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Substitution methods allowed us to determine some antiderivatives that involve trigonometric functions, such as $\displaystyle{\int \cos(x) \sin(x) dx}$. Integration by parts allowed us to determine antiderivatives of more complicated functions involving trigonometric functions, such as $\displaystyle{\int x^2\sin(x)dx}$. In this section we will develop more general techniques for determining antiderivatives that involve trigonometric functions. The next section will develop techniques for using trigonometric identities in conjunction with the substitution of variables method.
All trigonometric definitions and identities are rooted in sine and cosine functions, which in turn are grounded in concepts of angle measure. See Chapter 2 for an introduction to angle measure and trigonometric functions. Figure 9.3.1 provides a reminder of the idea of angle measure in radians and the ways that the sine, cosine, and tangent functions are defined as functions of an angle measure in radians as the angle measure varies.
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$\csc(\theta)=\dfrac{1}{\sin(\theta)}$ |
$\sec(\theta)=\dfrac{1}{\cos(\theta)}$ |
$\begin{align}\tan(\theta)&=\dfrac{\sin(\theta)}{\cos(\theta)}\\[1ex] &=\dfrac{\sec(\theta)}{\csc(\theta)}\end{align}$ |
$\begin{align}\cot(\theta)&=\dfrac{1}{\tan(\theta)}\\[1ex] &=\dfrac{\cos(\theta)}{\sin(\theta)}\end{align}$ |
$(\sin\theta)^2+(\cos\theta)^2=1$ |
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Identities based on $(\sin\theta)^2+(\cos\theta)^2=1$ |
How to Get Them |
| $(\tan\theta)^2+1=(\sec\theta)^2$ | Divide both sides of $(\sin\theta)^2+(\cos\theta)^2=1$ by $(\cos\theta)^2$
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| $1+(\cot\theta)^2=(\csc\theta)^2$ | Divide both sides of $(\sin\theta)^2+(\cos\theta)^2=1$ by $(\sin\theta)^2$ |
Angle sum and part-angle identities |
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| $\sin(\theta+\phi)=\sin(\theta)\cos(\phi)+\sin(\phi)\cos(\theta)$ | $\text{When }\theta=\phi\\[1ex]$ : $\sin(2\theta)=2\sin(\theta)\cos(\theta)$ |
| $\cos(\theta+\phi)=\cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)$ | $\text{When }\theta=\phi\\[1ex]$ : $\cos(2\theta)=(\cos\theta)^2-(\sin\theta)^2$ |
| $(\cos\theta)^2=\dfrac{1+\cos(2\theta)}{2}$ | Use $(\sin\theta)^2=1-(\cos\theta)^2$ in $\cos(2\theta)=(\cos\theta)^2-(\sin\theta)^2$, then solve for $(\cos\theta)^2$. |
| $(\sin\theta)^2=\dfrac{1-\cos(2\theta)}{2}$ | Use $(\cos\theta)^2=1-(\sin\theta)^2$ in $\cos(2\theta)=(\cos\theta)^2-(\sin\theta)^2$, then solve for $(\sin\theta)^2$. |
| $\sin\left(\dfrac{\theta}{n}\right)=2\sin\left(\dfrac{\theta}{2n}\right)\cos\left(\dfrac{\theta}{2n}\right)$ | When $n=2\\[1ex]$: $\sin\left(\dfrac{\theta}{2}\right)=2\sin\left(\dfrac{\theta}{4}\right)\cos\left(\dfrac{\theta}{4}\right)$ |
| $\cos\left(\dfrac{\theta}{n}\right)=2\cos^2\left(\dfrac{\theta}{2n}\right)-1\\[1ex]$ $\cos\left(\dfrac{\theta}{n}\right)=1-2\sin^2\left(\dfrac{\theta}{2n}\right)$ |
When $n=2\\[1ex]$: $\cos\left(\dfrac{\theta}{2}\right)=2\cos^2\left(\dfrac{\theta}{4}\right)-1\\[1ex]$ $\cos\left(\dfrac{\theta}{2}\right)=1-2\sin^2\left(\dfrac{\theta}{4}\right)$ |
Some antiderivatives of trigonometric functions can be determined using substitution techniques, a trigonometric identity, or a combination of the two.
