Using complex exponential to show the indefinite integration of sin(x)sinh(x) dx

Question

Use the complex exponential to evaluate the indefinite integral of $\sin x \sinh x$.

Express your answer in terms of trigonometric and/or hyperbolic functions

The attached photo is what I have tried so far

Answer 1

Method I

Using integration by parts twice,

\begin{align} I &=\int \sin ax \sinh bx \, dx \\ &= \frac{\sin ax \cosh bx}{b}-\int \frac{\cosh bx}{b} d(\sin ax) \\ &= \frac{\sin ax \cosh bx}{b}-\frac{a}{b} \int \cos ax \cosh bx \, dx \\ &= \frac{\sin ax \cosh bx}{b}-\frac{a}{b} \left[ \frac{\cos ax \sinh bx}{b}-\int \frac{\sinh bx}{b} d(\cos ax) \right] \\ &= \frac{\sin ax \cosh bx}{b}- \frac{a\cos ax \sinh bx}{b^2}- \frac{a^2}{b^2} \int \sin ax \sinh bx \, dx \\ I &= \frac{\sin ax \cosh bx}{b}- \frac{a\cos ax \sinh bx}{b^2}-\frac{a^2}{b^2}I+C \\ I &= \frac{b\sin ax \cosh bx-a\cos ax \sinh bx}{a^2+b^2}+C \end{align}

Method II

\begin{align} \cos (u+iv) &= \cos u \cosh v-i\sin u \sinh v \\ \sin (u+iv) &= \sin u \cosh v+i\cos u \sinh v \\ \int \cos [(a+ib)x] \, dx &= \int \cos ax \cosh bx \, dx- i\int \sin ax \sinh bx \, dx \\ \frac{\sin [(a+ib)x]}{a+ib} &= \int \cos ax \cosh bx \, dx-i\int \sin ax \sinh bx \, dx \\ \frac{(a-ib)(\sin ax \cosh bx+i\cos ax \sinh bx)}{a^2+b^2} &= \int \cos ax \cosh bx \, dx-i\int \sin ax \sinh bx \, dx \end{align}

Equating the imaginary parts,

$$\fbox{$ \int \sin ax \sinh bx \, dx= \frac{b\sin ax \cosh bx-a\cos ax \sinh bx}{a^2+b^2}+C $}$$

Method III

By brute force,

\begin{align} & \quad \int \sin ax \sinh bx \, dx \\[10pt] &= \int \left[ \frac{(e^{iax}-e^{-iax})(e^{bx}-e^{-bx})}{4i} \right] \, dx \\[10pt] &= \int \left[ \frac{e^{(b+ia)x}+e^{-(b+ia)x}-e^{(b-ia)x}-e^{(-b+ia)x}}{4i} \right] \, dx \\[10pt] &= \frac{e^{(b+ia)x}-e^{-(b+ia)x}}{4i(b+ia)} +\frac{-e^{(b-ia)x}+e^{-(b-ia)x}}{4i(b-ia)} \\[10pt] &= \frac{e^{(b+ia)x}-e^{-(b+ia)x}}{4(-a+ib)} +\frac{-e^{(b-ia)x}+e^{-(b-ia)x}}{4(a+ib)} \\[10pt] &= \frac{(-a-ib)[e^{(b+ia)x}-e^{-(b+ia)x}]}{4(a^2+b^2)} +\frac{(a-ib)[-e^{(b-ia)x}+e^{-(b-ia)x}]}{4(a^2+b^2)} \\[10pt] &= \frac{bi}{4(a^2+b^2)} [-e^{(b+ia)x}+e^{-(b+ia)x}+e^{(b-ia)x}-e^{-(b-ia)x}] \\ & \quad + \frac{a}{4(a^2+b^2)} [-e^{(b+ia)x}+e^{-(b+ia)x}-e^{(b-ia)x}+e^{-(b-ia)x}] \\[10pt] &= \frac{bi}{4(a^2+b^2)} [e^{bx}(-e^{iax}+e^{-iax})+e^{-bx}(e^{-iax}-e^{iax})] \\ & \quad + \frac{a}{4(a^2+b^2)} [e^{bx}(-e^{iax}-e^{-iax})+e^{-bx}(e^{-iax}+e^{iax})] \\[10pt] &= \frac{b}{a^2+b^2} \times \frac{e^{bx}+e^{-bx}}{2} \times \frac{e^{iax}-e^{-iax}}{2i}- \frac{a}{a^2+b^2} \times \frac{e^{bx}-e^{-bx}}{2} \times \frac{e^{iax}+e^{-iax}}{2} \\[10pt] &= \frac{b\sin ax \cosh bx-a\cos ax \sinh bx}{a^2+b^2} \end{align}

Answer 2

Start with $\sin x=\Im e^{ix}.$ Then your integral is the imaginary part of the following integral: $$\int e^{ix}\cdot\frac{e^x-e^{-x}}2\, dx=\frac 12 e^{ix}\cdot\left(\frac {e^x}{i+1}-\frac{e^{-x}}{i-1}\right)=\frac 12e^{ix}\cdot \frac{-(e^{-x}+e^x) + i(e^x-e^{-x})}{-2}=\\=\frac 12e^{ix}\cdot (\cosh x-i\sinh x).$$ As we are interested in the imaginary part, we can take use of the fact that $\Im ab=\Im a\cdot \Re b+\Re a\cdot \Im b$ and get the final result: $$\frac 12\left(\sin x\cosh x-\cos x\sinh x\right)$$

It agrees with the result from WolframAlpha.

Disclaimer: I omitted $+C$ for clarity. Many people dislike it, so consider adding it.

