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Pattern - 34 of Indefinite Integration

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⋆ Pattern – 34

$I_n=\ \int x^n$ sin mx dx, then $I_n=\ \frac{x^n\sin{mx}}{m}$ + $\ \frac{nx^{n-1}\cos{mx}}{m^2}$ – $\ \frac{n\left(n-1\right)}{m^2}I_{n-2}$ : n ≥ 2

Here, $I_3 = x^n$ ∫ cos mx dx – ∫ $\left(\frac{d}{dx}\ x^n\ cos\int{mx\ dx}\right)$ dx

$=\ \frac{x^n\sin{mx}}{m}-\ \int n\ x^{n-1}\times\frac{\sin{mx}}{m}\ dx$

$=\ \frac{x^n\sin{mx}}{m}-\frac{n}{m}\ \int{x^{n-1}\sin{mx\ dx}}$

$=\ \frac{x^n\sin{mx}}{m}-\frac{n}{m}\ \left[x^{n-1}\ sin\int{mx\ }dx-\ \int\left(\frac{d}{dx}\ x^{n-1}\int\sin{mx\ dx}\right)dx\right]$

$=\ \frac{x^n\sin{mx}}{m}+\frac{n}{m}\ x^{n-1}\frac{\cos{mx}}{m}-\frac{n}{m}\int{\left(n-1\right)x^{n-2}}\ \frac{\cos{mx}}{m}\ dx$

$=\ \frac{x^n\sin{mx}}{m}+\ \frac{nx^{n-1\ }\cos{mx}}{m^2}+\ \frac{n\left(n-1\right)}{m^2}\ I_{n-2}\ \ :n-2\geq0\ \therefore\ \ n\geq2$

1) Evaluate : ∫ $x^4$ cos 5x dx

Solution. Here, m = 5 &

$I_4=\ \int x^4\cos{5x}\ dx$

$=\ \frac{x^4\sin{5x}}{5}+\ \frac{4x^3\cos{5x}}{25}+\frac{12}{25}\ I_2$

$=\ \frac{x^4\sin{5x}}{5}+\ \frac{4x^3\cos{5x}}{25}+\frac{12}{25}\left[\frac{x^2\sin{5x}}{5}+\frac{2x\cos{5x}}{25}+\frac{2}{25}\ I_0\right]$

$=\ \frac{x^4\sin{5x}}{5}+\ \frac{4x^3\cos{5x}}{25}+\frac{12x^2\sin{5x}}{125}+\ \frac{24\ x\cos{5x}}{625}+\frac{24}{625}\times\frac{\sin{5x}}{5}+C$

$\because I_0=\ \int \cos{5x}\ \ dx=\ \frac{\sin{5x\ }}{5}$