| Start With |
Transform Integrand |
End With |
Method |
Principal
Antiderivative |
| $\displaystyle{\int \tan(x)\,dx}$ |
$\tan(x)=\dfrac{\sin(x)}{\cos(x)}$ |
$\displaystyle{\int\dfrac{\sin(x)}{\cos(x)}\,dx}$ |
$u=\cos(x)\\[1ex]du=-\sin(x)\,dx$ |
$-\ln|\mathrm{cos}(x)|$ |
| $\displaystyle{\int \cot(x)\,dx}$ | $\cot(x)=\dfrac{\cos(x)}{\sin(x)}$ | $\displaystyle{\int\dfrac{\cos(x)}{\sin(x)}}\,dx$ |
$u=\sin(x)\\[1ex]du=\cos(x)\,dx$ |
$\ln\left|\mathrm{sin}(x)\right|$ |
| $\displaystyle{\int\csc(x)\,dx}$ | Multiply integrand by $\dfrac{\csc(x)+\cot(x)}{\csc(x)+\cot(x)}$ |
$\displaystyle{\int\dfrac{(\csc x)^2+\csc(x)\cot(x)}{\csc(x)+\cot(x)}}\,dx$ | $u=\csc(x)+\cot(x)\\[1ex]du=-\left(\cot(x)\csc(x)+\csc^2(x)\right)\,dx$ | $-\ln\left|\csc(x)+\cot(x)\right|$ |
| $\displaystyle{\int(\sin x)^5\,dx}$ | $(\sin x)^5=(\sin x)^4\sin(x)\\[1ex](\sin x)^4=\left((\sin x)^2\right)^2\\[1ex](\sin x)^2=1-(\cos x)^2\\[1ex](\sin x)^5=\left(1-(\cos x)^2\right)^2\sin(x)$ | $\displaystyle{\int\left(1-(\cos x)^2\right)^2\sin(x)}\,dx$ | $u=\cos(x)\\[1ex] du=-\sin(x)\,dx\\[1ex] -(1-u^2)^2=-u^4+2u^2-1$ |
$\dfrac{-(\cos x)^5}{5}+\dfrac{2(\cos x)^3}{3}-\cos(x)$ |
| $\displaystyle{\int(\sin x)^n\,dx}$, $n$ a positive integer | Rewrite $(\sin x)^n$ as $(\sin x)^{n-1}\sin(x)$ | $\displaystyle{\int(\sin x)^{n-1}\sin(x)\,dx}$ | $u=(\sin x)^{n-1},v=-\cos(x)\\[1ex] du=(n-1)(\sin x)^{n-2}\cos(x)\,dx\\[1ex] dv=\sin(x)\,dx$ |
$-\dfrac{(\sin x)^{n-1}\cos(x)}{n} +\dfrac{n-1}{n}\displaystyle{\int(\sin x)^{n-2}\,dx}\\[1ex]$ Apply integration by parts as often as necessary.$\\[2ex]$ This is called a reduction formula because each application of integration by parts reduces the power of the integrand. |
| $\displaystyle{\int(\cos x)^n\,dx}$, $n$ a positive integer | Rewrite $(\cos x)^n$ as $(\cos x)^{n-1}\cos(x)$ | $\displaystyle{\int(\cos x)^{n-1}\cos(x)\,dx}$ | $u=(\cos x)^{n-1},v=\sin(x)\\[1ex] du=-(n-1)(\cos x)^{n-2}\sin(x)\,dx\\[1ex] dv=\cos(x)\,dx$ |
$\dfrac{(\cos x)^{n-1}\sin(x)}{n} +\dfrac{n-1}{n}\displaystyle{\int(\sin x)^{n-2}\,dx}\\[1ex]$ Apply integration by parts as often as necessary.$\\[2ex]$ This is called a reduction formula because each application of integration by parts reduces the power of the integrand. |
| $\displaystyle{\int_0^x\sqrt{a^2-t^2}}\,dt,0≤x≤a$ | Substitute $t=a\sin(\theta),\,dt=a\cos(\theta)\,d\theta$ Rewrite $\begin{align}\sqrt{a^2-t^2}&=\sqrt{a^2-(a\sin\theta)^2}\\[1ex]&=\sqrt{a^2(1-(\sin\theta)^2)}\\[1ex]&=\sqrt{a^2(\cos \theta)^2}\\[1ex]&=a\cos(\theta)\text{ for }t≥0.\end{align}$ |
