Let’s first use the substitution to eliminate the root.

\[\sqrt {1 - 7{w^2}} = \sqrt {1 - {{\sin }^2}\left( \theta \right)} = \sqrt {{{\cos }^2}\left( \theta \right)} = \left| {\cos \left( \theta \right)} \right|\]

Next, because we are doing an indefinite integral we will assume that the cosine is positive and so we can drop the absolute value bars to get,

\[\sqrt {1 - 7{w^2}} = \cos \left( \theta \right)\]

For a final substitution preparation step let’s also compute the differential so we don’t forget to use that in the substitution!

\[dw = \frac{1}{{\sqrt 7 }}\cos \left( \theta \right)\,d\theta \] Show Step 3

Now let’s do the actual substitution.

\[\int{{\sqrt {1 - 7{w^2}} \,dw}} = \int{{\cos \left( \theta \right)\,\left( {\frac{1}{{\sqrt 7 }}\cos \left( \theta \right)} \right)d\theta }} = \frac{1}{{\sqrt 7 }}\int{{{{\cos }^2}\left( \theta \right)\,d\theta }}\]

Do not forget to substitute in the differential we computed in the previous step. This is probably the most common mistake with trig substitutions. Forgetting the differential can substantially change the problem, often making the integral very difficult to evaluate.

Show Step 4

We now need to evaluate the integral. Here is that work.

\[\int{{\sqrt {1 - 7{w^2}} \,dw}} = \frac{1}{{\sqrt 7 }}\int{{\frac{1}{2}\left[ {1 + \cos \left( {2\theta } \right)} \right]\,d\theta }} = \frac{1}{{2\sqrt 7 }}\left[ {\theta + \frac{1}{2}\sin \left( {2\theta } \right)} \right] + c\]

Don’t forget all the “standard” manipulations of the integrand that we often need to do in order to evaluate integrals involving trig functions. If you don’t recall them you’ll need to go back to the previous section and work some practice problems to get good at them.

Every trig substitution problem reduces down to an integral involving trig functions and the majority of them will need some manipulation of the integrand in order to evaluate.

Show Step 5

As the final step we just need to go back to \(w\)’s.

To eliminate the the first term (i.e. the \(\theta \)) we can use any of the inverse trig functions. The easiest is to probably just use the original substitution and get a formula involving inverse sine but any of the six trig functions could be used if we wanted to. Using the substitution gives us,

\[\sin \left( \theta \right) = \sqrt 7 \,w\,\,\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}\theta = {\sin ^{ - 1}}\left( {\sqrt 7 \,w} \right)\]

Eliminating the \(\sin \left( {2\theta } \right)\) requires a little more work. We can’t just use a right triangle as we normally would because that would only give trig functions with an argument of \(\theta \) and we have an argument of \(2\theta \). However, we could use the double angle formula for sine to reduce this to trig functions with arguments of \(\theta \) . Doing this gives,

\[\int{{\sqrt {1 - 7{w^2}} \,dw}} = \frac{1}{{2\sqrt 7 }}\left[ {\theta + \sin \left( \theta \right)\cos \left( \theta \right)} \right] + c\]

We can now do the right triangle work.

From the substitution we have, \[\sin \left( \theta \right) = \frac{{\sqrt 7 w}}{1}\,\,\,\,\,\left( { = \frac{{{\mbox{opp}}}}{{{\mbox{hyp}}}}} \right)\] From the right triangle we get, \[\cos \left( \theta \right) = \sqrt {1 - 7{w^2}} \]

The integral is then,

\[\int{{\sqrt {1 - 7{w^2}} \,dw}} = \frac{1}{{2\sqrt 7 }}\left[ {{{\sin }^{ - 1}}\left( {\sqrt 7 w} \right) + \sqrt 7 \,w\sqrt {1 - 7{w^2}} } \right] + c\]