# Hasil perhitungan rumus

Rumus
Hitunglah integralnya
Jawaban
$$\displaystyle\int { \dfrac { 1 } { \cos\left( x \right) } } d { x }$$
$- 1 \left ( \dfrac { 1 } { 2 } \ln { \left( | \sin\left( x \right) - 1 | \right) } - \dfrac { 1 } { 2 } \ln { \left( | \sin\left( x \right) + 1 | \right) } \right )$
Hitunglah integralnya
$\displaystyle\int { \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ \cos\left( \color{#FF6800}{ x } \right) } } } } d { \color{#FF6800}{ x } }$
 Substitusikanlah menjadi $u = \sin\left( x \right)$ dan hitunglah integralnya. 
$\left [ \displaystyle\int { \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ u } ^ { \color{#FF6800}{ 2 } } } } } d { \color{#FF6800}{ u } } \right ] _ { \color{#FF6800}{ u } = \color{#FF6800}{ \sin\left( \color{#FF6800}{ x } \right) } }$
$\left [ \displaystyle\int { \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 1 } \color{#FF6800}{ - } \color{#FF6800}{ u } ^ { \color{#FF6800}{ 2 } } } } } d { \color{#FF6800}{ u } } \right ] _ { u = \sin\left( x \right) }$
 Buatlah koefisien dari suku tertinggi pada penyebut menjadi 1. 
$\left [ \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } } \displaystyle\int { \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ u } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 } } } } d { \color{#FF6800}{ u } } \right ] _ { u = \sin\left( x \right) }$
$\left [ \frac { 1 } { - 1 } \displaystyle\int { \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ u } ^ { \color{#FF6800}{ 2 } } \color{#FF6800}{ - } \color{#FF6800}{ 1 } } } } d { \color{#FF6800}{ u } } \right ] _ { u = \sin\left( x \right) }$
 Hitunglah integralnya menggunakan integral parsial 
$\left [ \frac { 1 } { - 1 } \left ( \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \left [ \displaystyle\int { \color{#FF6800}{ v } ^ { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } } d { \color{#FF6800}{ v } } \right ] _ { \color{#FF6800}{ v } = \color{#FF6800}{ u } \color{#FF6800}{ - } \color{#FF6800}{ 1 } } \color{#FF6800}{ - } \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \left [ \displaystyle\int { \color{#FF6800}{ v } ^ { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } } d { \color{#FF6800}{ v } } \right ] _ { \color{#FF6800}{ v } = \color{#FF6800}{ u } \color{#FF6800}{ + } \color{#FF6800}{ 1 } } \right ) \right ] _ { u = \sin\left( x \right) }$
$\left [ \frac { 1 } { - 1 } \left ( \frac { 1 } { 2 } \left [ \displaystyle\int { \color{#FF6800}{ v } ^ { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } } d { \color{#FF6800}{ v } } \right ] _ { v = u - 1 } - \frac { 1 } { 2 } \left [ \displaystyle\int { v ^ { - 1 } } d { v } \right ] _ { v = u + 1 } \right ) \right ] _ { u = \sin\left( x \right) }$
 Gunakanlah rumus $\int x^{-1} dx = \ln(|x|)$ dan hitunglah integralnya. 
$\left [ \frac { 1 } { - 1 } \left ( \frac { 1 } { 2 } \left [ \ln { \left( | \color{#FF6800}{ v } | \right) } \right ] _ { v = u - 1 } - \frac { 1 } { 2 } \left [ \displaystyle\int { v ^ { - 1 } } d { v } \right ] _ { v = u + 1 } \right ) \right ] _ { u = \sin\left( x \right) }$
$\left [ \frac { 1 } { - 1 } \left ( \frac { 1 } { 2 } \left [ \ln { \left( | \color{#FF6800}{ v } | \right) } \right ] _ { \color{#FF6800}{ v } = \color{#FF6800}{ u } \color{#FF6800}{ - } \color{#FF6800}{ 1 } } - \frac { 1 } { 2 } \left [ \displaystyle\int { v ^ { - 1 } } d { v } \right ] _ { v = u + 1 } \right ) \right ] _ { u = \sin\left( x \right) }$
 Kembalikanlah nilai yang telah disubstitusi. 