Answer 3

This can be not the only way to proceed, but it is convenient to avoid working with $e^{(i \pm 1) x}$ terms too much. Don't be afraid by the amount of lines, I simply did almost all the steps.

$$I = \int \sin x \sinh x \ \mathrm{d}x =\\ \int \frac{e^{ix} - e^{-ix}}{2i} \cdot \frac{e^{x} - e^{-x}}{2} \ \mathrm{d}x =\\ \frac{1}{4i} \int \left( e^{ix} e^x - e^{-ix}e^x - e^{ix}e^{-x} + e^{-ix}e^{-x} \right) \ \mathrm{d}x =\\ \frac{1}{4i} \int \left[ e^{(i + 1)x} - e^{(1 - i)x} - e^{(i - 1)x} + e^{-(i + 1)x} \right] \ \mathrm{d}x =\\ \frac{1}{4i} \int \left[ e^{(i + 1)x} - e^{-(i - 1)x} - e^{(i - 1)x} + e^{-(i + 1)x} \right] \ \mathrm{d}x =\\ \frac{1}{4i} \int \left[ e^{(i + 1)x} + e^{-(i + 1)x} - \left( e^{(i - 1)x} + e^{-(i - 1)x} \right) \right] \ \mathrm{d}x $$

What can be recognized here is a hyperbolic cosine (not a trigonometric cosine, because it would be $\cos \alpha = (e^{i \alpha} + e^{-i \alpha})/2$ with $\alpha$ real: here, instead, the exponents contain complex quantities, $(i \pm 1)x$): simply applying the definition

$$\cosh z = \frac{e^{z} + e^{- z}}{2}$$

The integral becomes:

$$I = \frac{1}{4i} \int \left[ 2 \cosh \left[ (i + 1)x \right] - 2 \cosh \left[ (i - 1)x \right] \right] \ \mathrm{d}x =\\ = \frac{1}{2i} \int \left[ \cosh \left[ (i + 1)x \right] - \cosh \left[ (i - 1)x \right] \right] \ \mathrm{d}x =\\$$

Hyperbolic functions, like trigonometric functions, have sum and subtraction identities:

$$\sinh(\alpha + \beta) = \sinh \alpha \cosh \beta + \cosh \alpha \sinh \beta\\ \sinh(\alpha - \beta) = \sinh \alpha \cosh \beta - \cosh \alpha \sinh \beta\\ \cosh(\alpha + \beta) = \cosh \alpha \cosh \beta + \sinh \alpha \sinh \beta\\ \cosh(\alpha - \beta) = \cosh \alpha \cosh \beta - \sinh \alpha \sinh \beta$$

The integrand function is currently of the type $\cosh(\alpha + \beta) - \cosh(\alpha - \beta)$, whose value would be (by side-by-side subtraction of the above identities):

$$\cosh(\alpha + \beta) - \cosh(\alpha - \beta) = 2 \sinh \alpha \sinh \beta$$

But this would lead you back to the starting point. So, it is not convenient to use those identities now. It is convenient to directly integrate:

$$I = \frac{1}{2i} \left\{ \frac{\sinh \left[ (i + 1)x \right]}{i + 1} - \frac{\sinh \left[ (i - 1)x \right]}{i - 1} \right\} + C = \\ \frac{1}{2i} \left\{ \frac{i - 1}{i - 1} \frac{\sinh \left[ (i + 1)x \right]}{i + 1} - \frac{i + 1}{i + 1} \frac{\sinh \left[ (i - 1)x \right]}{i - 1} \right\} + C = \\ \frac{1}{2i} \left\{ \frac{i \sinh \left[ (i + 1)x \right] - \sinh \left[ (i + 1)x \right]}{-2} - \frac{i \sinh \left[ (i - 1)x \right] + \sinh \left[ (i - 1)x \right]}{-2} \right\} + C = \\ \frac{1}{4i} \left\{ \sinh \left[ (i + 1)x \right] - i \sinh \left[ (i + 1)x \right] + i \sinh \left[ (i - 1)x \right] + \sinh \left[ (i - 1)x \right] \right\} + C = \\ \left\{ \frac{\sinh \left[ (i + 1)x \right]}{4i} - \frac{ \sinh \left[ (i + 1)x \right] }{4} + \frac{ \sinh \left[ (i - 1)x \right] }{4} + \frac{ \sinh \left[ (i - 1)x \right] }{4i} \right\} + C = \\ \left\{ \frac{\sinh (ix + x) + \sinh (ix - x)}{4i} - \frac{ \sinh (ix + x) - \sinh (ix - x) }{4} \right\} + C = \\ $$

Now, the sum/subtraction identities for the hyperbolic sine

$$\sinh(\alpha + \beta) + \sinh(\alpha - \beta) = 2 \sinh \alpha \cosh \beta\\ \sinh(\alpha + \beta) - \sinh(\alpha - \beta) = 2 \cosh \alpha \sinh \beta$$

can be applied.

$$I = \frac{2 \sinh (ix) \cosh (x)}{4i} - \frac{2 \cosh (ix) \sinh (x)}{4} + C$$

Applying the definitions,

$$\sinh(ix) = \frac{e^{ix} + e^{-ix}}{2} = i \sin x$$

and

$$\cosh(ix) = \frac{e^{ix} + e^{-ix}}{2} = \cos x$$

So,

$$I = \frac{2 i \sin (x) \cosh (x)}{4i} - \frac{2 \cos (x) \sinh (x)}{4} + C =\\ \frac{1}{2} \left[ \sin (x) \cosh (x) - \cos (x) \sinh (x) \right] + C$$

which is the expected result.