Soal dan Jawaban IntegraL

INTEGRAL
∫▒x^2 cos⁡〖x dx=⋯〗
Answer :
Pakai rumus integral parsial : ∫▒〖u dv=uv- ∫▒〖v dv〗〗
Misal : u = x^2 →du=2x dx
Dv = cos x dx →v=cos⁡x dx=sin⁡x
Sehingga :
∫▒x^2 cos⁡〖x dx=〗 x^2 .sinx-2 ∫▒〖x sin⁡〖x dx〗 〗
∫▒〖x sin⁡〖x dx〗 〗 perlu di parsialkan lagi tersendiri :
Misal : u = x→du=dx
dv = sin⁡〖x dx〗 →v = ∫▒sin⁡〖x dx〗 = -cos x

Sehingga :

∫▒〖x sin⁡〖x dx〗 〗 = x . (-cosx) - ∫▒〖- cosx dx〗
= - x cos x + ∫▒〖 cosx dx〗
= - x cos x + sin⁡〖x+c〗


∫▒x^2 √(2x^3+3) dx = …
Answer :

Cara Subtitusi :
Misal : u = 2x^3+3
du/dx = 6x^2 →dx= du/(dx^2 )
Sehingga :
∫▒x^2 √(2x^3+3) dx = ∫▒x^2 u^(1/2) du/(dx^2 )
=∫▒1/6 u^(1/2) du = 1/6 1/(1+1/2) u^(1/2) + c
= 1/6 2/3 u^(3/2) + c = 1/9 (2x^3+3) √(2x^3+3)+ c


∫_(π/6)^(π/2)▒〖sin〗^2 x cosx dx = …
Answer :
Pakai rumus : ∫▒〖sin〗^n (ax+b) cos⁡〖(ax+b)〗 dx= 1/(a (n+1)) 〖sin〗^(n+1) (ax+b)+ c

∫_(π/6)^(π/2)▒〖sin〗^2 x cosx dx = 1/3 〖sin〗^3 x ∫_(π/6)^(π/2)
= 1/3 (1^3- 〖(1/2)〗^3) = 1/3 . 7/(8 )= 7/24
∫▒〖(2x^3+ 〗 3x^2+ x+7) dx=⋯
Answer :
Pakai Rumus :
∫▒〖kx^n dx= k/(n+1) x^(n+1 )+ c〗

∫▒〖(2x^3+ 〗 3x^2+ x+7) dx= 2/4 x^n+3/3 x^3+ 1/2 x^2+ 7x+c
∫▒〖sin 〗 3xsin 2xdx = …
Answer :
Ingat rumus trigonometri :
-2 sin α sin β = cos (α + β) – cos (α - β)
Sin α sin β = - 1/2 (cos (α + β) - cos (α - β)

= 1/2 (cos (α - β) - cos (α + β)

∫▒〖sin 〗 3xsin 2xdx = ∫▒〖1/2 cos⁡(3x-2x)dx-〗 ∫▒〖1/2 cos⁡(3x- 2x)dx 〗
= ∫▒〖1/2 cos⁡〖x dx〗- ∫▒〖1/2 cos⁡〖5x dx〗→ pakai rumus 〗〗 ∫▒〖cos⁡(ax-b)dx= 1/a sin⁡〖(ax+b)〗 〗+ c
Sehingga menjadi = 1/2 sin x - 1/2 1/5 sin 5x + c
= 1/2 sin x - 1/10 sin 5x + c