Compute the indefinite integrals $$\int \cos ^5 3x \sin ^2 3x \, dx \ \text { and } \ \ \int \cos ^2 3x \sin ^5 3x \, dx.$$

• When dealing with products of sines and cosines of the same quantity (in this case, the $3x$ is the same inside both the sine and the cosine), we can use a substitution. We look for one with an odd power and build a substitution using the other. In this case, we would use the substitution $u = \cos 3x$$u = \sin 3x$ in the first integral and $u = \cos 3x$$u = \sin 3x$ in the second.
• Using the substitutions just identified, we have $du = \answer {3 \cos 3x} \, dx$ in the former case and $du = \answer {-3 \sin 3x}\, dx$ in the latter. This means the integrals become $$\frac {1}{3} \int \cos ^4 3x \sin ^2 3x \, du \ \ \text { and } \ \ - \frac {1}{3} \int \cos ^2 3x \sin ^4 3x \, du.$$
• We continue to simplify, using the trig identity $\cos ^2 \theta + \sin ^2 \theta = 1$ to completely eliminate all references to the variable $x$ in the integrand: $$\frac {1}{3} \int (\answer {1-u^2})^2 u^2 \, du \ \ \text { and } \ \ - \frac {1}{3} \int u^2 (\answer {1-u^2})^2 \, du.$$
• Now we calculate the integral, giving $$\frac {1}{3} \int (\answer {1-u^2})^2 u^2 \, du = \answer {\frac {u^3}{9} - \frac {2 u^5}{15} + \frac {u^7}{21}} + C$$ in the first case and $$-\frac {1}{3} \int u^2 (\answer {1-u^2})^2 \, du = \answer {-\frac {u^3}{9} + \frac {2 u^5}{15} - \frac {u^7}{21}} + C$$ in the second.
• To conclude, we reverse the substitution, so that $$\int \cos ^5 3x \sin ^2 3x \, dx = \answer {\frac {(\sin 3x)^3}{9} - \frac {2 (\sin 3x)^5}{15} + \frac {(\sin 3x)^7}{21}} + C$$ and $$\int \cos ^2 3x \sin ^5 3x \, dx = \answer {-\frac {(\cos 3x)^3}{9} + \frac {2 (\cos 3x)^5}{15} - \frac {(\cos 3x)^7}{21}} + C.$$

Compute the indefinite integral $$\int \sin ^2 3x \cos ^2 3x \, dx.$$

• When both powers are even, your only option is to use a trigonometric identity to reduce the power. In this case, the identities are $$\sin ^2 \theta = \frac {1 - \cos 2 \theta }{2} \ \ \text { and } \ \ \cos ^2 \theta = \frac {1 + \cos 2 \theta }{2}.$$
• This means that $$\sin ^2 3x \cos ^2 3 x = \frac {1}{4} - \answer { \frac {(\cos 6 x)^2}{4}}.$$
• Because we still have only even powers, we should use power reduction again: $$\sin ^2 3x \cos ^2 3x = \frac {1}{4} - \frac {1}{4} \left [ \frac {1}{2} + \answer {\frac {1}{2} \cos 12x} \right ] = \frac {1}{8} - \frac {1}{8} \answer {\cos 12x}.$$
• Integrating this last expression gives $$\int \sin ^2 3x \cos ^2 3x \, dx = \answer {\frac {x}{8} - \frac {1}{96} \sin 12x} + C.$$

Compute the indefinite integrals $$\int \sec ^2 3x \tan ^2 3x \, dx \ \ \text { and } \ \ \int \sec ^3 3x \tan ^3 3x \, dx.$$

• In this case we look for either an even power of $secant$, which indicates a tangent substitution, or an odd number of both secant and tangent, which indicates a secant substitution. So we substitute $u = \tan 3x$ in the first integral and $u = \sec 3x$ in the second. To simplify and write the integrands in terms of $u$ only, use the identity $\sec ^2 \theta = \tan ^2 \theta + 1$. This gives $$\int \answer {\frac {1}{3} u^2} \, du \ \ \text { and } \ \ \int \answer {\frac {1}{3} u^2 (u^2 - 1)} \, du$$ for the two integrals (don’t forget the extra factor of $1/3$ coming from the chain rule).
• Integrating and reversing the substitution gives $$\int \sec ^2 3x \tan ^2 3x \, dx = \answer { \frac {(\tan 3x)^3}{9}} + C$$ and $$\int \sec ^3 3x \tan ^3 3x \, dx = \answer { \frac {(\sec 3x)^5}{15} - \frac {(\sec 3x)^3}{9}} + C.$$

When there are are odd powers of secant and even (or just zero) powers of tangent, there’s a trick: integrate by parts letting $dv = \sec ^2 x$ and $u$ being everything else. $$\begin {aligned} \int \sec ^7 x \, dx & = \int \sec ^{\answer {5}} x \sec ^2 x \, dx \\ & = \sec ^5 x \answer {\tan x} - \int \answer {5 (\sec x)^4 \sec x (\tan x)^2} dx \\ & = \sec ^5 x \answer {\tan x} - 5 \int \sec ^7 x \, dx + 5 \int \sec ^5 x \, dx. \end {aligned}$$ Just like for other integration-by-parts examples, we can now solve for the integral in terms of a simpler integral: $$\int \sec ^7 x \, dx = \answer {\frac {1}{6}} \left [ \answer {(\sec x)^5 \tan x} + 5 \int \sec ^5 x \, dx \right ].$$ This is an example of a reduction formula, because the integral on the right-hand side is similar to the one we started with, but simpler. We could then compute the integral of $\sec ^5 x$ in terms of the integral of $\sec ^3 x$ and ultimately to $\sec x$, which is one which we can evaluate (or look up in a table).