$\left [ \frac { 1 } { - 1 } \left ( \frac { 1 } { 2 } \ln { \left( | \color{#FF6800}{ u } \color{#FF6800}{ - } \color{#FF6800}{ 1 } | \right) } - \frac { 1 } { 2 } \left [ \displaystyle\int { v ^ { - 1 } } d { v } \right ] _ { v = u + 1 } \right ) \right ] _ { u = \sin\left( x \right) }$
$\left [ \frac { 1 } { - 1 } \left ( \frac { 1 } { 2 } \ln { \left( | u - 1 | \right) } - \frac { 1 } { 2 } \left [ \displaystyle\int { \color{#FF6800}{ v } ^ { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } } d { \color{#FF6800}{ v } } \right ] _ { v = u + 1 } \right ) \right ] _ { u = \sin\left( x \right) }$
 Gunakanlah rumus $\int x^{-1} dx = \ln(|x|)$ dan hitunglah integralnya. 
$\left [ \frac { 1 } { - 1 } \left ( \frac { 1 } { 2 } \ln { \left( | u - 1 | \right) } - \frac { 1 } { 2 } \left [ \ln { \left( | \color{#FF6800}{ v } | \right) } \right ] _ { v = u + 1 } \right ) \right ] _ { u = \sin\left( x \right) }$
$\left [ \frac { 1 } { - 1 } \left ( \frac { 1 } { 2 } \ln { \left( | u - 1 | \right) } - \frac { 1 } { 2 } \left [ \ln { \left( | \color{#FF6800}{ v } | \right) } \right ] _ { \color{#FF6800}{ v } = \color{#FF6800}{ u } \color{#FF6800}{ + } \color{#FF6800}{ 1 } } \right ) \right ] _ { u = \sin\left( x \right) }$
 Kembalikanlah nilai yang telah disubstitusi. 
$\left [ \frac { 1 } { - 1 } \left ( \frac { 1 } { 2 } \ln { \left( | u - 1 | \right) } - \frac { 1 } { 2 } \ln { \left( | \color{#FF6800}{ u } \color{#FF6800}{ + } \color{#FF6800}{ 1 } | \right) } \right ) \right ] _ { u = \sin\left( x \right) }$
$\left [ \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } } \left ( \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \ln { \left( | \color{#FF6800}{ u } \color{#FF6800}{ - } \color{#FF6800}{ 1 } | \right) } \color{#FF6800}{ - } \color{#FF6800}{ \frac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \ln { \left( | \color{#FF6800}{ u } \color{#FF6800}{ + } \color{#FF6800}{ 1 } | \right) } \right ) \right ] _ { \color{#FF6800}{ u } = \color{#FF6800}{ \sin\left( \color{#FF6800}{ x } \right) } }$
 Kembalikanlah nilai yang telah disubstitusi. 
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } } \left ( \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \ln { \left( | \color{#FF6800}{ \sin\left( \color{#FF6800}{ x } \right) } \color{#FF6800}{ - } \color{#FF6800}{ 1 } | \right) } \color{#FF6800}{ - } \color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ 2 } } } \ln { \left( | \color{#FF6800}{ \sin\left( \color{#FF6800}{ x } \right) } \color{#FF6800}{ + } \color{#FF6800}{ 1 } | \right) } \right )$
$\color{#FF6800}{ \dfrac { \color{#FF6800}{ 1 } } { \color{#FF6800}{ - } \color{#FF6800}{ 1 } } } \left ( \dfrac { 1 } { 2 } \ln { \left( | \sin\left( x \right) - 1 | \right) } - \dfrac { 1 } { 2 } \ln { \left( | \sin\left( x \right) + 1 | \right) } \right )$
 Hitung nilainya 
$\color{#FF6800}{ - } \color{#FF6800}{ 1 } \left ( \dfrac { 1 } { 2 } \ln { \left( | \sin\left( x \right) - 1 | \right) } - \dfrac { 1 } { 2 } \ln { \left( | \sin\left( x \right) + 1 | \right) } \right )$
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