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Section 5.3 : Substitution Rule for Indefinite Integrals

16. Evaluate \( \displaystyle \int{{\frac{1}{{\sqrt {4 - 9{w^2}} }}\,dw}}\).

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Hint : Be careful with this substitution. The integrand should look somewhat familiar, so maybe we should try to put it into a more familiar form.
Start Solution

The integrand looks an awful lot like the derivative of the inverse sine.

\[\frac{d}{{du}}\left( {{{\sin }^{ - 1}}\left( u \right)} \right) = \frac{1}{{\sqrt {1 - {u^2}} }}\]

So, let’s do a little rewrite to make the integrand look more like this.

\[\int{{\frac{1}{{\sqrt {4 - 9{w^2}} }}\,dw}} = \int{{\frac{1}{{\sqrt {4\left( {1 - \frac{9}{4}{w^2}} \right)} }}\,dw}} = \frac{1}{2}\int{{\frac{1}{{\sqrt {1 - \frac{9}{4}{w^2}} }}\,dw}}\]
Hint : One more little rewrite of the integrand should make this look almost exactly like the derivative the inverse sine and the substitution should then be fairly obvious.
Show Step 2

Let’s do one more rewrite of the integrand.

\[\int{{\frac{1}{{\sqrt {4 - 9{w^2}} }}\,dw}} = \frac{1}{2}\int{{\frac{1}{{\sqrt {1 - {{\left( {\frac{{3w}}{2}} \right)}^2}} }}\,dw}}\]

At this point we can see that the following substitution will work for us.

\[u = \frac{{3w}}{2}\hspace{0.5in} \to \hspace{0.5in}du = \frac{3}{2}dw\hspace{0.5in} \to \hspace{0.5in}\,dw = \frac{2}{3}du\] Show Step 3

Doing the substitution and evaluating the integral gives,

\[\int{{\frac{1}{{\sqrt {4 - 9{w^2}} }}\,dw}} = \frac{1}{2}\left( {\frac{2}{3}} \right)\int{{\frac{1}{{\sqrt {1 - {u^2}} }}\,du}} = \frac{1}{3}{\sin ^{ - 1}}\left( u \right) + c\]
Hint : Don’t forget that the original variable in the integrand was not \(u\)!
Show Step 4

Finally, don’t forget to go back to the original variable!

\[\int{{\frac{1}{{\sqrt {4 - 9{w^2}} }}\,dw}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{3}{{\sin }^{ - 1}}\left( {\frac{{3w}}{2}} \right) + c}}\]

Substitutions for inverse trig functions can be a little tricky to spot when you are first start doing them. Once you do enough of them however they start to become a little easier to spot.