Question

Integral tentu fungsi trigonometri berpangkat
[tex]\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^4 x~dx[/tex]

Answer 1

Verified answer

[tex] = \displaystyle \int_{0}^{\frac{\pi}{2}}\sin^4 x~dx[/tex]

[tex] = \displaystyle \int_{0}^{\frac{\pi}{2}}( {sin}^{2} x) {}^{2} \: dx [/tex]

[tex]= \displaystyle \int_{0}^{\frac{\pi}{2}}( \frac{1}{2} (1 - cos(2x))) {}^{2} \: dx[/tex]

[tex]= \displaystyle \int_{0}^{\frac{\pi}{2}} \frac{1}{4} (1 - cos(2x)) {}^{2} \: dx [/tex]

[tex] = \frac{1}{4} \displaystyle \int_{0}^{\frac{\pi}{2}}(1 - cos(2x)) {}^{2} \: dx[/tex]

[tex]= \frac{1}{4} \displaystyle \int_{0}^{\frac{\pi}{2}}1 - 2cos(2x) + cos {}^{2} x \: dx[/tex]

[tex]= \frac{1}{4} \displaystyle \int_{0}^{\frac{\pi}{2}}1 - 2cos(2x) + \frac{1}{2} (1 + cos(4x)) \: dx \\ [/tex]

[tex]= \frac{1}{4} \displaystyle \int_{0}^{\frac{\pi}{2}}1 - 2cos(2x) + \frac{1}{2} + \frac{1}{2} cos(4x) \: dx \\ [/tex]

[tex]= \frac{1}{4} \displaystyle \int_{0}^{\frac{\pi}{2}} \frac{3}{2} - 2cos(2x) + \frac{1}{2} cos(4x) \: dx \\ [/tex]

[tex] = \frac{1}{4} ( \frac{3}{2} x - sin(2x) + \frac{1}{8} sin(4x))_{0}^{ \frac{\pi}{2} } [/tex]

[tex] = \frac{1}{4} (( \frac{3}{2} ( \frac{\pi}{2} ) - sin(2( \frac{\pi}{2} )) + \frac{1}{8} sin(4( \frac{\pi}{2} ))) - ( \frac{3}{2} (0) - sin(2(0)) + \frac{1}{8} sin(4(0)))) \\ [/tex]

[tex] = \frac{1}{4} (( \frac{3}{4} \pi - 0 + 0) - (0 - 0 + 0))[/tex]

[tex] = \frac{1}{4} ( \frac{3}{4} \pi)[/tex]

[tex] = \frac{3}{16} \pi[/tex]

Answer 2

Penjelasan dengan langkah-langkah:

Integral Trigonometri

Cara 1

sin²x = ½(1-cos2x)

cos²x = ½(1+cos2x)

ʃ sin⁴x dx [ 0 π/2 ]

sin⁴x => (sin²x)² => (½(1-cos2x))²

= ʃ (½(1-cos2x))² dx [ 0 π/2 ]

= ʃ ¼ . (1-cos2x)² dx [ 0 π/2 ]

= ¼ ʃ (1 - 2cos2x + cos²2x dx) [ 0 π/2]

= ¼ʃ dx - ½ʃ cos 2x dx + ¼ʃ cos²2x dx [ 0 π/2 ]

= ¼x - ½(½ sin 2x) + ¼ʃ (½(1+cos4x) dx [ 0 π/2 ]

= (¼(π/2)- (¼(sin 2(π/2) + ⅛(x+¼sin4x) [ 0 π/2 ]

= ⅛π - ¼sin π + ⅛(x+¼sin4x [ 0 π/2])

= ⅛π - 0 + (⅛(π/2 + ¼sin4(π/2))

= ⅛π + (⅛(π/2 + 0)

= ⅛π + π/16

= (2π+π)/16

= 3π/16

Cara 2

ʃ sin⁴x dx

= -¼(sin³x.cosx) + ¾.ʃ sin²x dx [ 0 π/2 ]

= -¼(sin³x.cosx) + ¾.ʃ ½(1-cos 2x) dx [ 0 π/2 ]

= -¼(sin³x.cosx) + ⅜.ʃ 1 - cos 2x dx [ 0 π/2 ]

= -¼(sin³x.cosx) + ⅜(x - ½sin2x) [ 0 π/2 ]

= -¼(sin³x.cosx) + ⅜x - 1/16.sin2x

= -¼(sin³(π/2).cos(π/2] + ⅜(π/2) - 1/16.sin(2.π/2)

= 0 + 3π/16 - 0

= 3π/16