Techniques for Trigonometric Integrals

7.2 Trig Integrals §

These integrals commonly (potentially always) will require use of some Trigonometric Identities.

$\int si{n}^{m}xco{s}^{n}xdx$

Cases for Trig Integrals: §

There are two cases when dealing with Trig Integrals: ==1. Case 1: ”m” & or ”n” is ODD. 2. Case 2: ”m” & ”n” are EVEN.==

Case 1: §

If power of “SIN” is ODD, keep 1 factor of “SIN” and use Pythagorean Identity to change ==$SI{N}^{2}x=1-CO{S}^{2}x$==.

• If power of “COS” is ODD, do Pythagorean Identity for “COS”.
• If both are ODD, pick one that will reduce our exponent to the power of 2 (if possible).

Case 2: §

If both powers are EVEN, use HALF-ANGLE Identity:

• $SI{N}^{2}x=\frac{1}{2}(1-COS2x)$ <— this is $cos2x$, not a typo for $co{s}^{2x}$
• $CO{S}^{2}x=\frac{1}{2}(1+COS2x)$

Examples: §

Example

$\int si{n}^{3}xco{s}^{2}xdx$ $\int 1-co{s}^{2}xco{s}^{2}x\ast sinxdx$ == $\leftarrow$ ==We factored out a sin, and $si{n}^{2}x=1-co{s}^{2}x$ per the Pythag. I.D== $u=cos(x)$ $du=-sin(x)dx$ $-du=sin(x)dx$ $\int (1-u^2)u^2\ast -du$

$-\int u^2-u^4\ast du$

$-[\frac{u^3}{3}-\frac{u^5}{5}]$

$-\frac{u^3}{3}+\frac{u^5}{5}$

$=\frac{co{s}^{5}x}{5}-\frac{co{s}^3}{3}$ $\leftarrow$ Final Answer!

Dealing with tan, cot, sec, csc §

$\int ta{n}^m(x)se{c}^m(x)dx$ or $\int co{t}^m(x)cs{c}^m(x)dx$

• If power of “tan” (“cot”) is ODD, keep ”$sec(x)tan(x)$“($csc(x)cot(x)$) & use ==$ta{n}^{2}x=se{c}^{2}x-1$. ($co{t}^{2}x=cs{c}^{2}x-1$)==

• If power of “sec” (“csc”) is EVEN, keep ”$se{c}^2$” (”$cs{c}^{2}x$”) & use ==$se{c}^{2}x=1+ta{n}^{2}x$. ($cs{c}^{2}x=1+co{t}^{2}x$)==

Example

$\int ta{n}^5(x)se{c}^3(x)dx$ $=\int ta{n}^4(x)se{c}^2(x)\ast sec(x)tan(x)dx$ $=\int (ta{n}^{2}x{)}^{2}se{c}^2(x)\ast sec(x)tan(x)dx$ $=\int (se{c}^{2}x-1{)}^{2}se{c}^2(x)\ast sec(x)tan(x)dx$ $\leftarrow$ apply identity $u=sec(x)$ $du=sec(x)tan(x)dx$

Example

${\int }_0^{\pi /4}\sqrt{tanx}se{c}^{6}xdx$ $={\int }_0^{\pi /4}ta{n}^{1/2}xse{c}^{4}x\ast se{c}^{2}xdx$ $={\int }_0^{\pi /4}ta{n}^{1/2}x(se{c}^{2}x{)}^2\ast se{c}^{2}xdx$ $={\int }_0^{\pi /4}ta{n}^{1/2}x(1+ta{n}^{2}x{)}^2\ast se{c}^{2}xdx$

$u=tan(x)\text{when }x=\frac{\pi }{4}\to u=tan(\frac{\pi }{4})=1$ $du=se{c}^2(x)dx\text{when }x=0\to u=tan(0)=0$ Note: Bounds of integration are changed above.

$={\int }_0^1{u}^{1/2}(1+u^2{)}^{2}du$ $={\int }_0^1{u}^{1/2}(1+2u^2+u^4)du$ $={\int }_0^1{u}^{1/2}(1+2u^{5/2}+u^{9/2})du$ $={\left[\frac{2}{3}{u}^{3/2}+\frac{4}{7}{u}^{7/2}+\frac{2}{11}{u}^{11/2}\right]}_0^1$ $=\frac{2}{3}+\frac{4}{7}+\frac{2}{11}-0$ $=\frac{328}{231}$ $\leftarrow$ Final Answer.

Footnotes / Tips: §

• Every function that begins with a "C" has a negative derivative. Conversely, this rule is flipped when doing integrals.
  • Derivative Example: $\text{if }u=cosx\text{ then}du=-sinxdx$
  • Integral Example: ==$\int sin(x)=-cos(x)+C$==

#sapling