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Is the following evaluation of correct?

\begin{align*} \int e^x \sin^2 x \mathrm{d}x &= e^x \sin^2 x -2\int e^x \sin x \cos x \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (\cos^2 x - \sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int e^x (1 - 2\sin^2x) \mathrm{d}x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 \int -2 e^x \sin^2x \mathrm{d}x + 2 e^x \\ &= e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x -4 \int e^x \sin^2x \mathrm{d}x \end{align*}

First two steps use integration by parts. In the first step we differentiate $\sin^2 x$. In the second step we differentiate $\sin x \cos x$. Using this, we reach $$5\int e^x \sin^2 x \mathrm{d}x = e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x$$

$$\int e^x \sin^2 x \mathrm{d}x = \frac{e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x}{5}+C$$

I can't reach the form that most integral calculators give, which has terms $\cos(2x)$ and $\sin(2x)$ by just using trig identities, so I wonder whether the result is correct. I would also be interested in a method that immediately gives the form $$-\frac{e^x[2 \sin(2x)+ \cos(2x)-5]}{10}+C$$

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    $\begingroup$ What if you start with $\sin^2x=(1-\cos(2x))/2$? $\endgroup$ – mickep Aug 16 '15 at 7:04
  • $\begingroup$ WolframAlpha agrees: wolframalpha.com/input/… $\endgroup$ – Krijn Aug 16 '15 at 7:05
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    $\begingroup$ "I wonder whether the result is correct" - Differentiate your result and see if you get the integrand. If doesn't matter if your result looks equal to some other result or not. So long as its derivative matches the integrand, it's correct and equivalent to any other result whose derivative also matches the integrand. $\endgroup$ – wltrup Aug 16 '15 at 7:05
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Notice

$$\int e^x\sin^2x\mathrm{d}x=$$ $$=\int e^x\left(\frac{1-\cos 2x}{2}\right)\mathrm{d}x$$ $$=\frac{1}{2}\int e^xdx-\frac{1}{2}\int e^x \cos 2x \mathrm{d}x$$

Using $\displaystyle \int e^{ax}\cos (bx) \mathrm{d}x=\frac{e^{ax}}{a^2+b^2}(a\cos bx+b\sin bx)$, we get

$$=\frac{1}{2}e^x-\frac{1}{2}\frac{e^x}{1^2+2^2}(\cos 2x+2\sin 2x)+C$$ $$=\frac{1}{2}e^x-\frac{1}{10}e^x(\cos 2x+2\sin 2x)+C$$ $$=-\frac{e^x(2\sin 2x+\cos 2x-5)}{10}+C$$

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We have $$\int e^{x}\sin^{2}\left(x\right)dx=\frac{1}{2}\int e^{x}dx-\frac{1}{2}\int e^{x}\cos\left(2x\right)dx $$ and $$\int e^{x}\cos\left(2x\right)dx=\textrm{Re}\left(\int e^{x+2ix}dx\right) $$ then $$\int e^{x+2ix}dx=\frac{e^{x+2ix}}{1+2i} $$ and so $$\int e^{x}\sin^{2}\left(x\right)dx=\frac{1}{2}e^{x}-\frac{1}{2}e^{x}\left(\frac{\cos\left(2x\right)+2\sin\left(2x\right)}{5}\right)+C=$$ $$=-\frac{e^{x}\left(\cos\left(2x\right)+2\sin\left(2x\right)-5\right)}{10}+C. $$

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Let $$\displaystyle I = \frac{1}{2}\int e^x 2\sin^2 xdx = \frac{1}{2}\int e^x-\frac{1}{2}\int e^x \cos 2x dx$$

Now Using $$\displaystyle \cos \phi +i\sin \phi = e^{i\phi}$$ and $$\cos \phi-i\sin \phi = e^{-i\phi}.$$

