Integration of a product of an exponential and a trigonometric/hyperbolic function

수학 모음 (Maths collection) 전체보기

This post has been partly motivated by

• P2 §11.6 Integration by parts
• CP2 §6.5 Integrating hyperbolic functions
• CP2 §7.1 First-order differential equations (obtaining the particular integrals using the integrating factor)

Question. Derive the following results. $$ \begin{align}     \textrm{(a)} &&& I_1(a,b) = \int \textrm e^{ ax } \cos b x \, \textrm dx = \frac{ \textrm e^{ ax } ( a \cos bx + b \sin bx ) }{ a^2 + b^2 } + C \\ \\   \textrm{(b)} &&& I_2(a,b) = \int \textrm e^{ ax } \sin b x \, \textrm dx = \frac{ \textrm e^{ ax } ( a \sin bx - b \cos bx ) }{ a^2 + b^2 } + C \\ \\   \textrm{(c)} &&& I_3(a,b) = \int \textrm e^{ ax } \cosh b x \, \textrm dx = \frac{ \textrm e^{ ax } ( a \cosh bx - b \sinh bx ) }{ a^2 - b^2 } + C, \quad a^2 \ne b^2 \\ &&& I_3(a,a) = \int \textrm e^{ ax } \cosh a x \, \textrm dx = \frac1{ 4a }\textrm e^{ 2ax } + \frac12 x + C \\ \\   \textrm{(d)} &&& I_4(a,b) = \int \textrm e^{ ax } \sinh b x \, \textrm dx = \frac{ \textrm e^{ ax } ( a \sinh bx - b \cosh bx ) }{ a^2 - b^2 } + C, \quad a^2 \ne b^2 \\ &&& I_4(a,a) = \int \textrm e^{ ax } \sinh a x \, \textrm dx = \frac1{ 4a }\textrm e^{ 2ax } - \frac12 x + C \end{align} $$

 

(a) For $$ \begin{align} I_1 = \int \textrm e^{ ax } \cos b x \, \textrm dx \end{align} $$ (i) Method 1. By integration by parts twice: $$ \begin{align} u &= \textrm e^{ ax } & v' &= \cos bx \\ u' &= a \textrm e^{ ax } & v &= \frac1b \sin bx \end{align} $$ We find: $$ \begin{align} \int uv' \, \textrm dx &= uv - \int u'v \, \textrm dx \\ \\ \Rightarrow \qquad I_1 = \int \textrm e^{ ax } \cos b x \, \textrm dx &= \frac1b \textrm e^{ ax } \sin bx - \frac{a}{b} \int \textrm e^{ ax } \sin bx \, \textrm dx \end{align} $$ For the last integral, we need to apply integration by parts again, $$ \begin{align} u &= \textrm e^{ ax } & v' &= \sin bx \\ u' &= a \textrm e^{ ax } & v &= - \frac1b \cos bx \end{align} $$ which gives $$ \begin{align} &\Rightarrow & I_1 = \int \textrm e^{ ax } \cos b x \, \textrm dx &= \frac1b \textrm e^{ ax } \sin bx - \frac{a}{b} \left( - \frac1b \textrm e^{ ax } \cos bx + \frac{a}{b} \int \textrm e^{ ax } \cos bx \, \textrm dx \right) \\ &&&= \frac1b \textrm e^{ ax } \sin bx + \frac{a}{b^2} \textrm e^{ ax } \cos bx - \frac{a^2}{b^2} \underbrace{ \int \textrm e^{ ax } \cos bx \, \textrm dx }_{ I_1 } \\ \\ &\Rightarrow & \left( 1 + \frac{a^2}{b^2} \right) I_1 &= \textrm e^{ ax } \left( \frac{ a \cos bx + b \sin bx }{ b^2 } \right) \\ \\ &\Rightarrow & I_1 &= \frac{ \textrm e^{ ax } ( a \cos bx + b \sin bx ) }{ a^2 + b^2 } + C \qquad \checkmark \end{align} $$

 

