高専の数学解説

第8章 8.11

In=∫xncos⁡x dx=xnsin⁡x−∫nxn−1sin⁡x dx=xnsin⁡x−n∫xn−1sin⁡x dx=xnsin⁡x−nJn−1I_n = \int x^n \cos x \, dx = x^n \sin x - \int nx^{n-1} \sin x \, dx = x^n \sin x - n\int x^{n-1} \sin x \, dx = x^n \sin x - nJ_{n-1}

Jn=∫xnsin⁡x dx=−xncos⁡x−∫nxn−1(−cos⁡x)dx=−xnsin⁡x+n∫xn−1cos⁡x dx=−xnsin⁡x+nJn−1J_n = \int x^n \sin x \, dx = -x^n \cos x - \int nx^{n-1} (-\cos x) dx = -x^n \sin x + n\int x^{n-1} \cos x \, dx = -x^n \sin x + nJ_{n-1}

(1)

I3=x3sin⁡x−3J2=x3sin⁡x−3(−x2cos⁡x+2I1)=x3sin⁡x+3x2cos⁡x−6(xsin⁡x−J0)I_3 = x^3 \sin x - 3J_2 = x^3 \sin x - 3(-x^2 \cos x + 2I_1) = x^3 \sin x + 3x^2 \cos x - 6(x\sin x - J_0)

=x3sin⁡x+3x2cos⁡x−6xsin⁡x+6∫x0sin⁡x dx=(x3−6x)sin⁡x+3x2cos⁡x+6(−cos⁡x)= x^3 \sin x + 3x^2 \cos x - 6x\sin x + 6\int x^0 \sin x \, dx = (x^3 - 6x)\sin x + 3x^2 \cos x + 6(-\cos x)

=(x3−6x)sin⁡x+3(x2−2)sin⁡x= (x^3 - 6x)\sin x + 3(x^2 - 2)\sin x

(2)

J3=−x3cos⁡x+3I2=−x3cos⁡x+3(x2sin⁡x−2J1)=−x3cos⁡x+3x2sin⁡x−6(−xcos⁡x+I0)J_3 = -x^3 \cos x + 3I_2 = -x^3 \cos x + 3(x^2 \sin x - 2J_1) = -x^3 \cos x + 3x^2 \sin x - 6(-x\cos x + I_0)

=−x3cos⁡x+3x2sin⁡x+6xcos⁡x−6∫x0cos⁡x dx=(−x3+6x)cos⁡x+3x2sin⁡x−6sin⁡x= -x^3 \cos x + 3x^2 \sin x + 6x\cos x - 6\int x^0 \cos x \, dx = (-x^3 + 6x)\cos x + 3x^2 \sin x - 6\sin x

=(−x3+6x)cos⁡x+3(x2−2)sin⁡x= (-x^3 + 6x)\cos x + 3(x^2 - 2)\sin x

解説: r31bn1z