Too many parts! (My fourteenth integral problem)

Using Euler's identity as Abhineet Nayyar , but perhaps slightly easier.

$\begin{aligned} \int_0^\pi e^x x^3 \cos x \space dx & = \Re \int_0^\pi e^x x^3 e^{ix} \space dx \\ & = \Re \int_0^\pi x^3 e^{\color{#3D99F6}{(1+i)x}} \space dx \quad \quad \small \color{#3D99F6}{\text{Let } z = (1+i)x \space \Rightarrow x = \frac{z}{1+i}} \\ & = \Re \int_0^{(1+i)\pi} \frac{\color{#3D99F6}{z}^3 e^{\color{#3D99F6}{z}}} {(1+i)^4} \space d\color{#3D99F6}{z} \\ & = \Re \left(- \frac{1}{4} \int_0^{(1+i)\pi} z^3e^z \space dz \right) \quad \quad \small \color{#3D99F6}{\text{By integration by parts}} \\ & = - \frac{1}{4} \Re \left[z^3e^z - 3z^2e^z + 6ze^z - 6e^z \right]_0^{(1+i)\pi} \\ & = - \frac{1}{4} \Re \left(e^{(1+i)\pi}[(1+i)^3\pi^3 - 3(1+i)^2\pi^2 + 6(1+i)\pi -6] + 6 \right) \\ & = - \frac{1}{4} \Re \left(-e^\pi[(-2+2i)\pi^3 - 6i\pi^2 + (6+6i)\pi -6] + 6 \right) \\ & = - \frac{1}{4} \left(-e^\pi[-2\pi^3 + 6\pi -6] + 6 \right) \\ & = - \frac{3}{2} - \frac{3e^\pi}{2} + \frac{3e^\pi\pi}{2} - \frac{e^\pi\pi^3}{2} \end{aligned}$

$\Rightarrow a+b+c+d+f+g+h+j+k = 3+2+3+2+3+2+1+3+2 = \boxed{21}$

I'll be using $\int{e^x(f(x)+f'(x))} = e^xf(x)+C$ to solve the integral.
Let $g(f(x)) = \int e^x(f(x)+f'(x)) dx = e^xf(x)+C$ $g(x^3\cos x)+g(x^3\sin x) = \int \color{#3D99F6}{e^x2x^3\cos x} + \color{#D61F06}{e^x3x^2(\cos x+\sin x)} dx = e^xx^3(\cos x+\sin x) + C$ We'll do the same process with terms involving $x^2 \text{ and } x$ .
The final result is obtained as:
$\int \color{#3D99F6}{2e^xx^3\cos x}dx = \color{#20A900}{g(x^3\cos x) + g(x^3 \sin x)} + \color{#624F41}{3(-g(x^2\sin x)+g(x\sin x)-g(x\cos x)+g(\cos x))} \\= e^x(\color{#20A900}{x^3\cos x + x^3 \sin x} + \color{#624F41}{3(-x^2\sin x+x\sin x-x\cos x+\cos x)})+ C$

Ookay! Loved solving it, btw! :)

$\int _{ 0 }^{ \pi }{ { e }^{ x }{ x }^{ 3 }\cos { x } dx }$

But, since, $\cos { x } = \Re ({ e }^{ ix })$

So, substituting this value in the original expression, and applying integration by parts once, we get:

$\quad \quad \Re (\int _{ 0 }^{ \pi }{ { e }^{ x }{ x }^{ 3 }{ e }^{ ix }dx } )\\ =\quad \Re (\int _{ 0 }^{ \pi }{ { e }^{ x(1+i) }{ x }^{ 3 }dx } )\\ =\quad \Re \{ \frac { { \pi }^{ 3 }{ e }^{ \pi (1+i) } }{ 1+i } -\frac { 3 }{ 1+i } \int _{ 0 }^{ \pi }{ { x }^{ 2 } } { e }^{ x(1+i) }dx\}$

......and again:

$\Re \{ \frac { { \pi }^{ 3 }{ e }^{ \pi (1+i) } }{ 1+i } -\frac { 3{ \pi }^{ 2 }{ e }^{ \pi (1+i) } }{ { (1+i) }^{ 2 } } +\frac { 6 }{ { (1+i) }^{ 2 } } \int _{ 0 }^{ \pi }{ { x } } { e }^{ x(1+i) }dx\}$

.....and again: (Promise, this is the last one! :))

$\Re \{ \frac { { \pi }^{ 3 }{ e }^{ \pi (1+i) } }{ 1+i } -\frac { 3{ \pi }^{ 2 }{ e }^{ \pi (1+i) } }{ { (1+i) }^{ 2 } } +\frac { 6\pi { e }^{ \pi (1+i) } }{ { (1+i) }^{ 3 } } -\frac { 6[{ e }^{ \pi (1+i) }-1] }{ { (1+i) }^{ 4 } } \}$

Now, putting ${ e }^{ \pi i }=-1$ and separating the real and complex terms:

$\quad \quad \Re \{ \frac { { \pi }^{ 3 }{ e }^{ \pi }(-1)(1-i) }{ 2 } -\frac { 3{ \pi }^{ 2 }{ e }^{ \pi }(-1) }{ 2i } +\frac { 3\pi { e }^{ \pi }(1+i) }{ 2 } -\frac { 6[{ e }^{ \pi }+1] }{ 4 } \} \\ =\quad \Re \{ \frac { -{ \pi }^{ 3 }{ e }^{ \pi } }{ 2 } +\frac { 3\pi { e }^{ \pi } }{ 2 } -\frac { 3{ e }^{ \pi } }{ 2 } -\frac { 3 }{ 2 } +i[\frac { { \pi }^{ 3 }{ e }^{ \pi } }{ 2 } -\frac { 3{ \pi }^{ 2 }{ e }^{ \pi } }{ 2 } +\frac { 3\pi { e }^{ \pi } }{ 2 } ]\}$

Since, we just need the real part, so our integral ends up as:

$\frac { -{ \pi }^{ 3 }{ e }^{ \pi } }{ 2 } +\frac { 3\pi { e }^{ \pi } }{ 2 } -\frac { 3{ e }^{ \pi } }{ 2 } -\frac { 3 }{ 2 }$

And, finally, comparing with the given variables: $a=3, b=2, c=3, d=2, f=3, g=2, h=1, k=2, j=3$

So, $a+b+c+d+f+g+h+k+j = 3+3+3+3+2+2+2+2+1 = 21$

Phew! Thanks for bearing with me! :P