(ii) Method 2. Using Euler's formula: $$ \begin{align} \textrm e^{ ix } = \cos x + i \sin x \end{align} $$ We may evaluate:   $$ \begin{align} \int \textrm e^{ ax } e^{ i b x } \, \textrm{d} x &= \int \textrm e^{ ( a + ib ) x } \, \textrm{d} x \\ &= \frac{ 1 }{ a + ib } \textrm e^{ ( a + ib ) x } + C \\ &= \frac{ a - i b }{ a^2 + b^2 } \textrm e^{ ax } \Big( \cos b x + i \sin b x \Big) + C \\ &= \textrm e^{ax} \left( \frac{ a \cos b x + b \sin b x }{ a^2 + b^2 } \right) + i \textrm e^{ax} \left( \frac{ a \sin b x - b \cos b x }{ a^2 + b^2 } \right) + C \end{align} $$ Taking the real and imaginary parts separately gives $I_1$ and $I_2$, respectively: $$ \begin{align} I_1 = \textrm{Re} \left( \int \textrm e^{ ax } e^{ i b x } \, \textrm{d} x \right) &= \frac{ \textrm e^{ax} ( a \cos b x + b \sin b x ) }{ a^2 + b^2 } + C \qquad \checkmark \\ \\ I_2 = \textrm{Im} \left( \int \textrm e^{ ax } e^{ i b x } \, \textrm{d} x \right) &= \frac{ \textrm e^{ax} ( a \sin b x - b \cos b x ) }{ a^2 + b^2 } + C \qquad \checkmark \end{align} $$

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(b) For $$ \begin{align} I_2 = \int \textrm e^{ ax } \sin b x \, \textrm dx \end{align} $$ (i) Method 1. By integration by parts twice: $$ \begin{align} u &= \textrm e^{ ax } & v' &= \sin bx \\ u' &= a \textrm e^{ ax } & v &= - \frac1b \cos bx \end{align} $$ We find: $$ \begin{align} \int uv' \, \textrm dx &= uv - \int u'v \, \textrm dx \\ \\ \Rightarrow \qquad I_1 = \int \textrm e^{ ax } \sin b x \, \textrm dx &= - \frac1b \textrm e^{ ax } \cos bx + \frac{a}{b} \int \textrm e^{ ax } \cos bx \, \textrm dx \end{align} $$ For the last integral, we need to apply integration by parts again, $$ \begin{align} u &= \textrm e^{ ax } & v' &= \cos bx \\ u' &= a \textrm e^{ ax } & v &= \frac1b \sin bx \end{align} $$ which gives $$ \begin{align} &\Rightarrow & I_2 = \int \textrm e^{ ax } \cos b x \, \textrm dx &= - \frac1b \textrm e^{ ax } \cos bx + \frac{a}{b} \left( \frac1b \textrm e^{ ax } \sin bx - \frac{a}{b} \int \textrm e^{ ax } \sin bx \, \textrm dx \right) \\ &&&= - \frac1b \textrm e^{ ax } \cos bx + \frac{a}{b^2} \textrm e^{ ax } \sin bx - \frac{a^2}{b^2} \underbrace{ \int \textrm e^{ ax } \sin bx \, \textrm dx }_{ I_2 } \\ \\ &\Rightarrow & \left( 1 + \frac{a^2}{b^2} \right) I_2 &= \textrm e^{ ax } \left( \frac{ a \sin bx - b \cos bx }{ b^2 } \right) \\ \\ &\Rightarrow & I_2 &= \frac{ \textrm e^{ ax } ( a \sin bx - b \cos bx ) }{ a^2 + b^2 } + C \qquad \checkmark \end{align} $$

 

(ii) Method 2. Using Euler's formula: See Method 2 for part (a).

