22. Riemann Sums, Integrals and the FTC
d. Integration Rules and Properties
2. Special Functions
The table on the previous page listed the properties of derivatives and integrals with respect to algebraic operations. The tables on this page gives the derivatives and integral for special functions.
| Derivative Rule | Integral Rule | |
|---|---|---|
| Natural Exponential |
\(\dfrac{d}{dx}(e^x) =e^x\) | \(\displaystyle \int e^x\,dx=e^x+C\) |
| General Exponential |
\(\dfrac{d}{dx}(b^x) =b^x\ln b\) | \(\displaystyle \int b^x\,dx=\dfrac{b^x}{\ln b}+C\) |
| Natural Logarithm\(^\text{1}\) |
\(\dfrac{d}{dx}(\ln x) =\dfrac{1}{x}\) | \(\displaystyle \int \dfrac{1}{x}\,dx=\ln|x|+C\) |
| General Logarithm\(^\text{2}\) |
\(\dfrac{d}{dx}(\log_b x) =\dfrac{1}{x\ln b}\) | \(\displaystyle \int \dfrac{1}{x}\,dx=\ln b\log_b|x|+C\) |
\(^\text{1}\)
This is the integral which produces \(\ln x\). We will
discuss \(\displaystyle \int \ln x\,dx\) and
\(\displaystyle \int \log_b x\,dx\) in Calculus 2 in the chapter on
Integration by Parts.
\(^\text{2}\)
This integral formula is not useful, since the formula using
\(\ln x\) is equivalent and easier.
Why is there an absolute value in the log integrals?
Like the function \(\sqrt{x}\), the function \(\ln{x}\) is only defined for positive values of \(x\), since it is the inverse function of \(e^x\) which is always positive. However, the function \(\dfrac{1}{x}\) is defined for all \(x\ne0\) and we might want to know \(\displaystyle \int \dfrac{1}{x}\,dx\) for negative as well as positive values of \(x\).
For \(x \gt 0\) we have \(\displaystyle \int \dfrac{1}{x}\,dx=\ln(x)+C\) without or with the absolute value.
However, for \(x \lt 0\), we need to recompute \(\displaystyle \int \dfrac{1}{x}\,dx\). We work backwards: For \(x \lt 0\), we have \(|x|=-x\) and so \(\ln|x|=\ln(-x)\). We differentiate using the Chain Rule: \[\begin{aligned} \dfrac{d}{dx}\ln|x|&=\dfrac{d}{dx}\ln(-x)=\dfrac{1}{(-x)}\dfrac{d}{dx}(-x) \\ &=\dfrac{1}{(-x)}(-1)=\dfrac{1}{x} \end{aligned}\] Reversing this, we have \(\displaystyle \int \dfrac{1}{x}\,dx=\ln|x|+C\). So this formula works for both positive or negative values of \(x\).
| 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 \(\displaystyle \int \tan x\,dx\), \(\displaystyle \int \cot x\,dx\), \(\displaystyle \int \sec x\,dx\) and \(\displaystyle \int \csc x\,dx\), in Calculus 2 in the chapter on Trig Integrals.
| 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 \(\displaystyle \int \arcsin x\,dx\), \(\displaystyle \int \arctan x\,dx\) and \(\displaystyle \int \mathrm{arcsec}\, x\,dx\), 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 exercises are the same as on previous pages on antiderivative rules on exponential and trig functions.
Compute \(\displaystyle \int 3x^2e^{\small x^{\scriptstyle 3}}\,dx\) and \(\displaystyle \int_0^1 3x^2e^{\small x^{\scriptstyle 3}}\,dx\).
\(\displaystyle \int 3x^2e^{\small x^{\scriptstyle 3}}\,dx=e^{\small x^{\scriptstyle 3}}+C\)
\(\displaystyle \int_0^1 3x^2e^{\small x^{\scriptstyle 3}}\,dx=e-1\)
Since \(\dfrac{d}{dx}x^3=3x^2\), the Chain Rule and Exponential Rule say the indefinite integral is: \[ \int 3x^2e^{\small x^{\scriptstyle 3}}\,dx=e^{\small x^{\scriptstyle 3}}+C \] Be sure to check by differentiating!
By the Fundamental Theorem of Calculus, the definite integral is: \[\begin{aligned} \int_0^1 &3x^2e^{\small x^{\scriptstyle 3}}\,dx =\left[e^{\small x^{\scriptstyle 3}}\right]_0^1 \\ &=\left(e^{\small 1^{\scriptstyle 3}}\right)-\left(e^{\small 0^{\scriptstyle 3}}\right) =e-1 \end{aligned}\]
We check the indefinite integral by differentiating. If \(F=e^{\small x^{\scriptstyle 3}}\) then: \[ F'(x)=e^{\small x^{\scriptstyle 3}}\cdot3x^2 \]
Compute \(\displaystyle \int 2x\sin(x)+x^2\cos(x)\,dx\) and \(\displaystyle \int_{\pi/2}^\pi 2x\sin(x)+x^2\cos(x)\,dx\).
\(\displaystyle \int 2x\sin(x)+x^2\cos(x)\,dx=x^2\sin(x)+C\)
\(\displaystyle \int_{\pi/2}^\pi 2x\sin(x)+x^2\cos(x)\,dx=-\dfrac{\pi^2}{4}\)
Since \(\dfrac{d}{dx}x^2=2x\) and \(\dfrac{d}{dx}\sin(x)=\cos(x)\), the Product Rule says \(\dfrac{d}{dx}x^2\sin(x)=2x\sin(x)+x^2\cos(x)\). So the indefinite integral is: \[ \int 2x\sin(x)+x^2\cos(x)\,dx=x^2\sin(x)+C \]
By the Fundamental Theorem of Calculus, the definite integral is: \[\begin{aligned} \int_{\pi/2}^\pi &2x\sin(x)+x^2\cos(x)\,dx =\left[\rule{0pt}{10pt}x^2\sin(x)\right]_{\pi/2}^\pi \\ &=\pi^2\sin(\pi)-\left(\dfrac{\pi}{2}\right)^2\sin\left(\dfrac{\pi}{2}\right) =-\dfrac{\pi^2}{4} \end{aligned}\]
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