22. Riemann Sums, Integrals and the FTC

d. Integration Rules and Properties

2. Trig Functions

The table on a previous page listed the properties of derivatives and integrals with respect to algebraic operations. The tables on this page gives the derivatives and integrals for trig functions and their inverses.

Trigonometric Functions
	Derivative Rule	Integral Rule
Sine	\(\dfrac{d}{dx}(\sin x) =\cos x\)	\(\displaystyle \int \cos x\,dx=\sin x+C\)
Cosine	\(\dfrac{d}{dx}(\cos x)=-\sin x\)	\(\displaystyle \int \sin x\,dx=-\cos x+C\)
Tangent\(^\text{3}\)	\(\dfrac{d}{dx}(\tan x) =\sec^2 x\)	\(\displaystyle \int \sec^2 x\,dx=\tan x+C\)
Cotangent\(^\text{3}\)	\(\dfrac{d}{dx}(\cot x)=-\csc^2 x\)	\(\displaystyle \int \csc^2 x\,dx=-\cot x+C\)
Secant\(^\text{3}\)	\(\dfrac{d}{dx}(\sec x) =\sec x\tan x\)	\(\displaystyle \int \sec x\tan x\,dx=\sec x+C\)
Cosecant\(^\text{3}\)	\(\dfrac{d}{dx}(\csc x)=-\csc x\cot x\)	\(\displaystyle \int \csc x\cot x\,dx=-\csc x+C\)

\(^\text{3}\) These are the integrals which produce the trig functions. We will discuss the integrals of \(\tan x\), \(\cot x\), \(\sec x\) and \(\csc x\), in Calculus 2 in the chapter on Trig Integrals.

Inverse Trigonometric Functions
	Derivative Rule	Integral Rule
Inverse Sine\(^\text{4}\)	\(\dfrac{d}{dx}(\arcsin x) =\dfrac{1}{\sqrt{1-x^2}}\)	\(\displaystyle \int \dfrac{1}{\sqrt{1-x^2}}\,dx=\arcsin x+C\)
Inverse Tangent\(^\text{4}\)	\(\dfrac{d}{dx}(\arctan x) =\dfrac{1}{1+x^2}\)	\(\displaystyle \int \dfrac{1}{1+x^2}\,dx=\arctan x+C\)
Inverse Secant\(^\text{4}\)	\(\dfrac{d}{dx}(\mathrm{arcsec}\,x) =\dfrac{1}{x\sqrt{x^2-1}}\)	\(\displaystyle \int \dfrac{1}{x\sqrt{x^2+1}}\,dx=\mathrm{arcsec}\, x+C\)
Inverse Cosine\(^\text{5}\)	\(\dfrac{d}{dx}(\arccos x)=-\dfrac{1}{\sqrt{1-x^2}}\)	\(\displaystyle \int \dfrac{1}{\sqrt{1-x^2}}\,dx=-\arccos x+C\)
Inverse Cotangent\(^\text{5}\)	\(\dfrac{d}{dx}(\mathrm{arccot}\,x)=-\dfrac{1}{1+x^2}\)	\(\displaystyle \int \dfrac{1}{1+x^2}\,dx=-\mathrm{arccot}\, x+C\)
Inverse Cosecant\(^\text{5}\)	\(\dfrac{d}{dx}(\mathrm{arccsc}\,x)=-\dfrac{1}{x\sqrt{x^2-1}}\)	\(\displaystyle \int \dfrac{1}{x\sqrt{x^2-1}}\,dx=-\mathrm{arccsc}\, x+C\)

\(^\text{4}\) These are the integrals which produce the inverse trig functions. We will discuss the integrals of \(\arcsin x\), \(\arctan x\) and \(\mathrm{arcsec}\, x\), in Calculus 2 in the chapter on Integration by Parts.

\(^\text{5}\) Do not memorize these 3 integrals. They are just alternate forms of the 3 integrals above them.

The following exercise is the same as on a previous page on antiderivative rules for trig functions.

Compute \(\displaystyle \int 2\theta\sin(\theta)+\theta^2\cos(\theta)\,d\theta\) and \(\displaystyle \int_{\pi/2}^\pi 2\theta\sin(\theta)+\theta^2\cos(\theta)\,d\theta\).

\(\displaystyle \int 2\theta\sin(\theta)+\theta^2\cos(\theta)\,d\theta =\theta^2\sin(\theta)+C\)
\(\displaystyle \int_{\pi/2}^\pi 2\theta\sin(\theta)+\theta^2\cos(\theta)\,d\theta =-\dfrac{\pi^2}{4}\)

Since \(\dfrac{d}{d\theta}\theta^2=2\theta\) and \(\dfrac{d}{d\theta}\sin(\theta)=\cos(\theta)\), the Product Rule says \(\dfrac{d}{d\theta}\theta^2\sin(\theta)=2\theta\sin(\theta)+\theta^2\cos(\theta)\). So the indefinite integral is: \[ \int 2\theta\sin(\theta)+\theta^2\cos(\theta)\,d\theta=\theta^2\sin(\theta)+C \]

By the Fundamental Theorem of Calculus, the definite integral is: \[\begin{aligned} \int_{\pi/2}^\pi &2\theta\sin(\theta)+\theta^2\cos(\theta)\,d\theta =\left[\rule{0pt}{10pt}\theta^2\sin(\theta)\right]_{\pi/2}^\pi \\ &=\pi^2\sin(\pi)-\left(\dfrac{\pi}{2}\right)^2\sin\left(\dfrac{\pi}{2}\right) =-\dfrac{\pi^2}{4} \end{aligned}\]

Compute \(\displaystyle \int \dfrac{x^2}{1+x^2}\,dx\) and \(\displaystyle \int_0^3 \dfrac{x^2}{1+x^2}\,dx\).

Add and subtract \(1\) from the numerator.

\(\displaystyle \int \dfrac{x^2}{1+x^2}\,dx=x-\arctan x+C\)
\(\displaystyle \int_0^3 \dfrac{\rule{0pt}{12pt}x^2}{1+x^2}\,dx=3-\arctan 3\).

We first use a trick of adding and subtracting \(1\) from the numerator. So the integrand becomes: \[ \dfrac{x^2}{1+x^2}=\dfrac{x^2+1-1}{1+x^2}=1-\dfrac{1}{1+x^2} \] So the indefinite integral is: \[ \int \dfrac{x^2}{1+x^2}\,dx =\int \left(1-\dfrac{1}{1+x^2}\right)\,dx =x-\arctan x+C \]

By the Fundamental Theorem of Calculus, the definite integral is: \[\begin{aligned} \int_0^3 \dfrac{x^2}{1+x^2}\,dx &=\left[\rule{0pt}{10pt}x-\arctan x\right]_0^3 \\ &=(3-\arctan 3)-(0-\arctan 0) \\ &=3-\arctan 3 \end{aligned}\] since \(\arctan 0=0\).

22. Riemann Sums, Integrals and the FTC

d. Integration Rules and Properties

2. Trig Functions

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