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(c) For $$ \begin{align} I_3 = \int \textrm e^{ ax } \cosh b x \, \textrm dx \end{align} $$ (i) Method 1. By integration by parts twice: $$ \begin{align} u &= \textrm e^{ ax } & v' &= \cosh bx \\ u' &= a \textrm e^{ ax } & v &= \frac1b \sinh bx \end{align} $$ We find: $$ \begin{align} \int uv' \, \textrm dx &= uv - \int u'v \, \textrm dx \\ \\ \Rightarrow \qquad I_3 = \int \textrm e^{ ax } \cosh b x \, \textrm dx &= \frac1b \textrm e^{ ax } \sinh bx - \frac{a}{b} \int \textrm e^{ ax } \sinh bx \, \textrm dx \end{align} $$ For the last integral, we need to apply integration by parts again, $$ \begin{align} u &= \textrm e^{ ax } & v' &= \sinh bx \\ u' &= a \textrm e^{ ax } & v &= \frac1b \cosh bx \end{align} $$ which gives $$ \begin{align} &\Rightarrow & I_3 = \int \textrm e^{ ax } \cosh b x \, \textrm dx &= \frac1b \textrm e^{ ax } \sinh bx - \frac{a}{b} \left( \frac1b \textrm e^{ ax } \cosh bx - \frac{a}{b} \int \textrm e^{ ax } \cosh bx \, \textrm dx \right) \\ &&&= \frac1b \textrm e^{ ax } \sinh bx - \frac{a}{b^2} \textrm e^{ ax } \cosh bx + \frac{a^2}{b^2} \underbrace{ \int \textrm e^{ ax } \cosh bx \, \textrm dx }_{ I_3 } \qquad (*) \\ \\ &\Rightarrow & \left( \frac{a^2}{b^2} - 1 \right) I_3 &= \textrm e^{ ax } \left( \frac{ a \cosh bx - b \sinh bx }{ b^2 } \right) \\ \\ &\Rightarrow & I_3 &= \frac{ \textrm e^{ ax } ( a \cosh bx - b \sinh bx ) }{ a^2 - b^2 } + C \qquad \left( \textrm{for} \quad a^2 \ne b^2 \right) \end{align} $$ For $a^2=b^2$, the equation (*) reads $$ \begin{align} I_3 &= \frac1a \textrm e^{ ax } \sinh ax - \frac{1}{a} \textrm e^{ ax } \cosh ax + \underbrace{ \int \textrm e^{ ax } \cosh ax \, \textrm dx }_{ I_3 } \\ &= - \frac1a \textrm e^{ ax } \underbrace{ ( \cosh ax - \sinh ax ) }_{ \textrm e^{-ax} } + I_3 + C \\ &= - \frac1a + I_3 + C \end{align} $$ and $I_3$ cancels from both sides, i.e. when $a^2=b^2$, applying integration by part twice does not yield a useful result, only leaving a condition on the integration constant, namely, $C = \frac1a$.

 

Aside. There are similar situations where integration by parts yields no definite answers.

For example: (See P2 §11.6 Integration by parts for more details) $$ \begin{align} I = \int \frac{ \ln x }{ x } \, \textrm dx \end{align} $$ By integration by parts, $$ \begin{align} u &= \frac1x, & v' &= \ln x \\ u' &= - \frac1{x^2}, & v &= x \ln x - x \end{align} $$ we find: $$ \begin{align} \Rightarrow \qquad &\Rightarrow& I &= \frac{ x \ln x - x }{ x } + \int \frac{ x \ln x - x }{ x^2 } \, \textrm dx \\ &&&= \ln x - 1 + \underbrace{ \int \frac{ \ln x }{ x } \, \textrm dx }_{ I } - \ln x + C \\ &&&= \ln x - 1 + I - \ln x + C \end{align} $$ which requires $ C = 1 $.

 

