8.3 | Trigonometric Integrals


For trigonometric integrals of the form
$$
\int \sin ^m x \cdot \cos ^n x \; dx,
$$
if

  • \(m\) is odd, then retain a factor of \(\sin x\) and use the identity \(\sin^2 x = 1- \cos^2 x\) to rewrite the remaining power of \(\sin x\) in terms of \(\cos x\)
  • \(n\) is odd, then retain a factor of \(\cos x\) and use the identity \(\cos^2 = 1 – \sin^2 x \) to rewrite the remaining power of \(\cos x\) in terms of \(\sin x\)
  • \(m\) and \(n\) are both even, then apply the identities
    $$
    \sin^2 x = \frac{1- \cos(2x)}{2} \text{ and } \cos^2 x = \frac{1+ \cos(2x)}{2}
    $$
    $$
    \sin^2 x = \frac{1- \cos(2x)}{2}
    $$
    and
    $$
    \cos^2 x = \frac{1+ \cos(2x)}{2}
    $$
    $$
    \sin^2 x = \frac{1- \cos(2x)}{2}
    $$
    and
    $$
    \cos^2 x = \frac{1+ \cos(2x)}{2}
    $$

    to reduce the powers.

Then apply \(u\)-substitution.

For trigonometric integrals of the form
$$
\int \sec^m x \cdot \tan^n x \; dx,
$$
if

  • \(n\) is odd, then retain a factor of \(\sec x \tan x\) and use the identity \(\tan^2 x = \sec^2x -1 \) to rewrite the remaining power of \(\tan x\) in terms of \( \sec x\).
  • \(m\) is even, then retain a factor of \(\sec^2 x\) and use the identity \(\sec^2x = \tan^2 x +1 \) to rewrite the remaining power of \(\sec x\) in terms of \(\tan x\).

Then apply \(u\)-substitution.

Trigonometric Identities
$$
\begin{aligned}
\sin m x \sin nx & = \frac{1}{2}[ \cos(m-n)x – \cos(m+n)x] \\
\sin m x \cos n x & = \frac{1}{2}[ \sin(m-n)x + \sin(m+n)x] \\
\cos mx \cos nx & = \frac{1}{2} [ \cos(m-n)x + \cos(m+n)x]
\end{aligned}
$$
Find
$$
\int \sin^7 x \cos^4 x \; dx.
$$


$$
\begin{aligned}
\int \sin^7 x \cos^4 x \; dx & = \int \sin x (\sin^6 x) \cos ^4 x \; dx \\
& = \int \sin x (\sin^2 x )^3 \cos^4 x \; dx \\
& = \int \sin x (1- \cos^2 x)^3 \cos^4 x \; dx \\
& = \int \sin x (1- u^2)^3 u^4 \frac{du}{- \sin x} \\
& = -\int (1- u^2)^3 u^4 \; du \\
& = -\int (1-3u^2+3u^4 +u^6) u^4 \; du \\
& = -\int u^4-3u^6+3u^8 +u^{10} \; du \\
& = -\frac{u^5}{5}+\frac{3u^7}{7}-\frac{u^9}{3} -\frac{u^{11}}{11}+C \\
& = -\frac{\cos^5x}{5}+\frac{3\cos^7x}{7}-\frac{\cos^9x}{3} -\frac{\cos^{11}x}{11}+C \\
\end{aligned}
$$
$$
\begin{aligned}
\int & \sin^7 x \cos^4 x \; dx \\
& = \int \sin x (\sin^6 x) \cos ^4 x \; dx \\
& = \int \sin x (\sin^2 x )^3 \cos^4 x \; dx \\
& = \int \sin x (1- \cos^2 x)^3 \cos^4 x \; dx \\
& = \int \sin x (1- u^2)^3 u^4 \frac{du}{- \sin x} \\
& = -\int (1- u^2)^3 u^4 \; du \\
& = -\int (1-3u^2+3u^4 +u^6) u^4 \; du \\
& = -\int u^4-3u^6+3u^8 +u^{10} \; du \\
& = -\frac{u^5}{5}+\frac{3u^7}{7}-\frac{u^9}{3} -\frac{u^{11}}{11}+C \\
& = -\frac{\cos^5x}{5}+\frac{3\cos^7x}{7}-\frac{\cos^9x}{3} -\frac{\cos^{11}x}{11}+C
\end{aligned}
$$
$$
\begin{aligned}
& \int \sin^7 x \cos^4 x \; dx \\
& = \int \sin x (\sin^6 x) \cos ^4 x \; dx \\
& = \int \sin x (\sin^2 x )^3 \cos^4 x \; dx \\
& = \int \sin x (1- \cos^2 x)^3 \cos^4 x \; dx \\
& = \int \sin x (1- u^2)^3 u^4 \frac{du}{- \sin x} \\
& = -\int (1- u^2)^3 u^4 \; du \\
& = -\int (1-3u^2+3u^4 +u^6) u^4 \; du \\
& = -\int u^4-3u^6+3u^8 +u^{10} \; du \\
& = -\frac{u^5}{5}+\frac{3u^7}{7}-\frac{u^9}{3} -\frac{u^{11}}{11}+C \\
& = -\frac{\cos^5x}{5}+\frac{3\cos^7x}{7}-\frac{\cos^9x}{3} \\
& \qquad -\frac{\cos^{11}x}{11}+C \\
\end{aligned}
$$

E 8.3 Exercises