (ktan x)2 = (ksec x)2

When an integrand contains x2k2, we may be able to use the trig identity,

sin2x + cos2x = 1 that is k2 − (ksin x)2 = (kcos x)2

### Examples

Consider the integral .

Note that substituting g(x) = x2 + 1 by u will not work, as g '(x) = 2x is not a factor of the integrand.

Let us make the substitution x = tan θ then and dx = sec2 θ dθ.

The integral becomes Consider this integral Substitute x = sin θ then dx = cos θ dθ.

Solution of the integral becomes Now a little more complex example: In order to use the first identity, we need 4x2 = 9tan2p. Solving for x gives x = tan p.

Hence dx = sec2p dp and, rearranging again, p = arctan( ). Substituting, simplifying, integrating and re-substituting gives: ### Exercise

Try some of these. dx = Working:

Now some closely related examples to point out the importance of signs and roots in substitutions.

### Examples We want x2 = 3sin2u so we can use an identity.

Let x = sin u and then dx = cos u du. Substituting, simplifying, integrating and resubstituting gives:  We want x2 = 3tan2u so we can use an identity.

Let x = tan u and then dx = sec2u du. Substituting, simplifying, integrating and resubstituting gives:  We need x2 = 3tan2u so we can substitute.

Let x = tan u and then dx = sec2u du. Substituting, simplifying, integrating and resubstituting gives: This integral is apparently simpler but is beyond the integration tools covered so far. We can try x2 = 3sin2u.

Let x = sin u and then dx = cos2u du. Substituting, simplifying, integrating and resubstituting gives: This problem is better approached with the clever use of some High School algebra. By splitting the reciprocal of a difference of two squares into a simpler pair of fractions we obtain an integrable expression. This is an example of the method of partial fractions. (The keen might want to show this by simplifying the right side. Show me now.)  Then the integration is simpler: ### Exercise

You should do plenty of exercises until you are sure you recognise each type of problem and its solution. dx = Working:

Look now at the more general situation where you have a fraction with a numeric term divided by a quadratic expression. If you are able to express the quadratic expression in the form (linear expression) squared plus some number, the tan substitution is possible.

### Example First see whether the quadratic fits the pattern by completing the square.

x2 + 12x + 45 = (x + 6)2 + 9.

It does. Set x = 3tan w − 6. then dx = 3sec2 w and w = arctan( ). Alternatively, if the expression is a fraction with a numeric term divided by the square root of a quadratic expression and the quadratic can be expressed as a positive number minus some linear expression squared, a sin substitution is possible.

### Example First see whether the quadratic fits the pattern by completing the square.

9 + 8xx2 = 25 − (x − 4)2.

It does. Substitute x = 5sin w + 4 , then dx = 5cos w and w = arcsin( ).   ### Exercises

Decide whether trigonometric substitution will be helpful for these expressions and integrate them if possible.

Evaluate the indefinite integral whenever possible: dx = Contact Us | About Massey University | Sitemap | Disclaimer | Last updated: November 21, 2012     © Massey University 2003