(ii) Method 2. Using the exponential definition for $\cosh bx$: $$ \begin{align} \cosh bx = \frac{ \textrm e^{ bx } + \textrm e^{ -bx } }{ 2 } \end{align} $$ We may evaluate: $$ \begin{align} I_3 &= \int \textrm e^{ ax } \cosh bx \, \textrm{d} x \\ &= \int \textrm e^{ ax } \left( \frac{ \textrm e^{ bx } + \textrm e^{ -bx } }{ 2 } \right) \textrm{d} x \\ &= \frac12 \int \Big( \textrm e^{ (a+b)x } + \textrm e^{ (a-b)x } \Big) \textrm dx \end{align} $$ (i) For $b=a$, $$ \begin{align} I_3 = \int \textrm e^{ ax } \cosh a x \, \textrm dx &= \frac12 \int \Big( \textrm e^{ 2ax } + 1 \Big) \textrm dx \\ &= \frac1{ 4a }\textrm e^{ 2ax } + \frac12 x + C \end{align} $$ (ii) For $b=-a$, $$ \begin{align} I_3 = \int \textrm e^{ ax } \cosh a x \, \textrm dx = &= \frac12 \int \Big( 1 + \textrm e^{ 2ax } \Big) \textrm dx \\ &= \frac12 x + \frac1{ 4a } \textrm e^{ 2ax } + C \end{align} $$ (iii) For $b^2 \ne a^2$, $$ \begin{align} I_3 = \int \textrm e^{ ax } \cosh b x \, \textrm dx &= \frac12 \int \Big( \textrm e^{ (a+b)x } + \textrm e^{ (a-b)x } \Big) \textrm dx \\ &= \frac12 \left( \frac{ \textrm e^{ (a+b)x } }{ a + b } + \frac{ \textrm e^{ (a-b)x } }{ a - b } \right) + C \\ &= \frac12 \textrm e^{ ax } \left( \frac{ (a-b) \textrm e^{ bx } + (a+b) \textrm e^{ -bx } }{ a^2 - b^2 } \right) + C \\ &= \frac12 \textrm e^{ ax } \left( \frac{ a \left( \textrm e^{ bx } + \textrm e^{ -bx } \right) - b \left( \textrm e^{ bx } - \textrm e^{ -bx } \right) }{ a^2 - b^2 } \right) + C \\ &= \frac{ \textrm e^{ ax } ( a \cosh bx - b \sinh bx ) }{ a^2 - b^2 } + C \\ \end{align} $$

The final result may be summarised as: $$ \begin{align} I_3 = \int \textrm e^{ ax } \cosh b x \, \textrm dx = \left\{ \begin{array}{lcl} \frac{ \textrm e^{ ax } ( a \cosh bx - b \sinh bx ) }{ a^2 - b^2 } + C & \textrm{for} & b^2 \ne a^2 \\ \frac1{ 4a }\textrm e^{ 2ax } + \frac12 x + C & \textrm{for} & b^2 = a^2 \end{array} \right. \end{align} $$

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(d) For $$ \begin{align} I_4 = \int \textrm e^{ ax } \sinh b x \, \textrm dx \end{align} $$

(i) Method 1. By integration by parts twice: $$ \begin{align} u &= \textrm e^{ ax } & v' &= \sinh bx \\ u' &= a \textrm e^{ ax } & v &= \frac1b \cosh bx \end{align} $$ We find: $$ \begin{align} \int uv' \, \textrm dx &= uv - \int u'v \, \textrm dx \\ \\ \Rightarrow \qquad I_4 = \int \textrm e^{ ax } \sinh b x \, \textrm dx &= \frac1b \textrm e^{ ax } \cosh bx - \frac{a}{b} \int \textrm e^{ ax } \cosh bx \, \textrm dx \end{align} $$ For the last integral, we need to apply integration by parts again, $$ \begin{align} u &= \textrm e^{ ax } & v' &= \cosh bx \\ u' &= a \textrm e^{ ax } & v &= \frac1b \sinh bx \end{align} $$ which gives $$ \begin{align} &\Rightarrow & I_4 = \int \textrm e^{ ax } \sinh b x \, \textrm dx &= \frac1b \textrm e^{ ax } \cosh bx - \frac{a}{b} \left( \frac1b \textrm e^{ ax } \sinh bx - \frac{a}{b} \int \textrm e^{ ax } \sinh bx \, \textrm dx \right) \\ &&&= \frac1b \textrm e^{ ax } \cosh bx - \frac{a}{b^2} \textrm e^{ ax } \sinh bx + \frac{a^2}{b^2} \underbrace{ \int \textrm e^{ ax } \sinh bx \, \textrm dx }_{ I_4 } \qquad (**) \\ \\ &\Rightarrow & \left( \frac{a^2}{b^2} - 1 \right) I_4 &= \textrm e^{ ax } \left( \frac{ a \sinh bx - b \cosh bx }{ b^2 } \right) \\ \\ &\Rightarrow & I_3 &= \frac{ \textrm e^{ ax } ( a \sinh bx - b \cosh bx ) }{ a^2 - b^2 } + C \qquad \left( \textrm{for} \quad a^2 \ne b^2 \right) \end{align} $$

For $b=a$, the equation (**) reads $$ \begin{align} I_4 &= \frac1a \textrm e^{ ax } \cosh ax - \frac{1}{a} \textrm e^{ ax } \sinh ax + \underbrace{ \int \textrm e^{ ax } \sinh ax \, \textrm dx }_{ I_4 } \\ &= \frac1a \textrm e^{ ax } \underbrace{ ( \cosh ax - \sinh ax ) }_{ \textrm e^{-ax} } + I_4 + C \\ &= \frac1a + I_4 + C \end{align} $$ and $I_4$ cancels from both sides leaving a condition, $C=-\frac1a$, in a similar fashion to the case (c).

