∫xcos⁡3x dx\int x\cos{3x} \, dx
1
Use Integration by Parts on ∫xcos⁡3x dx\int x\cos{3x} \, dx.
Let u=xu=x, dv=cos⁡3xdv=\cos{3x}, du=dxdu=dx, v=sin⁡3x3v=\frac{\sin{3x}}{3}

2
Substitute the above into uv−∫v duuv-\int v \, du.
xsin⁡3x3−∫sin⁡3x3 dx\frac{x\sin{3x}}{3}-\int \frac{\sin{3x}}{3} \, dx

3
Use Constant Factor Rule: ∫cf(x) dx=c∫f(x) dx\int cf(x) \, dx=c\int f(x) \, dx.
xsin⁡3x3−13∫sin⁡3x dx\frac{x\sin{3x}}{3}-\frac{1}{3}\int \sin{3x} \, dx

4
Use Integration by Substitution on ∫sin⁡3x dx\int \sin{3x} \, dx.
Let u=3xu=3x, du=3 dxdu=3 \, dx, then dx=13 dudx=\frac{1}{3} \, du

5
Using uu and dudu above, rewrite ∫sin⁡3x dx\int \sin{3x} \, dx.
∫sin⁡u3 du\int \frac{\sin{u}}{3} \, du

6
Use Constant Factor Rule: ∫cf(x) dx=c∫f(x) dx\int cf(x) \, dx=c\int f(x) \, dx.
13∫sin⁡u du\frac{1}{3}\int \sin{u} \, du

7
Use Trigonometric Integration: the integral of sin⁡u\sin{u} is −cos⁡u-\cos{u}.
−cos⁡u3-\frac{\cos{u}}{3}

8
Substitute u=3xu=3x back into the original integral.
−cos⁡3x3-\frac{\cos{3x}}{3}

9
Rewrite the integral with the completed substitution.
xsin⁡3x3+cos⁡3x9\frac{x\sin{3x}}{3}+\frac{\cos{3x}}{9}

10
Add constant.
xsin⁡3x3+cos⁡3x9+C\frac{x\sin{3x}}{3}+\frac{\cos{3x}}{9}+C

Done
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