22. Riemann Sums, Integrals and the FTC

d. Integration Rules and Properties

2. Hyperbolic Trig Functions (Optional)

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 hyperbolic trig functions and their inverses.

Hyperbolic Trig Functions
Derivative Rule Integral Rule
Sine \(\dfrac{d}{dx}(\sinh x)=\cosh x\) \(\displaystyle \int \cosh x\,dx=\sinh x+C\)
Cosine \(\dfrac{d}{dx}(\cosh x)=\sinh x\) \(\displaystyle \int \sinh x\,dx=\cosh x+C\)
Tangent\(^\text{3}\) \(\dfrac{d}{dx}(\tanh x)=\,\mathrm{sech}\,^2 x\) \(\displaystyle \int \mathrm{sech}^2 x\,dx=\tanh x+C\)
Cotangent\(^\text{3}\) \(\dfrac{d}{dx}(\coth x)=-\,\mathrm{csch}^2 x\) \(\displaystyle \int \mathrm{csch}^2 x\,dx=-\coth x+C\)
Secant\(^\text{3}\) \(\dfrac{d}{dx}(\mathrm{sech}\,x) =-\,\mathrm{sech}\,x\tanh x\) \(\displaystyle \int \mathrm{sech}\,x\tanh x\,dx=-\,\mathrm{sech}\,x+C\)
Cosecant\(^\text{3}\) \(\dfrac{d}{dx}(\mathrm{csch}\,x)=-\,\mathrm{csch}\,x\coth x\) \(\displaystyle \int \mathrm{csch}\,x\coth x\,dx=-\,\mathrm{csch}x+C\)

\(^\text{3}\) These are the integrals which produce the hyperbolic trig functions. We will discuss the integrals of \(\tanh x\), \(\coth x\), \(\mathrm{sech}\,x\) and \(\mathrm{csch}\,x\), in Calculus 2 in the chapter on Trig Integrals.

Inverse Hyperbolic Trig Functions
Derivative Rule Integral Rule
Inverse
Hyperbolic
Sine\(^\text{4}\)		 \(\dfrac{d}{dx}(\mathrm{arcsinh}\,x) =\dfrac{1}{\sqrt{1+x^2}}\) \(\displaystyle \int \dfrac{1}{\sqrt{1+x^2}}\,dx=\,\mathrm{arcsinh}\,x+C\)
Inverse
Hyperbolic
Tangent\(^\text{4}\)		 \(\dfrac{d}{dx}(\mathrm{arctanh}x) =\dfrac{1}{1-x^2}\) \(\displaystyle \int \dfrac{1}{1-x^2}\,dx=\,\mathrm{arctanh}\,x+C\)
Inverse
Hyperbolic
Secant\(^\text{4}\)		 \(\dfrac{d}{dx}(\mathrm{arcsech}\,x) =\dfrac{-1}{\sqrt{x^2(1-x^2)}}\) \(\displaystyle \int \dfrac{1}{\sqrt{x^2(1-x^2)}}\,dx=-\mathrm{arcsech}\, x+C\)
Inverse
Hyperbolic
Cosine\(^\text{4}\)		 \(\dfrac{d}{dx}(\mathrm{arccosh}x)=\dfrac{1}{\sqrt{x^2-1}}\) \(\displaystyle \int \dfrac{1}{\sqrt{x^2-1}}\,dx=\mathrm{arccosh}\,+C\)
Inverse
Hyperbolic
Cotangent\(^\text{4}\)		 \(\dfrac{d}{dx}(\mathrm{arccoth}\,x)=\dfrac{-1}{x^2-1}\) \(\displaystyle \int \dfrac{1}{x^2-1}\,dx=-\mathrm{arccoth}\, x+C\)
Inverse
Hyperbolic
Cosecant\(^\text{4}\)		 \(\dfrac{d}{dx}(\mathrm{arccsch}\,x)=\dfrac{-1}{\sqrt{x^2(x^2-1)}}\) \(\displaystyle \int \dfrac{1}{\sqrt{x^2(x^2-1)}}\,dx=-\mathrm{arccsch}\, x+C\)

\(^\text{4}\) These are the integrals which produce the inverse hyprebolic trig functions. We will discuss the integrals of \(\mathrm{arcsinh}\,x\), \(\mathrm{arctanh}\,x\) \(\mathrm{arcsech}\,x\), \(\mathrm{arccosh}\,x\), \(\mathrm{arccoth}\,x\) and \(\mathrm{arccsch}\,x\) in Calculus 2 in the chapter on Integration by Parts.

