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Surely it's easier to use substitution and trig identities, but I wonder if it's possible to use integration by parts. Here's what I tried

$$\int \sin^3(x) \cos^2(x)dx$$

Let: $u'=\cos (x), u = \sin (x), v = \sin (x) , v'=\cos (x)$

$$\int v^3 u'^2$$

And here's where I'm stuck, how do I go about this? I'm not sure how to do integration by parts with such exponents

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    $\begingroup$ dont forget the integration symbol dx, you must replace it with some d(new variable) for the integral to make any sense $\endgroup$ – mathreadler Jan 31 '18 at 20:16
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split your integral and write $$\sin(x)^3(1-\sin(x)^2)=\sin(x)^3-\sin(x)^5$$ furthere use that $$\sin(x)^3=\frac{1}{4} (3 \sin (x)-\sin (3 x))$$ and $$\sin(x)^5=\frac{1}{16} (10 \sin (x)-5 \sin (3 x)+\sin (5 x))$$

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$$\int \sin ^{ 3 }{ x } \cos ^{ 2 }{ x } dx=-\int { \sin ^{ 2 }{ x } \cos ^{ 2 }{ x } d\left( \cos { x } \right) } =\\ =-\int { \left( 1-\cos ^{ 2 }{ x } \right) \cos ^{ 2 }{ x } d\left( \cos { x } \right) } =\int { \cos ^{ 4 }{ x } -\cos ^{ 2 }{ x } d\left( \cos { x } \right) } =\\ =\frac { \cos ^{ 5 }{ x } }{ 5 } -\frac { \cos ^{ 3 }{ x } }{ 3 } +C$$

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$$\int\sin^3x\cos^2xdx=\frac{\sin^4x}{4}\cdot\cos{x}-\frac{1}{4}\int\sin^4x(-\sin{x})dx=$$ $$=\frac{\sin^4x}{4}\cdot\cos{x}-\frac{1}{4}\int(1-\cos^2x)^2d(\cos{x})=$$ $$=\frac{\sin^4x\cos{x}}{4}-\frac{\cos{x}}{4}+\frac{\cos^3x}{6}-\frac{\cos^5x}{20}+C.$$

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Since you acknowledge that you can do this with more straightforward substitutions, but are curious about parts, let's show that in can be done with parts.

$\int \sin^3x \cos^2 x \ dx$

pick something easy to integrate to be $v'$

$v' = \sin x\cos^2 x$ or $v' = \sin^3x\cos x$ would be my suggestions.

lets go with the first one.

$u = \sin^2 x, dv = \sin x\cos^2 x\ dx\\ du = 2\sin x\cos x\ dx, v = -\frac 13 \cos^3x$

$-\frac 13 \cos^3 x\sin^2 x + \frac 23 \int \cos^4 x\sin x \ dx\\ -\frac 13 \cos^3 x\sin^2 x - \frac 2{15} \cos^5 x\\ \frac 15\cos^5 x - \frac 13 \cos^3 x$

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Integration by parts is unnecessary. Do this $$\int \sin^3(x)\cos^2(x)\,dx = \int (1-\cos^2(x))\cos^2(x)\sin(x)\,dx $$ Let $u = \cos(x), du = -\sin(x)\,dx$ and this resolves very simply.

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