$\displaystyle{\int_0^x(a\cos\theta)\,a\cos(\theta)\,\,d\theta,\\[1ex]\qquad 0≤x≤\pi/2}\\[1ex]$
The first term, $a\cos(\theta)$, is $\sqrt{a^2-t^2}$ transformed by the substitution $t=a\sin(\theta)$. The second term, $a\cos(\theta)\,d \theta$ is $dt=a\cos(\theta)\,d\theta$. x now varies from 0 to $\pi/2$ because as t varies from 0 to $a$, $a\sin(\theta)$ varies from 0 to $a$. For $a\sin(\theta)$ to vary from 0 to $a$, $\theta$ must vary from 0 to $\pi/2$. |
Substitute $(\cos\theta)^2=\dfrac{1-\cos(2\theta)}{2}$ in $a^2\displaystyle{\int (\cos\theta)^2\,d\theta}$ |
$\displaystyle{a^2 \left.\left(\frac{\theta}{2} + \frac{\sin(2\theta)}{4} \right)\right|_0^x},\,0≤x≤\pi/2$ |
Confirm, by determining its derivative, that each principal antiderivative listed in the table of antiderivatives given above is indeed an antiderivative of the original function.
$\displaystyle{f(x)=\int_0^xt^2\cos(t)\,dt}$. $F(x)=???$
$\displaystyle{h(x)=\int_1^x(\tan t)^2\,dt,\,-\pi/2 \lt x \lt \pi/2}$. $H(x)=???$
Why is $H(x)$ negative for $x\lt 1$ when $(\tan t)^2$ is always greater than or equal to 0?
$\displaystyle{k(y)=\int_1^y(\sin t)^2(2\cos t)^4\,dt}$. $K(y)=???$
Show that the reduction formula for $\displaystyle{\int(\sin x)^n\,dx}$ applied to $\displaystyle{\int(\sin x)^5\,dx}$ produces the same result as the derivation of $\displaystyle{\int(\sin x)^5\,dx}$ given in the table.
Derive the principal antiderivative of $\displaystyle{\int(\tan x)^3\,dx}$ and confirm that what you derive is indeed an antiderivative of $(\tan x)^3$.
The table of antiderivatives given above sketches the derivation of the reduction formula $\displaystyle{\int(\sin x)^n\,dx=-\dfrac{(\sin x)^{n-1}\cos(x)}{n}
+\dfrac{n-1}{n}\int(\sin x)^{n-2}\,dx}$. Fill in the details of this derivation. You will need to use a trig identity along the way. Also, don't be alarmed when you get $\displaystyle{\int(\sin x)^n\,dx}$ on both sides of the equation. Just solve for $\displaystyle{\int(\sin x)^n\,dx}$.
The table of antiderivatives given above sketches the derivation of $\displaystyle{
\left.\left(\frac{\theta}{2}
+
\frac{\sin(2\theta)}{4}
\right)\right|_0^x},\,0≤x≤\pi/2$ as defining the same function as $\displaystyle{\int_0^x\sqrt{a^2-t^2}}\,dt,0≤x≤1$. Fill in the details.
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