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 ddx(sinh⁡x)=cosh⁡x\dfrac{d}{dx}(\sinh x)=\cosh x ∫cosh⁡xdx=sinh⁡x+C\displaystyle \int \cosh x\,dx=\sinh x+C
Cosine ddx(cosh⁡x)=sinh⁡x\dfrac{d}{dx}(\cosh x)=\sinh x ∫sinh⁡xdx=cosh⁡x+C\displaystyle \int \sinh x\,dx=\cosh x+C
Tangent3^\text{3} ddx(tanh⁡x)=sech2x\dfrac{d}{dx}(\tanh x)=\,\mathrm{sech}\,^2 x ∫sech2xdx=tanh⁡x+C\displaystyle \int \mathrm{sech}^2 x\,dx=\tanh x+C
Cotangent3^\text{3} ddx(coth⁡x)=−csch2x\dfrac{d}{dx}(\coth x)=-\,\mathrm{csch}^2 x ∫csch2xdx=−coth⁡x+C\displaystyle \int \mathrm{csch}^2 x\,dx=-\coth x+C
Secant3^\text{3} ddx(sechx)=−sechxtanh⁡x\dfrac{d}{dx}(\mathrm{sech}\,x) =-\,\mathrm{sech}\,x\tanh x ∫sechxtanh⁡xdx=−sechx+C\displaystyle \int \mathrm{sech}\,x\tanh x\,dx=-\,\mathrm{sech}\,x+C
Cosecant3^\text{3} ddx(cschx)=−cschxcoth⁡x\dfrac{d}{dx}(\mathrm{csch}\,x)=-\,\mathrm{csch}\,x\coth x ∫cschxcoth⁡xdx=−cschx+C\displaystyle \int \mathrm{csch}\,x\coth x\,dx=-\,\mathrm{csch}x+C

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

Inverse Hyperbolic Trig Functions
Derivative Rule Integral Rule
Inverse
Hyperbolic
Sine4^\text{4}		 ddx(arcsinhx)=11+x2\dfrac{d}{dx}(\mathrm{arcsinh}\,x) =\dfrac{1}{\sqrt{1+x^2}} ∫11+x2dx=arcsinhx+C\displaystyle \int \dfrac{1}{\sqrt{1+x^2}}\,dx=\,\mathrm{arcsinh}\,x+C
Inverse
Hyperbolic
Tangent4^\text{4}		 ddx(arctanhx)=11−x2\dfrac{d}{dx}(\mathrm{arctanh}x) =\dfrac{1}{1-x^2} ∫11−x2dx=arctanhx+C\displaystyle \int \dfrac{1}{1-x^2}\,dx=\,\mathrm{arctanh}\,x+C
Inverse
Hyperbolic
Secant4^\text{4}		 ddx(arcsechx)=−1x2(1−x2)\dfrac{d}{dx}(\mathrm{arcsech}\,x) =\dfrac{-1}{\sqrt{x^2(1-x^2)}} ∫1x2(1−x2)dx=−arcsechx+C\displaystyle \int \dfrac{1}{\sqrt{x^2(1-x^2)}}\,dx=-\mathrm{arcsech}\, x+C
Inverse
Hyperbolic
Cosine4^\text{4}		 ddx(arccoshx)=1x2−1\dfrac{d}{dx}(\mathrm{arccosh}x)=\dfrac{1}{\sqrt{x^2-1}} ∫1x2−1dx=arccosh+C\displaystyle \int \dfrac{1}{\sqrt{x^2-1}}\,dx=\mathrm{arccosh}\,+C
Inverse
Hyperbolic
Cotangent4^\text{4}		 ddx(arccothx)=−1x2−1\dfrac{d}{dx}(\mathrm{arccoth}\,x)=\dfrac{-1}{x^2-1} ∫1x2−1dx=−arccothx+C\displaystyle \int \dfrac{1}{x^2-1}\,dx=-\mathrm{arccoth}\, x+C
Inverse
Hyperbolic
Cosecant4^\text{4}		 ddx(arccschx)=−1x2(x2−1)\dfrac{d}{dx}(\mathrm{arccsch}\,x)=\dfrac{-1}{\sqrt{x^2(x^2-1)}} ∫1x2(x2−1)dx=−arccschx+C\displaystyle \int \dfrac{1}{\sqrt{x^2(x^2-1)}}\,dx=-\mathrm{arccsch}\, x+C

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

Compute ∫e2x(e2x+1)2dx\displaystyle \int \dfrac{e^{2x}}{(e^{2x}+1)^2}\,dx and ∫01e2x(e2x+1)2dx\displaystyle \int_0^1 \dfrac{e^{2x}}{(e^{2x}+1)^2}\,dx.

Hint

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

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Answer

∫e2x(e2x+1)2dx=14tanh⁡x+C=14ex−e−xex+e−x+C\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
∫01e2x(e2x+1)2dx=14tanh⁡1=14e2−1e2+1\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}

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Solution

We write the integrand as a perfect square. Then we divide numerator and denominator by exe^x, and identify the integrand as the square of a hyperbolic trig function. ∫e2x(e2x+1)2dx=∫(exe2x+1)2dx=∫(1ex+e−x)2dx=∫(12cosh⁡x)2dx=∫(sechx2)2dx=∫14sech2xdx=14tanh⁡x+C=14ex−e−xex+e−x+C\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: ∫01e2x(e2x−1)2dx=[14tanh⁡x]01=14tanh⁡1−14tanh⁡0=14tanh⁡1\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\tanh 0=0. In terms of exponentials, ∫01e2x(e2x−1)2dx=14e−e−1e+e−1=14e2−1e2+1 \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}

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Check

We check the indefinite integral by differentiating. If F=14tanh⁡xF=\dfrac{1}{4}\tanh x then: F′(x)=14sech2x=14(2ex+e−x)2=(1ex+e−x)2=(exe2x+1)2=e2x(e2x+1)2\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.

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Compute ∫−x21−x2dx\displaystyle \int \dfrac{-x^2}{1-x^2}\,dx and ∫24−x21−x2dx\displaystyle \int_2^4 \dfrac{-x^2}{1-x^2}\,dx.

Hint

Add and subtract 11 from the numerator.

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Answer

∫−x21−x2dx=x−arctanhx+C\displaystyle \int \dfrac{-x^2}{1-x^2}\,dx=x-\mathrm{arctanh}\,x+C
∫24−x21−x2dx=2−arctanh4+arctanh2\displaystyle \int_2^4 \dfrac{\rule{0pt}{12pt}-x^2}{1-x^2}\,dx =2-\mathrm{arctanh}\,4+\mathrm{arctanh}\,2.

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Solution

We first use a trick of adding and subtracting 11 from the numerator. So the integrand becomes: −x21−x2=1−x2−11−x2=1−11−x2 \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: ∫−x21−x2dx=∫(1−11−x2)dx=x−arctanhx+C \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: ∫24−x21−x2dx=[x−arctanhx]24=(4−arctanh4)−(2−arctanh2)=2−arctanh4+arctanh2\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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