We have used substitution on problems similar to this problem in Section 6.1 . If we let \(u=\cos(x)\text{,}\) then \(du=-\sin(x)\,dx\text{.}\) Unfortunately, we have \(\sin^3(x)\) as part of our integrand, not just \(\sin(x)\text{.}\) The solution to this problem is to replace some of our powers of sine (two of them to be exact) with expressions that involve cosine. We will use the Pythagorean Identity \(\sin^2(x)=1-\cos^2(x)\text{.}\) \begin{align*} \ds\int\sin^3(x) \cos(x) \ dx \amp =\int\sin(x)\cdot \sin^2(x) \cos(x) \ dx\\ \amp =\int\sin(x)\left(1-\cos^2(x)\right) \cos(x) \ dx \end{align*}

Now we let \(u=\cos(x)\) so that \(-du=\sin(x)\,dx\text{.}\) \begin{align*} \ds\int\sin^3(x) \cos(x) \ dx \amp =\int\sin(x)\left(1-\cos^2(x)\right) \cos(x) \ dx\\ \amp =\int -\left(1-u^2\right)u\ du\\ \amp =\int -\left(u-u^3\right)\ du\\ \amp =\frac{u^2}{2}-\frac{u^4}{4}+C\\ \amp =\frac{\cos^2(x)}{2}-\frac{\cos^4(x)}{4}+C \end{align*}

The power of the sine term is odd, so we rewrite \(\sin^5(x)\) as \begin{align*} \sin^5(x) \amp = \sin^4(x) \sin(x) \\ \amp = (\sin^2(x) )^2\sin(x)\\ \amp = (1-\cos^2(x) )^2\sin(x) \text{.} \end{align*}

Our integral is now \(\ds \int (1-\cos^2(x) )^2\cos^8(x) \sin(x) \ dx\text{.}\) Let \(u = \cos(x)\text{,}\) hence \(du = -\sin(x) \ dx\text{.}\) Making the substitution and expanding the integrand gives \begin{align*} \int (1-\cos^2)^2\cos^8(x) \sin(x) \ dx \amp = -\int (1-u^2)^2u^8\ du \\ \amp = -\int \big(1-2u^2+u^4\big)u^8\ du\\ \amp = -\int \big(u^8-2u^{10}+u^{12}\big)\ du\text{.} \end{align*}

This final integral is not difficult to evaluate, giving \begin{align*} -\int \big(u^8-2u^{10}+u^{12}\big)\ du \amp = -\frac19u^9 + \frac2{11}u^{11} - \frac1{13}u^{13} + C\\ \amp =-\frac19\cos^9(x) + \frac2{11}\cos^{11}(x) - \frac1{13}\cos^{13}(x) + C. \end{align*}

The powers of both the sine and cosine terms are odd, therefore we can apply the techniques of Key Idea 6.3.2 to either power. We choose to work with the power of the cosine term since the previous example used the sine term's power.

We rewrite \(\cos^9(x)\) as \begin{align*} \cos^9(x) \amp = \cos^8(x) \cos(x) \\ \amp = \left(\cos^2(x)\right)^4\cos(x) \\ \amp = \left(1-\sin^2(x)\right)^4\cos(x) \text{.} \end{align*}

We rewrite the integral as \begin{equation*} \int\sin^5(x) \cos^9(x) \ dx = \int\sin^5(x) \left(1-\sin^2(x)\right)^4\cos(x) \ dx. \end{equation*}

Now substitute and integrate, using \(u = \sin(x)\) and \(du = \cos(x) \ dx\text{.}\) Expand the binomial using algebra.

