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Could someone show me how to solve this integral? $$\int_0^{\frac{\pi}{2}} e^{x+2}\sin(x) \,dx$$ I think that it's improper, but I'm not sure.
I tried to solve by parts, but first I sobstitute $$e^{x+2} = u$$ And this is what I obtained: $$\int{u\sin(\log(u) - 2)\frac{1}{u}}\,du$$ And at this point, integrating by part seems easy, but...
Could you help me? Maybe avoiding to follow the way that uses WolframAlpha because I don't know that method. Thanks in advance

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You can simplify matters a tad by writing $e^{x+2}=e^x\cdot e^2$ and factoring out $e^2$ from the integral. To find the antiderivative of $e^x\sin x$, integrate by parts twice. This will give you an equation with the desired antiderivative as the variable. Once you find the antiderivative, you can then compute the definite integral. – David Mitra Jun 16 '12 at 16:38
What is the "antiderivative"? – Overflowh Jun 16 '12 at 16:41
This question may help. – David Mitra Jun 16 '12 at 16:44
My usual comment: We're talking about evalutating an integral, not about "solving" an integral. One solves equations; one solves problems; one evaluates expressions. – Michael Hardy Jun 16 '12 at 16:52
@MichaelHardy I'm sorry.. I'm not familiar with mathematical english.. – Overflowh Jun 16 '12 at 17:18
up vote 3 down vote accepted

We can compute the indefinite integral first, and then apply Barrow's rule. We start by factoring out $e^2$, since it's a constant factor.

$\displaystyle \int e^{x+2}\sin x \,dx = e^2 \int e^x \sin x \,dx $

To evaluate $\int e^x \sin x \,dx$, we perform integration by parts $\int u \,dv = uv - \int v \,du$, with $u=e^x$ and $dv = \sin x \,dx$ first, and then again with $u = e^x$, $dv = \cos x \,dx$.

$\begin{align*} \int e^x \sin x \,dx &= - e^x \cos x - \int e^x (-\cos x) \,dx = -e^x \cos x + \int e^x \cos x \,dx = \\ &= - e^x \cos x + \big( e^x \sin x - \int e^x \sin x \,dx \big) = - e^x \cos x + e^x \sin x - \int e^x \sin x \,dx \end{align*}$

Note that we get back the original integral, so we can solve for it. This is sometimes called a cyclic integral.

$\displaystyle 2 \int e^x \sin x \,dx = e^x (\sin x - \cos x)$, and then $\displaystyle \int e^x \sin x \,dx = \frac{1}{2} e^x(\sin x - \cos x)$.

Finally, we have, with Barrow's rule:

$\begin{align*} \int_0^{\pi/2} e^{x+2}\sin x \,dx &= e^2 \int_0^{\pi/2} e^x \sin x \,dx = e^2\big(\frac{1}{2} e^x(\sin x - \cos x)\big|_0^{\pi/2}\big) \\ &= e^2 \big( \frac{1}{2}e^{\pi/2}(\sin {\frac{\pi}{2}} - \cos {\frac{\pi}{2}}) - \frac{1}{2}e^0(\sin 0 - \cos 0)\big) \\ &= \frac{1}{2}e^2 \big(e^{\pi/2}(1 - 0) - (0 - 1)\big) = \frac{1}{2}e^2(1+e^{\pi/2}) \end{align*}$

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It's very clear and helpful! I was worried about recursive integration, but now It's really clear :) Thank you very much! – Overflowh Jun 16 '12 at 18:00

$\sin(x) = \Im(e^{ix})$, so $$\begin{aligned} \int_0^{\pi/2} e^{x+2}\sin(x) \mathrm d x &= \Im\left(\int _0^{\pi/2} e^{2+(1+i)x} \mathrm d x\right)\\ &=e^2 \cdot \Im\left(\frac{i e^{\pi/2} - 1}{1+i}\right)\\ &= \frac{e^2}{2}(e^{\pi/2} +1 ). \end{aligned}$$

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The other answers do not have your minus sign. – GEdgar Jun 16 '12 at 17:35
@NateEldredge — Thanks, corrected ! – Lierre Jun 16 '12 at 19:54
@GEdgar — Sorry, my fault, that's corrected. – Lierre Jun 16 '12 at 19:54
+1, I think this is by far the simplest solution. – Nate Eldredge Jun 16 '12 at 20:02


Then $$\int^\frac{\pi}{2}_0e^x\sin(x)dx=\frac{e^\frac{\pi}{2}+1}{2}$$

And so $$\int^\frac{\pi}{2}_0e^{x+2}\sin(x)dx=e^2\left(\frac{e^\frac{\pi}{2}+1}{2}\right)$$

(The first step is an integration per parts with $u=e^x$ and $v'=\sin(x)$ and the second step is an other ipp with $u=e^x$ and $v'=\cos(x)$)

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Integrating by parts, we let $u = e^{x+2}$, $dv = \sin(x) \ dx$. Then $du = e^{x+2} \ dx$ and $v = -\cos(x)$, and so

$$ \int_0^{\frac\pi2}e^{x+2}\sin(x)\ dx = \left.-e^{x+2}\cos(x)\right|_0^{\frac\pi2} - \int_0^{\frac\pi2}e^{x+2}(-\cos(x)) \ dx $$ $$ = \left.-e^{x+2}\cos(x)\right|_0^{\frac\pi2} + \int_0^{\frac\pi2}e^{x+2}\cos(x) \ dx = \left.-e^{x+2}\cos(x)\right|_0^{\frac\pi2} + I $$

Now we concentrate on $I = \int_0^{\frac\pi2}e^{x+2}\cos(x) \ dx$. Again, letting $u = e^{x+2}$ and $dv = \cos(x) \ dx$, we have that $du = e^{x+2} \ dx$ and $v = \sin(x)$.

$$ I = \int_0^{\frac\pi2}e^{x+2}\cos(x) \ dx = \left.e^{x+2}\sin(x)\right|_0^{\frac\pi2} - \int_0^{\frac\pi2}e^{x+2}\sin(x) \ dx $$

Substituting what we have so far,

$$ \int_0^{\frac\pi2}e^{x+2}\sin(x)\ dx = \left.-e^{x+2}\cos(x)\right|_0^{\frac\pi2} + I $$ $$= \left.\left(-e^{x+2}\cos(x) + e^{x+2}\sin(x)\right)\right|_0^{\frac\pi2} - \int_0^{\frac\pi2}e^{x+2}\sin(x) \ dx $$

Now rearranging the terms, we have that

$$ 2\int_0^{\frac\pi2}e^{x+2}\sin(x)\ dx = \left.\left(-e^{x+2}\cos(x) + e^{x+2}\sin(x)\right)\right|_0^{\frac\pi2} $$

and so

$$ \int_0^{\frac\pi2}e^{x+2}\sin(x)\ dx = \left.\frac12e^{x+2}\left(\sin(x) - \cos(x)\right)\right|_0^{\frac\pi2} =\frac12e^{\frac\pi2 + 2} + \frac12e^2 $$

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