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

Write the integrand as a perfect square. Then identify it as the square of a hyperbolic trig function.

\(\displaystyle \int \dfrac{e^{2x}}{(e^{2x}+1)^2}\,dx =\dfrac{1}{4}\tanh x+C =\dfrac{1}{4}\dfrac{e^x-e^{-x}}{e^x+e^{-x}}+C \)
\(\displaystyle \int_0^1 \dfrac{e^{2x}}{(e^{2x}+1)^2}\,dx =\dfrac{1}{4}\tanh 1 =\dfrac{1}{4}\dfrac{e^2-1}{e^2+1} \)

We write the integrand as a perfect square. Then we divide numerator and denominator by \(e^x\), and identify the integrand as the square of a hyperbolic trig function. \[\begin{aligned} \int &\dfrac{e^{2x}}{(e^{2x}+1)^2}\,dx =\int \left(\dfrac{e^{x}}{e^{2x}+1}\right)^2\,dx \\ &=\int \left(\dfrac{1}{e^x+e^{-x}}\right)^2\,dx =\int \left(\dfrac{1}{2\cosh x}\right)^2\,dx \\ &=\int \left(\dfrac{\mathrm{sech}\,x}{2}\right)^2\,dx =\int \dfrac{1}{4}\,\mathrm{sech}^2x\,dx \\ &=\dfrac{1}{4}\tanh x+C =\dfrac{1}{4}\dfrac{e^x-e^{-x}}{e^x+e^{-x}}+C \end{aligned}\]

By the Fundamental Theorem of Calculus, the definite integral is: \[\begin{aligned} \int_0^1 &\dfrac{e^{2x}}{(e^{2x}-1)^2}\,dx =\left[\rule{0pt}{10pt}\dfrac{1}{4}\tanh x\right]_0^1 \\ &=\dfrac{1}{4}\tanh 1-\dfrac{1}{4}\tanh 0 =\dfrac{1}{4}\tanh 1 \end{aligned}\] since \(\tanh 0=0\). In terms of exponentials, \[ \int_0^1 \dfrac{e^{2x}}{(e^{2x}-1)^2}\,dx =\dfrac{1}{4}\dfrac{e-e^{-1}}{e+e^{-1}} =\dfrac{1}{4}\dfrac{e^2-1}{e^2+1} \]

We check the indefinite integral by differentiating. If \(F=\dfrac{1}{4}\tanh x\) then: \[\begin{aligned} F'(x)&=\dfrac{1}{4}\,\mathrm{sech}^2\,x =\dfrac{1}{4}\left(\dfrac{2}{e^x+e^{-x}}\right)^2 \\ &=\left(\dfrac{1}{e^x+e^{-x}}\right)^2 =\left(\dfrac{e^x}{e^{2x}+1}\right)^2 =\dfrac{e^{2x}}{(e^{2x}+1)^2} \end{aligned}\] which is the original integrand.

Compute \(\displaystyle \int \dfrac{-x^2}{1-x^2}\,dx\) and \(\displaystyle \int_2^4 \dfrac{-x^2}{1-x^2}\,dx\).

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

\(\displaystyle \int \dfrac{-x^2}{1-x^2}\,dx=x-\mathrm{arctanh}\,x+C\)
\(\displaystyle \int_2^4 \dfrac{\rule{0pt}{12pt}-x^2}{1-x^2}\,dx =2-\mathrm{arctanh}\,4+\mathrm{arctanh}\,2\).

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

By the Fundamental Theorem of Calculus, the definite integral is: \[\begin{aligned} \int_2^4 \dfrac{-x^2}{1-x^2}\,dx &=\left[\rule{0pt}{10pt}x-\mathrm{arctanh}\,x\right]_2^4 \\ &=(4-\mathrm{arctanh}\,4)-(2-\mathrm{arctanh}\,2) \\ &=2-\mathrm{arctanh}\,4+\mathrm{arctanh}\,2 \end{aligned}\]

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