高専の数学解説

第8章 8.13

(1)

左辺=Iとおくと

I=∫eax⋅cos⁡bx dx=1aeax⋅cos⁡bx+∫1aeax⋅b⋅sin⁡bx dxI = \int e^{ax} \cdot \cos bx \, dx = \frac{1}{a}e^{ax}\cdot\cos bx + \int \frac{1}{a}e^{ax}\cdot b \cdot \sin bx\, dx

=1aeax⋅cos⁡bx++ba⋅1a⋅eax⋅sin⁡bx−∫ba⋅1a⋅b⋅eaxcos⁡bx dx+C= \frac{1}{a}e^{ax}\cdot\cos bx + + \frac{b}{a}\cdot\frac{1}{a}\cdot e^{ax}\cdot\sin bx - \int \frac{b}{a}\cdot\frac{1}{a}\cdot b \cdot e^{ax}\cos bx\,dx + C

=1aeax⋅cos⁡bx+1a⋅ba⋅eax⋅sin⁡bx−b2a2I+C= \frac{1}{a}e^{ax}\cdot\cos bx + \frac{1}{a}\cdot\frac{b}{a}\cdot e^{ax}\cdot\sin bx - \frac{b^2}{a^2}I + C

I(1+b2a2)=1a2⋅eax(a⋅cos⁡bx+b⋅sin⁡xb)+CI\left(1+\frac{b^2}{a^2}\right) = \frac{1}{a^2}\cdot e^{ax}(a\cdot\cos bx + b\cdot\sin xb) + C

I(a2+b2)=eax(a⋅cos⁡bx+b⋅sin⁡bx)+CI(a^2+b^2) = e^{ax}(a\cdot\cos bx + b\cdot\sin bx) + C

I=1a2+b2⋅eax(a⋅cos⁡bx+b⋅sin⁡bx)+C=(右辺)I = \frac{1}{a^2+b^2}\cdot e^{ax}(a\cdot\cos bx + b\cdot\sin bx) + C = (\text{右辺})

(2)

左辺=Iとおくと

I=∫eax⋅sin⁡bx dx=1a⋅eax⋅sin⁡bx−1a∫eax⋅b⋅cos⁡bx dxI = \int e^{ax}\cdot\sin bx\,dx = \frac{1}{a}\cdot e^{ax}\cdot\sin bx - \frac{1}{a}\int e^{ax}\cdot b\cdot\cos bx\,dx

=1a⋅eax⋅sin⁡bx−1a2⋅b⋅eax⋅cos⁡bx+1a2⋅b∫eax(−b⋅sin⁡bx) dx= \frac{1}{a}\cdot e^{ax}\cdot\sin bx - \frac{1}{a^2}\cdot b\cdot e^{ax}\cdot\cos bx + \frac{1}{a^2}\cdot b\int e^{ax}(-b\cdot\sin bx)\,dx

=1a⋅eax⋅sin⁡bx−ba2⋅eax⋅cos⁡bx−b2a2I+C= \frac{1}{a}\cdot e^{ax}\cdot\sin bx - \frac{b}{a^2}\cdot e^{ax}\cdot\cos bx - \frac{b^2}{a^2}I + C

I(1+b2a2)=1a⋅eax⋅sin⁡bx−ba2⋅eax⋅cos⁡bx+CI\left(1+\frac{b^2}{a^2}\right) = \frac{1}{a}\cdot e^{ax}\cdot\sin bx - \frac{b}{a^2}\cdot e^{ax}\cdot\cos bx + C

I(a2+b2)=eax(a⋅sin⁡bx−b⋅cos⁡bx)+CI(a^2+b^2) = e^{ax}(a\cdot\sin bx - b\cdot\cos bx) + C

I=1a2+b2⋅eax(a⋅sin⁡bx−b⋅cos⁡bx)+C=(右辺)I = \frac{1}{a^2+b^2}\cdot e^{ax}(a\cdot\sin bx - b\cdot\cos bx) + C = (\text{右辺})

解説: r31bn1z