So $$\displaystyle \cos \phi = \frac{e^{i\phi}+e^{-i\phi}}{2}$$

So we get $$\displaystyle \cos 2x = \left(\frac{e^{i2x}+e^{-i2x}}{2}\right)$$

so we get $$\displaystyle J = \bf{Re}\left[\int e^{x}\cdot \left(\frac{e^{i2x}+e^{-i2x}}{2}\right)dx\right] = \frac{1}{2}\bf{Re}\int \left[e^{(1+2i)x}+e^{(1-2i)x}]\right]dx$$

So we get $$\displaystyle J = \frac{1}{2}\bf{Re}\left[\frac{e^{(1+2i)x}}{(1+2i)}+\frac{e^{1-2i}}{(1-2i)}\right]$$

$$\displaystyle = \frac{e^x}{10}\bf{Re}\left[\left(\cos 2x+i\sin 2x\right)\cdot (1-2i)+\left(\cos 2x-i\sin 2x\right)\cdot (1+2i)\right]$$

So $$\displaystyle J = \frac{e^x}{10}\left[2\cos 2x+4\sin 2x\right] = \frac{e^x}{5}\left[\cos 2x+\sin 2x\right]$$

So $$\displaystyle I = \frac{e^x}{2}-\frac{e^x}{10}\left[\cos 2x+2\sin 2x\right]+\mathcal{C} = -\frac{e^x}{10}\left[\cos 2x+2\sin 2x-5\right]+\mathcal{C}$$

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You can use the Reduction formula:

$$I_n=\int e^{ax}\sin^n bx\mathrm. dx\\ =\frac{e^{ax}\sin^{n-1} bx (a\sin bx-nb\cos bx)}{a^2+n^2b^2}+\frac{n(n-1)b^2}{a^2+n^2b^2}I_{n-2}$$

Use $n=2,a=1,b=1$.

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Your solution is correct. To reach to required form See here,You allready Got this

$$\int e^x \sin^2 x \mathrm{d}x = \frac{e^x \sin^2 x -2e^x \sin x \cos x + 2 e^x}{5}+C$$

Now multiply and divide your result by $2$, you will get $$\frac{2e^x \sin^2 x -4e^x \sin x \cos x +4e^x}{10}+C\\ =\frac{2e^x \sin^2 x -4e^x \sin x \cos x +4e^x+e^x-e^x}{10}+C\\ =\frac{2e^x \sin^2 x-e^x -4e^x \sin x \cos x +5e^x}{10}+C\\ =\frac{e^x(2 \sin^2 x-1) -4e^x \sin x \cos x +5e^x}{10}+C\\ =-\frac{e^x[2 \sin(2x)+ \cos(2x)-5]}{10}+C$$

Where we used the identities

1) $2 \sin^2 x-1=-\cos 2x$

2)$2\sin x\cos x= \sin 2x $

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$$\int \left(e^x\sin^2(x)\right)\text{d}x =$$ $$\int \left(e^x\left(\frac{1}{2}(1-\cos(2x))\right)\right)\text{d}x =$$ $$\frac{1}{2}\int \left(e^x-e^x\cos(2x)\right)\text{d}x =$$ $$\frac{1}{2} \left(\int \left(e^x\right) \text{d}x-\int \left(e^x\cos(2x)\right) \text{d}x\right) =$$ $$\frac{1}{2} \left(\int e^x \text{d}x-\int e^x\cos(2x) \text{d}x\right) =$$ $$\frac{1}{2} \left(e^x-\int e^x\cos(2x) \text{d}x\right) =$$


For the integrand $e^x\cos(2x)$, use the formula:

$$\int\exp(\alpha x)\cos(\beta x)\text{d}x=\frac{\exp(\alpha x)(\alpha \cos(\beta x))+\beta\sin(\beta x)}{\alpha^2+\beta^2}$$


$$\frac{1}{2} \left(e^x-\frac{e^x(2\sin(2x)+\cos(2x))}{5}\right) + C =$$ $$-\frac{e^x(2\sin(2x)+\cos(2x)-5)}{10} + C $$

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