高専の数学解説

第9章 9.9

(1)

∫0π2cos⁡7x dx=67⋅45⋅23=1635\int_0^{\frac{\pi}{2}} \cos^7 x \, dx = \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3} = \frac{16}{35}

(2)

∫0π2sin⁡10x dx=910⋅78⋅56⋅34⋅12⋅π2=63512π\int_0^{\frac{\pi}{2}} \sin^{10} x \, dx = \frac{9}{10} \cdot \frac{7}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{63}{512}\pi

(3)

∫0π2sin⁡6x⋅cos⁡2x dx=∫0π2sin⁡6x⋅(1−sin⁡2x) dx=∫0π2sin⁡6x−sin⁡8x dx\int_0^{\frac{\pi}{2}} \sin^6 x \cdot \cos^2 x \, dx = \int_0^{\frac{\pi}{2}} \sin^6 x \cdot (1-\sin^2 x) \, dx = \int_0^{\frac{\pi}{2}} \sin^6 x - \sin^8 x \, dx

=56⋅34⋅12⋅π2−78⋅56⋅34⋅12⋅π2=5256π= \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} - \frac{7}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{5}{256}\pi

(4)

∫0π2sin⁡2x⋅cos⁡7x dx=∫0π2(1−cos⁡2x)⋅cos⁡7x dx=∫0π2cos⁡7x−cos⁡9x dx\int_0^{\frac{\pi}{2}} \sin^2 x \cdot \cos^7 x \, dx = \int_0^{\frac{\pi}{2}} (1-\cos^2 x) \cdot \cos^7 x \, dx = \int_0^{\frac{\pi}{2}} \cos^7 x - \cos^9 x \, dx

=67⋅45⋅23−89⋅67⋅45⋅23=16315= \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3} - \frac{8}{9} \cdot \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3} = \frac{16}{315}

(5)

∫0πcos⁡6x dx=2∫0π2cos⁡6x dx=2⋅56⋅34⋅12⋅π2=516π\int_0^{\pi} \cos^6 x \, dx = 2\int_0^{\frac{\pi}{2}} \cos^6 x \, dx = 2 \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{5}{16}\pi

(6)

∫0πcos⁡4x⋅sin⁡2x dx=2∫0π2cos⁡4x⋅sin⁡2x dx=2∫0π2cos⁡4x⋅(1−cos⁡2x) dx=2∫0π2cos⁡4x−cos⁡6x dx\int_0^{\pi} \cos^4 x \cdot \sin^2 x \, dx = 2\int_0^{\frac{\pi}{2}} \cos^4 x \cdot \sin^2 x \, dx = 2\int_0^{\frac{\pi}{2}} \cos^4 x \cdot (1-\cos^2 x) \, dx = 2\int_0^{\frac{\pi}{2}} \cos^4 x - \cos^6 x \, dx

=2(34⋅12⋅π2−56⋅34⋅12⋅π2)=116π= 2\left(\frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} - \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\right) = \frac{1}{16}\pi

(7)

∫0π4sin⁡32x dx2x=t とおくと2dx=dtdx=12dtx:0→π4t:0→π2これから\int_0^{\frac{\pi}{4}} \sin^3 2x\,dx \qquad 2x = t \text{ とおくと} \qquad 2dx = dt \qquad dx = \frac{1}{2}dt \qquad x : 0 \to \frac{\pi}{4} \qquad t : 0 \to \frac{\pi}{2} \qquad \text{これから}

∫0π212sin⁡3t dt=12⋅23=13\int_0^{\frac{\pi}{2}} \frac{1}{2}\sin^3 t\,dt = \frac{1}{2}\cdot\frac{2}{3} = \frac{1}{3}

(8)

∫0π41cos⁡4x dxtan⁡x=t とおくと1cos⁡2xdx=dtx:0→π4t:0→1これから\int_0^{\frac{\pi}{4}} \frac{1}{\cos^4 x}\,dx \qquad \tan x = t \text{ とおくと} \qquad \frac{1}{\cos^2 x}dx = dt \qquad x : 0 \to \frac{\pi}{4} \qquad t : 0 \to 1 \qquad \text{これから}

∫01(1+t2) dt=[t+13t3]01=1+13=43\int_0^1 (1+t^2)\,dt = \left[ t + \frac{1}{3}t^3 \right]_0^1 = 1 + \frac{1}{3} = \frac{4}{3}

解説: r31bn1z