Integral Trigonometri: Bentuk Parsial

Konsep Dasar
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INTEGRAL PARSIAL

$\int u . d v = u . v - \int v d u$

Contoh

$\int (2 x + 3) . \sin (4 x - 1) d x$

$\begin{aligned} u & = 2 x + 3 \\ d u & = 2 d x \end{aligned}$

$\begin{aligned} d v & = \sin (4 x - 1) d x \\ \int d v & = \int \sin (4 x - 1) d x \\ v & = \int \sin (4 x - 1) \frac{d (4 x - 1)}{4} \\ v & = \frac{1}{4} \int \sin (4 x - 1) d (4 x - 1) \\ v & = - \frac{1}{4} \cos (4 x - 1) \end{aligned}$

$\begin{aligned} \int u d v = u . v - \int v d u \\ \int (2 x + 3) . \sin (4 x - 1) d x \\ (2 x + 3) . - \frac{1}{4} \cos (4 x - 1) - \int - \frac{1}{4} \cos (4 x - 1) . 2 d x \\ - \frac{1}{4} (2 x + 3) . \cos (4 x - 1) + \frac{1}{2} \int \cos (4 x - 1) d x \\ - \frac{1}{4} (2 x + 3) . \cos (4 x - 1) + \frac{1}{8} \sin (4 x - 1) + c \end{aligned}$

SOAL LATIHAN

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