MATH 152 Chapter 8.2

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Trig Integrals and Identities

Key Identities:

$\sin ^{2}{x}+\cos ^{2}{x}=1\,\!$
$1+\cot ^{2}{x}=\csc ^{2}{x}\,\!$
$\tan ^{2}{x}+1=\sec ^{2}{x}\,\!$

Trig Integrals	Trig² Integrals
$\int {\sin {x}\,\mathrm {d} x}=-\cos {x}+C$	$\int {\sin ^{2}{x}\,\mathrm {d} x}=\int {{\frac {1}{2}}(1-\cos(2x))\,\mathrm {d} x}={\frac {1}{2}}(x-{\frac {1}{2}}\sin {2x})+C$
$\int {\cos {x}\,\mathrm {d} x}=\sin {x}+C$	$\int {\cos ^{2}{x}\,\mathrm {d} x}=\int {{\frac {1}{2}}(1+\cos(2x))\,\mathrm {d} x}={\frac {1}{2}}(x+{\frac {1}{2}}\sin {2x})+C$
$\int {\tan {x}\,\mathrm {d} x}=\ln {\|\sec {x}\|}+C$	$\int {\tan ^{2}{x}\,\mathrm {d} x}=\int {\sec ^{2}{x}-1\,\mathrm {d} x}=\tan {x}-x+C$
$\int {\cot {x}\,\mathrm {d} x}=\ln {\|\sin {x}\|}+C$	$\int {\cot ^{2}{x}\,\mathrm {d} x}=\int {\csc ^{2}{x}-1\,\mathrm {d} x}=-\cot {x}-x+C$
$\int {\sec {x}\,\mathrm {d} x}=\ln {\|\sec {x}+\tan {x}\|}+C$	$\int {\sec ^{2}{x}\,\mathrm {d} x}=\tan {x}+C$
$\int {\csc {x}\,\mathrm {d} x}=\ln {\|\csc {x}-\cot {x}\|}+C$	$\int {\csc ^{2}{x}\,\mathrm {d} x}=-\cot {x}+C$

Strategy

If possible, "save" derivative $\mathrm {d} u$ of a trig function $u$ (in the numerator) and rewrite the rest of the integral in terms of $u$ . (for u-substitution)

Example 1

$\int {{\frac {\sin ^{3}{x}}{\cos {x}}}\,\mathrm {d} x}\;=\;\int {{\frac {\sin ^{2}{x}}{\cos {x}}}(\sin {x}\,\mathrm {d} x)}\;=\;\int {{\frac {1-\cos ^{2}{x}}{\cos {x}}}(\sin {x}\,\mathrm {d} x)}$

${\text{Let }}u=\cos {x}\quad \therefore \quad \mathrm {d} u=-\sin {x}\,\mathrm {d} x$

$-\int {{\frac {1-u^{2}}{u}}\,\mathrm {d} u}\;=\;-\ln {|\cos {x}|}+{\frac {1}{2}}\cos ^{2}{x}+C$

Example 2

$\int {\tan ^{7}{x}\sec ^{3}{x}\,\mathrm {d} x}$

NOTE: If we save $\sec ^{2}{x}$ , we're left with $\int {\tan ^{7}{x}\sec {x}\,(\sec ^{2}{x}\,\mathrm {d} x)}$ . We want even powers; $\sec {x}$ is an odd power, so it's no good.

Let's save a $\sec {x}\tan {x}$ instead:

$\int {\tan ^{6}{x}\sec ^{2}{x}\,(\sec {x}\tan {x}\,\mathrm {d} x)}\;=\;\int {(\sec ^{2}{x}-1)^{3}\,\sec ^{2}{x}\,(\sec {x}\tan {x}\,\mathrm {d} x)}$

${\text{Let }}u=\sec {x}\quad \therefore \quad \mathrm {d} u=\sec {x}\tan {x}\,\mathrm {d} x$

$\int {u^{2}\,(u^{2}-1)^{3}\,\mathrm {d} u}$

Expand, Integrate, substitute, etc.

Example 3

$\int {\csc ^{4}{4x}\,\mathrm {d} x}\;=\;\int {\csc ^{2}{4x}\,(\csc ^{2}{4x}\,\mathrm {d} x)}$

${\text{Let }}u=\cot {4x}\quad \therefore \quad \mathrm {d} u=-4\csc ^{2}{4x}\,\mathrm {d} x$

$-{\frac {1}{4}}\int {1+u^{2}\,\mathrm {d} u}\;=\;-{\frac {1}{4}}(\cot {4x}+{\frac {1}{3}}\cot ^{3}{4x})+C$

Warm-up

$\int {\sin ^{4}{x}\,\mathrm {d} x}$

The strategy does not work!

$\int {(\sin ^{2}{x})^{2}\,\mathrm {d} x}\;=\;\int {({\frac {1}{2}}(1-\cos {2x}))^{2}\,\mathrm {d} x}\;=\;{\frac {1}{4}}\int {1-2\cos {2x}+\cos ^{2}{2x}\,\mathrm {d} x}$

Now this gets a little tricky...

${\frac {1}{4}}\,{\bigg (}x-\sin {2x}+\int {cos^{2}{2x}\,\mathrm {d} x}{\bigg )}$