Pembuktian Integral csc^2 x dx = -cot x + C


CARA I :

\int csc2x dx = -cot x + C

\frac{d}{dx} \quad \int csc2 x dx = \frac{d}{dx} -cot x + C

csc2 x dx = \frac{d}{dx} -cot x

jadi disini saya akan membuktikan \frac{d}{dx} -cot x = csc2 x dx

\frac{d}{dx} -cot x = \frac{d}{dx} \frac{-cos \quad x}{sin \quad x}

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

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

\frac{d}{dx} \frac{u}{v} = \frac{u'v-uv'}{v^2} (Turunan Aturan Pembagian)

= \frac{(sin \quad x)(sin \quad x)-(-cos \quad x)(cos\quad x)}{(sin \quad x)^2}

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

= \frac{1}{sin^2 \quad x}

= csc2 x \blacksquare

CARA II :

\int csc2 x dx = \int (1 + cot2 x) dx

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

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

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

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

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

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

v = \frac{1}{sin \quad x} \Rightarrow dv = \frac{-cos \quad x}{sin^2 \quad x} dx (menggunakan Turunan Aturan Pembagian)

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

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

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

= x – cot x – x + C

= -cot x + C \blacksquare

8 comments on “Pembuktian Integral csc^2 x dx = -cot x + C

    • coba kita pake langkah mundur.
      \int sin u du = -cos u + C [turun-kan kedua ruas]
      \frac{d}{du} \int sin u du = \frac{d}{du} -cos u + C
      sin u = –\frac{d}{du} cos u
      – sin u = \frac{d}{du} cos u

      dari pers terakhir ini kita bisa buktikan turunan cos u = -sin u
      kita buktikan dengan menggunakan Definisi Turunan
      misal f(x) := cos x
      f'(x) = lim_{h \to 0} \frac{f(x+h)-f(x)}{h}
      \frac{d}{du} cos x = lim_{h \to 0} \frac{cos(x+h)-cos(x)}{h}
      = lim_{h \to 0} \frac{cos(x).cos(h)-sin(x).sin(h)-cos(x)}{h}
      = lim_{h \to 0} \frac{cos(x).cos(h)-cos(x)}{h} lim_{h \to 0} \frac{sin(x).sin(h)}{h}
      = cos x lim_{h \to 0} \frac{cos(h)-1}{h} lim_{h \to 0} \frac{sin(x).sin(h)}{h}
      = cos x lim_{h \to 0} \frac{cos^2(h)-1}{h.(cos(h)+1)} – sin x lim_{h \to 0} \frac{sin(h)}{h}
      = cos x lim_{h \to 0} \frac{sin^2(h)}{h.(cos(h)+1)} – sin x lim_{h \to 0} \frac{sin(h)}{h}
      = cos x lim_{h \to 0} \frac{sin(h).sin(h)}{h.(cos(h)+1)} – sin x lim_{h \to 0} \frac{sin(h)}{h}
      = cos x . lim_{h \to 0} sin h . lim_{h \to 0} \frac{sin(h)}{h} . lim_{h \to 0} \frac{1}{cos(h)+1} – sin x . lim_{h \to 0} \frac{sin(h)}{h}
      = cos x . 0 . 1 . \frac{1}{2} – sin x . 1
      = -sin x

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