Integrate: \int \frac {du}{\sqrt {8 - u^2}}

Question:

Integrate:

{eq}\displaystyle \int \frac {du}{\sqrt {8 - u^2}} {/eq}

Integration using trigonometric substitution:

Depending on the function that we need to integrate, we will substitute one of the following trigonometric expressions to simplify the integration:

For {eq}\displaystyle \sqrt {{a}^{2}-{x}^{2}} {/eq}, use {eq}\displaystyle {x}={a} \sin {\theta} {/eq};

For {eq}\displaystyle \sqrt {{a}^{2}+{x}^{2}} {/eq}, use {eq}\displaystyle {x}={a} \tan {\theta} {/eq};

For {eq}\displaystyle \sqrt {{x}^{2}-{a}^{2}} {/eq}, use {eq}\displaystyle {x}={a} \sec {\theta} {/eq};

Answer and Explanation:

We have,

{eq}\displaystyle \int \frac {du}{\sqrt {8 - u^2}} {/eq}

Apply trig substitution,

Let {eq}u = 2\sqrt 2 \sin v \Rightarrow du = 2\sqrt 2 \cos v \ dv {/eq}


{eq}\displaystyle \begin{align*} \Rightarrow \int \frac {du}{\sqrt {8 - u^2}} &= \int \frac {2\sqrt 2 \cos v \ dv}{\sqrt {8 - (2\sqrt 2\sin v)^2}} \\ &= \int \frac {2\sqrt 2 \cos v \ dv}{\sqrt {8 - 8\sin^2 v}} \\ &= \int \frac {2\sqrt 2 \cos v \ dv}{\sqrt {8\left ( 1 - \sin^2 v \right )}} \\ &= \int \frac {2\sqrt 2 \cos v \ dv}{2\sqrt 2 \ \sqrt {1 - \sin^2 v}} \\ &= \int \frac {\cos v \ dv}{\sqrt {\cos^2 v}} \\ &= \int \frac {\cos v \ dv}{{\cos v}} \\ &= \int dv \\ &= v \\ \end{align*} {/eq}


Substitute back,

{eq}\displaystyle \begin{align*} u &= 2 \sqrt 2 \sin v \\ \sin v &= \dfrac {u}{2 \sqrt 2} \\ v &= \arcsin \left ( \dfrac {u}{2 \sqrt 2} \right ) \\ \end{align*} {/eq}


{eq}\Rightarrow \displaystyle \int \frac {du}{\sqrt {8 - u^2}} = \arcsin \left ( \dfrac {u}{2 \sqrt 2} \right ) + C {/eq}

Where {eq}C {/eq} is the constant of integration.


Learn more about this topic:

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How to Use Trigonometric Substitution to Solve Integrals

from Math 104: Calculus

Chapter 13 / Lesson 11
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