# Use trigonometric substitution to find the following integrals. a. x^3 {16 - x^2 } dx b. ...

## Question:

Use trigonometric substitution to find the following integrals.

a.{eq}\displaystyle \ \int \frac {x^3}{ \sqrt {16 - x^2} } dx {/eq}

b.{eq}\displaystyle \ \int \frac {x^2}{(1 + x^2)^2} \ dx {/eq}

c.{eq}\displaystyle \ \int \sqrt {5x^2 - 1}\ dx {/eq}

## Integration by Trigonometric Substitution

Trigonometric substitution consists of performing changes of variables whenever such change will result in a expression that could be easily simplified using trigonometric identities. Such situations usually occur with functions that have terms of the form {eq}\sqrt{a^2 + x^2} {/eq} and {eq}\sqrt{a^2- x^2} {/eq}. For positive sign try to substitute {eq}x= a tan(t) {/eq} or {eq}x= a sec(t) {/eq} and for negative sign, substitute {eq}x= a sin(t) {/eq} or {eq}x= a cos (t) {/eq}

We need to simplify the expressions in order to apply a proper change of variable.

(a)

I ={eq}\displaystyle \ \int \frac {x^3}{ \sqrt {16 - x^2} } dx {/eq}........................................(1)

Take {eq}x= 4 sin(t) {/eq}, to...

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