Integral of $\,e^x\!\sin(x)\,$ with substitution possible?

Question

In university we are currently learning about Integration by parts and with substitution and one of the Integrals we have to solve is the Integral of the function $\,e^x\!\sin(x)\,$.

From an exercise prior to this one, I know that the derivative of $\,e^x\!\big(\!\cos(x)-\sin(x)\big)$ is $\,e^x(-2)\sin(x)\,.$

So I could potentially show by differentiating the above function twice, that multiplying the second derivative with $\,-\dfrac12\,$ is the integral for $\,e^x\sin(x)\,.\,$ However we have to use Integration by parts (It is not permitted however to use some other technique that parts after integrating by parts at least once)

So the idea I came up with, after I integrated the function already once by parts and the result is

$\displaystyle\cos(x)\!\cdot\!e^x-e^x\!\cdot\!\sin(x)+\int\cos(x)\!\cdot\!e^x\mathrm dx$

to substitute $\,\cos(x)\,$ with $\,-\dfrac32\!\cdot\!\dfrac1{x^2}\,.$

This would result in $\,\displaystyle\int-\dfrac32\left(\dfrac1{x^2}\!\cdot\!e^x-\dfrac1x\!\cdot\!e^x\right)\!\mathrm dx=e^x\!\left(-\dfrac32\!\cdot\!\dfrac1{x^2}+\dfrac32\!\cdot\!\dfrac1x\right).$

Now my question is:

I can put cosine back as the identity without any problems, however, arithmetically, I could not figure out a way to show that $\,-\dfrac32\!\cdot\!\dfrac1x=\sin(x)\,$ based on the substitution I've chosen for cosine.

So is it possible to substitute $\,\dfrac32\!\cdot\!\dfrac1x\,$ back to $\,\sin(x)\,$ just by saying that $\,\dfrac32\!\cdot\!\dfrac1x\,$ is (one of) the stem function(s) for $\,-\dfrac32\!\cdot\!\dfrac1{x^2}\,$ and because $\,\sin(x)\,$ is the stem function for $\,\cos(x)\,,\,$ $\dfrac32\!\cdot\!\dfrac1x\,$ can be written as $\,\sin(x)\,?$

Or would that be considered mathematically faulty?

Thank you in advance for your answers.

Answer 1

Using the integration by parts we have

$u = e^x$ and $dv = \sin(x) dx$. Then, we find $du$ and $v$:

$$ \begin{align*} u &= e^x & \quad &dv = \sin(x) \,dx \\ du &= e^x \,dx & \quad &v = -\cos(x) \end{align*} $$

Now, apply the integration by parts formula:

$$ \int e^x \sin(x) \,dx = -e^x \cos(x) - \int (-\cos(x)) \cdot e^x \,dx$$

Solving the new integral:

Let $u = e^x$ and $dv = -\cos(x) \,dx$. Then, we find $du$ and $v$:

$$ \begin{align*} u &= e^x & \quad &dv = -\cos(x) \,dx \\ du &= e^x \,dx & \quad &v = -\sin(x) \end{align*} $$

Substitute these values into the integration by parts formula:

$$\int -\cos(x) \cdot e^x \,dx = -e^x \sin(x) - \int (-\sin(x)) \cdot e^x \,dx $$

Now, substitute this result back into the initial integration by parts formula:

$$\int e^x \sin(x) \,dx = -e^x \cos(x) + e^x \sin(x) - \int e^x \sin(x) \,dx $$

Add $\int e^x \sin(x) \,dx$ to both sides:

$$2 \int e^x \sin(x) \,dx = -e^x \cos(x) + e^x \sin(x)$$

Divide both sides by $2$:

$$ \int e^x \sin(x) \,dx = -\frac{1}{2}e^x \cos(x) + \frac{1}{2}e^x \sin(x) + C $$

where $C$ is the constant of integration.

Answer 2

You cannot substitute one function of $x$ with another (because they will not be equivalent for all $x$ in the domain of integration, right?). You would need to use another variable (like $t$), and change everything else in the integral, including the differential ($dt$) and the bounds if applicable.

As far as integrals of the form $\int e^{ax}\cos(bx) dx$, they can be directly computed without IBP by using Euler's theorem (substitute $\cos(x)$ with $(e^{ix}+e^{-ix})/2$ or $\sin(x)$ with $(e^{ix}-e^{-ix})/2i$).

Answer 3

I'm not sure if you have studied complex variables yet, but if so, you will recognize that

$$ \begin{align} I&=\int e^x\sin x dx \\ &= \mathcal{Im}\int e^x e^{ix} dx\\ &=\mathcal{Im}\frac{e^x e^{ix}}{1+i}\\ &=\mathcal{Im}\frac{e^x e^{ix} (1-i)}{2}\\ &=\frac{e^x (\sin x- \cos x)}{2}+\text{const.}\\ \end{align} $$

This is somewhat simpler than substituting $\sin x=\frac{e^{ix}-e^{-ix}}{2i}$.

Answer 4

You have to use IBP twice.

The first time gives you, as you have gotten, $\text{something} + \int e^x \cos(x) dx$.

Then, apply IBP to $\int e^x \cos(x) dx$ which will give you $\text{something else} + \int e^x \sin(x) dx$.

From this, solve for $\int e^x \sin(x) dx$.

Answer 5

You want to find

$$I=\int e^x \sin x dx.$$

We integrate by parts twice:, first set $u=\sin x, du = \cos x dx, v = e^x, dv = e^x dx$. Thus

\begin{align} I&=\int e^x \sin x dx\\ &=uv - \int v du\\ &=e^x \sin x - \int e^x \cos x dx \end{align}

Again, set $u=\cos x, du= - \sin x dx, v = e^x, dv = e^x dx.$

Thus

\begin{align} e^x \sin x - \int e^x \cos x dx&=e^x \sin x - \bigg(e^x\cos x + \int e^x \sin x dx \bigg)\\ \end{align}

Thus

$$I=e^x\sin x - e^x \cos x -I \Rightarrow 2I = e^x \sin x - e^x \cos x \Rightarrow I = \frac{1}{2}(e^x \sin x - e^x \cos x )+ C.$$

And we are done.