\begin{align*} \amp \int u^5(1-u^2)^4\ du \\ \amp = \int u^5(1-4u^2+6u^4-4u^6+u^8)\ du\\ \amp = \int\big(u^5-4u^7+6u^9-4u^{11}+u^{13}\big)\ du\\ \amp = \frac16u^6-\frac12u^8+\frac35u^{10}-\frac13u^{12}+\frac{1}{14}u^{14}+C\\ \amp = \frac16\sin^6(x) -\frac12\sin^8(x) +\frac35\sin^{10}(x)-\frac13\sin^{12}(x) +\frac{1}{14}\sin^{14}(x) +C. \end{align*}

The powers of sine and cosine are both even, so we employ the power–reducing formulas and algebra as follows. \begin{align*} \int \cos^4(x) \sin^2(x) \ dx \amp = \int\left(\frac{1+\cos(2x)}{2}\right)^2\left(\frac{1-\cos(2x)}2\right)\ dx\\ \amp = \int\frac{1+2\cos(2x)+\cos^2(2x)}4\cdot\frac{1-\cos(2x)}2\ dx\\ \amp = \int \frac18\big(1+\cos(2x)+\cos^2(2x)-\cos^3(2x)\big)\ dx\\ \amp = \frac18 \left(\underbrace{\int 1 \ dx}_{a}+\underbrace{\int \cos(2x)\ dx}_{b}-\underbrace{\int \cos^2(2x)\ dx}_{c}-\underbrace{\int \cos^3(2x)\ dx}_{d}\right) \end{align*}

The first integral labelled a is easy to integrate. The \(\cos(2x)\) term is also easy to integrate, especially with Key Idea 6.1.4. The \(\cos^2(2x)\) term is another trigonometric integral with an even power, requiring the power–reducing formula again. The \(\cos^3(2x)\) term is a cosine function with an odd power, requiring a substitution as done before. We integrate each in turn below. \begin{align*} \underbrace{\int\cos(2x)\ dx}_{b} \amp = \frac12\sin(2x)+C\\ \underbrace{\int\cos^2(2x)\ dx}_{c} \amp = \int \frac{1+\cos(4x)}2\ dx \\ \amp = \frac12\big(x+\frac14\sin(4x)\big)+C\text{.} \end{align*}

Finally, we rewrite \(\cos^3(2x)\) as \begin{align*} \cos^3(2x) \amp = \cos^2(2x)\cos(2x)\\ \amp = \big(1-\sin^2(2x)\big)\cos(2x)\text{.} \end{align*}

Letting \(u=\sin(2x)\text{,}\) we have \(du = 2\cos(2x)\ dx\text{,}\) hence \begin{align*} \underbrace{\int \cos^3(2x)\ dx}_{d} \amp = \int\big(1-\sin^2(2x)\big)\cos(2x)\ dx\\ \amp = \int \frac12(1-u^2)\ du\\ \amp = \frac12\Big(u-\frac13u^3\Big)+C\\ \amp = \frac12\Big(\sin(2x)-\frac13\sin^3(2x)\Big)+C\text{.} \end{align*}

Putting all the pieces together, we have \begin{align*} \amp \int \cos^4(x) \sin^2(x) \ dx\\ \amp =\int \frac18\big(1+\cos(2x)-\cos^2(2x)-\cos^3(2x)\big)\ dx\\ \amp = \frac18\Big[x+\frac12\sin(2x)-\frac12\big(x+\frac14\sin(4x)\big)-\frac12\Big(\sin(2x)-\frac13\sin^3(2x)\Big)\Big]+C\\ \amp =\frac18\Big[\frac12x-\frac18\sin(4x)+\frac16\sin^3(2x)\Big]+C\text{.} \end{align*}

The application of the formula and subsequent integration are straightforward: \begin{align*} \int\sin(5x)\cos(2x)\ dx \amp = \int \frac12\Big[\sin((5-2)x)+\sin((5+2)x)\Big]\ dx\\ \amp= \int \frac12\Big[\sin(3x)+\sin(7x)\Big]\ dx\\ \amp = -\frac16\cos(3x) - \frac1{14}\cos(7x) + C \end{align*}

