Theoretical Recall

Integration by parts formula:

\[\int f(x)g'(x)\,dx = f(x)g(x) - \int f'(x)g(x)\,dx.\]

It derives directly from the product rule $(fg)’=f’g+fg’$.
A common strategy:

  • Choose $f(x)$ to simplify upon differentiation.
  • Choose $g’(x)$ so that its integral $g(x)$ is elementary.

Author’s note: In most cases, the choice is guided by the LIATE rule (Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, Exponential).

Exercises

Exercise 1

Evaluate \(\int x e^x \, dx.\)

Solution:
Take $f(x)=x$, $g’(x)=e^x$. Then $f’(x)=1$, $g(x)=e^x$.

\[\int x e^x dx = x e^x - \int e^x dx = x e^x - e^x + C.\]

Final Result: \(\int x e^x dx = (x-1)e^x + C\)

Exercise 2

Evaluate \(\int x \cos x \, dx.\)

Solution:
$f=x$, $g’=\cos x$. Then $f’=1$, $g=\sin x$.

\[\int x\cos x dx = x\sin x - \int \sin x dx = x\sin x + \cos x + C.\]

Final Result: \(\int x\cos x dx = x\sin x + \cos x + C\)

Exercise 3

Evaluate \(\int x \sin x \, dx.\)

Solution:
$f=x$, $g’=\sin x$, $f’=1$, $g=-\cos x$.

\[\int x\sin x dx = -x\cos x + \int \cos x dx = -x\cos x + \sin x + C.\]

Final Result: \(\int x\sin x dx = -x\cos x + \sin x + C\)

Exercise 4

Evaluate \(\int x e^{2x}\,dx.\)

Solution:
$f=x$, $g’=e^{2x}$, $f’=1$, $g=\tfrac{1}{2}e^{2x}$.

\[\int x e^{2x} dx = \tfrac{x}{2}e^{2x} - \int \tfrac{1}{2} e^{2x} dx = \tfrac{x}{2}e^{2x} - \tfrac{1}{4} e^{2x} + C.\]

Final Result: \(\int x e^{2x} dx = \tfrac{1}{2}(x-\tfrac{1}{2}) e^{2x} + C\)

Exercise 5

Evaluate \(\int \ln x \, dx.\)

Solution:
Take $f=\ln x$, $g’=1$. Then $f’=1/x$, $g=x$.

\[\int \ln x dx = x\ln x - \int 1 dx = x\ln x - x + C.\]

Final Result: \(\int \ln x dx = x\ln x - x + C\)

Exercise 6

Evaluate \(\int x^2 e^x \, dx.\)

Solution:
First integration by parts: $f=x^2$, $g’=e^x$, then $f’=2x$, $g=e^x$.

\[\int x^2 e^x dx = x^2 e^x - \int 2x e^x dx.\]

Apply integration by parts again on $\int 2x e^x dx$. Final expression:

\[x^2 e^x - (2x e^x - 2e^x) = (x^2 - 2x + 2) e^x + C.\]

Final Result: \(\int x^2 e^x dx = (x^2 - 2x + 2) e^x + C\)

Exercise 7

Evaluate \(\int x \ln x \, dx.\)

Solution:
$f=\ln x$, $g’=x$, $f’=1/x$, $g=x^2/2$.

\[\int x \ln x dx = \tfrac{x^2}{2}\ln x - \int \tfrac{x^2}{2}\cdot \tfrac{1}{x} dx = \tfrac{x^2}{2}\ln x - \tfrac{1}{2}\int x dx = \tfrac{x^2}{2}\ln x - \tfrac{x^2}{4} + C.\]

Final Result: \(\int x \ln x dx = \tfrac{x^2}{2}\ln x - \tfrac{x^2}{4} + C\)

Exercise 8

Evaluate \(\int e^x \cos x \, dx.\)

Solution:
$f=\cos x$, $g’=e^x$, $f’=-\sin x$, $g=e^x$.

\[\int e^x \cos x dx = e^x \cos x + \int e^x \sin x dx.\]

Apply integration by parts again to $\int e^x \sin x dx$. Solving the resulting equation for $I=\int e^x\cos x dx$, we obtain:

\[I=\frac{e^x}{2}(\sin x + \cos x) + C.\]

Final Result: \(\int e^x \cos x dx = \tfrac{1}{2} e^x (\sin x + \cos x) + C\)

Exercise 9

Evaluate \(\int e^x \sin x \, dx.\)

Solution:
Similar to previous. Final result:

\[\int e^x \sin x dx = \tfrac{1}{2} e^x (\sin x - \cos x) + C.\]

Final Result: \(\int e^x \sin x dx = \tfrac{1}{2} e^x (\sin x - \cos x) + C\)

Exercise 10

Evaluate \(\int x^3 e^x \, dx.\)

Solution:
Apply parts iteratively. Eventually:

\[\int x^3 e^x dx = (x^3 - 3x^2 + 6x - 6) e^x + C.\]

Final Result: \(\int x^3 e^x dx = (x^3 - 3x^2 + 6x - 6) e^x + C\)