 

For $b=-a$, the equation (**) reads $$ \begin{align} I_4 = - \int \textrm e^{ ax } \sinh a x \, \textrm dx &= - \frac1a \textrm e^{ ax } \cosh ax + \frac{1}{a} \textrm e^{ ax } \sinh ax - \underbrace{ \int \textrm e^{ ax } \sinh ax \, \textrm dx }_{ = -I_4 } \\ &= - \frac1a \textrm e^{ ax } \underbrace{ ( \cosh ax - \sinh ax ) }_{ \textrm e^{-ax} } + I_4 + C \\ &= - \frac1a + I_4 + C \end{align} $$ and $I_4$ cancels from both sides leaving a condition, $C=\frac1a$. i.e. As in (c), when $a^2=b^2$, applying integration by part twice does not yield a useful result, only leaving a condition on the integration constant, namely, $C = \pm\frac1a$.

 

(ii) Method 2. Using the exponential definition for $\sinh bx$: $$ \begin{align} \sinh bx = \frac{ \textrm e^{ bx } - \textrm e^{ -bx } }{ 2 } \end{align} $$ We may evaluate: $$ \begin{align} I_4 &= \int \textrm e^{ ax } \sinh bx \, \textrm{d} x \\ &= \int \textrm e^{ ax } \left( \frac{ \textrm e^{ bx } - \textrm e^{ -bx } }{ 2 } \right) \textrm{d} x \\ &= \frac12 \int \Big( \textrm e^{ (a+b)x } - \textrm e^{ (a-b)x } \Big) \textrm dx \end{align} $$ (i) For $a=b$, $$ \begin{align} I_4 = \int \textrm e^{ ax } \sinh a x \, \textrm dx &= \frac12 \int \Big( \textrm e^{ 2ax } + 1 \Big) \textrm dx \\ &= \frac1{ 4a }\textrm e^{ 2ax } - \frac12 x + C \end{align} $$ (ii) For $a=-b$, $$ \begin{align} I_4 = - \int \textrm e^{ ax } \sinh a x \, \textrm dx &= \frac12 \int \Big( 1 - \textrm e^{ 2ax } \Big) \textrm dx \\ &= \frac12 x - \frac1{ 4a } \textrm e^{ 2ax } + C \end{align} $$ (iii) For $a^2 \ne b^2$, $$ \begin{align} I_4 &= \frac12 \int \Big( \textrm e^{ (a+b)x } - \textrm e^{ (a-b)x } \Big) \textrm dx \\ &= \frac12 \left( \frac{ \textrm e^{ (a+b)x } }{ a + b } - \frac{ \textrm e^{ (a-b)x } }{ a - b } \right) + C \\ &= \frac12 \textrm e^{ ax } \left( \frac{ (a-b) \textrm e^{ bx } - (a+b) \textrm e^{ -bx } }{ a^2 - b^2 } \right) + C \\ &= \frac12 \textrm e^{ ax } \left( \frac{ a \left( \textrm e^{ bx } - \textrm e^{ -bx } \right) - b \left( \textrm e^{ bx } + \textrm e^{ -bx } \right) }{ a^2 - b^2 } \right) + C \\ &= \frac{ \textrm e^{ ax } ( a \sinh bx - b \cosh bx ) }{ a^2 - b^2 } + C \\ \end{align} $$

The final result may be summarised as: $$ \begin{align} I_4 = \int \textrm e^{ ax } \sinh b x \, \textrm dx = \left\{ \begin{array}{lcl} \frac{ \textrm e^{ ax } ( a \sinh bx - b \cosh bx ) }{ a^2 - b^2 } + C & \textrm{for} & b^2 \ne a^2 \\ \frac1{ 4a }\textrm e^{ 2ax } - \frac12 x + C & \textrm{for} & b = a \end{array} \right. \end{align} $$