Agu 18 2012

Pembuktian Integral sec^2 x dx = tan x + c


Pada pembuktian \int sec2 x dx = tan x, saya akan membuktikan dengan dua jalan, langsung saja :

CARA I :

\int sec2x dx = tan x

\frac{d}{dx} \quad \int sec2 x dx = \frac{d}{dx} tan x

sec2 x dx = \frac{d}{dx} tan x

jadi disini saya akan membuktikan \frac{d}{dx} tan x = sec2 x dx

\frac{d}{dx} tan x = \frac{d}{dx} \frac{sin \quad x}{cos \quad x}

misal : u(x) = sin x \Rightarrow u'(x) = cos x

v(x) = cos x \Rightarrow v'(x) = -sin x

\frac{d}{dx} \frac{u}{v} = \frac{u'v-uv'}{v^2}

= \frac{cosx.cosx-sinx(-sinx)}{(cosx)^2}

= \frac{cos^2x+sin^2x}{cos^2x}

= \frac{1}{cos^2x}

= sec2 x \blacksquare

CARA II :

\int sec2 x dx = \int (1 + tan2 x)dx

= \int (1 + \frac{sin^2x}{cos^2x}) dx

= \int 1 dx + \int \frac{sin^2x}{cos^2x} dx

= \int 1 dx + \int sin x \frac{sin \quad x }{cos^2x} dx

misal : u = sin x \Rightarrow du = cos x dx

v = \frac{1}{cos \quad x} \Rightarrow dv = \frac{sin \quad x}{cos^2x} dx

gunakan integral parsial (\int u dv = uv – \int v du), sehingga diperoleh :

= \int 1 dx + (sin x \frac{1}{cos \quad x} – \int \frac{1}{cos \quad x } cos x dx)

= \int 1 dx + (\frac{sin \quad x}{cos \quad x} – \int \frac{cos \quad x }{cos \quad x} dx)

= x + tan x – x + c

= tan x + c \blacksquare

11 comments on “Pembuktian Integral sec^2 x dx = tan x + c”

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