Since the power of secant is even, we use Rule 1from Key Idea 6.3.8 and pull out a \(\sec^2(x)\) in the integrand. We convert the remaining powers of secant into powers of tangent. \begin{align*} \int \tan^2(x) \sec^6(x) \ dx \amp = \int\tan^2(x) \sec^4(x) \sec^2(x) \ dx\\ \amp = \int \tan^2(x) \big(1+\tan^2(x) \big)^2\sec^2(x) \ dx\\ \end{align*} Now substitute, with \(u=\tan(x)\text{,}\) with \(du = \sec^2(x) \ dx\text{.}\) \begin{align*} \amp =\int u^2\big(1+u^2\big)^2\ du\\ \end{align*} We leave the integration and subsequent substitution to the reader. The final answer is \begin{align*} \amp =\frac13\tan^3(x) +\frac25\tan^5(x) +\frac17\tan^7(x) +C. \end{align*}

We apply Rule 3 from Key Idea 6.3.8 as the power of secant is odd and the power of tangent is even (0 is an even number). We use Integration by Parts; the rule suggests letting \(dv = \sec^2(x) \ dx\text{,}\) meaning that \(u = \sec(x)\text{.}\)

Employing Integration by Parts, we have \begin{align*} \int \sec^3(x) \ dx \amp = \int \underbrace{\sec(x) }_u\cdot\underbrace{\sec^2(x) \ dx}_{dv}\\ \amp = \sec(x) \tan(x) - \int \sec(x) \tan^2(x) \ dx.\\ \end{align*} This new integral also requires applying Rule 3 of Key Idea 6.3.8: \begin{align*} \int \sec^3(x)\ dx\amp = \sec(x) \tan(x) - \int \sec(x) \big(\sec^2(x) -1\big)\ dx\\ \amp = \sec(x) \tan(x) - \int \sec^3(x) \ dx + \int \sec(x) \ dx\\ \amp = \sec(x) \tan(x) -\int \sec^3(x) \ dx + \ln\abs{\sec(x) +\tan(x) }\\ \end{align*} In previous applications of Integration by Parts, we have seen where the original integral has reappeared in our work. We resolve this by adding \(\int \sec^3(x) \ dx\) to both sides, giving: \begin{align*} 2\int \sec^3(x) \ dx \amp = \sec(x) \tan(x) + \ln\abs{\sec(x) +\tan(x) }\\ \int \sec^3(x) \ dx \amp = \frac12\Big(\sec(x) \tan(x) + \ln\abs{\sec(x) +\tan(x) }\Big)+C \end{align*}

We employ Rule 3of Key Idea 6.3.8. \begin{align*} \int \tan^6(x) \ dx \amp = \int \tan^4(x) \tan^2(x) \ dx\\ \amp = \int\tan^4(x) \big(\sec^2(x) -1\big)\ dx\\ \amp = \int\tan^4(x) \sec^2(x) \ dx - \int\tan^4(x) \ dx\\ \end{align*} Integrate the first integral with substitution, \(u=\tan(x)\text{;}\) integrate the second by employing rule Rule 4 again. \begin{align*} \amp = \frac15\tan^5(x) -\int\tan^2(x) \tan^2(x) \ dx\\ \amp = \frac15\tan^5(x) -\int\tan^2(x) \big(\sec^2(x) -1\big)\ dx\\ \amp = \frac15\tan^5(x) -\underbrace{\int\tan^2(x) \sec^2(x) \ dx}_{a} + \underbrace{\int\tan^2(x) \ dx}_b\\ \end{align*} Again, use substitution (\(u=\tan(x)\)) for the first integral (a) and Rule 4 for the second (b). \begin{align*} \amp = \frac15\tan^5(x) -\frac13\tan^3(x) +\int\big(\sec^2(x) -1\big)\ dx\\ \int \tan^6(x)\ dx\amp = \frac15\tan^5(x) -\frac13\tan^3(x) +\tan(x) - x+C\text{.} \end{align*}