Integral Sinus
$$\bbox[lightgreen, 5px, border: 2px solid green] {\int \sin ax \text{ dx} = -\dfrac{1}{a} \cos ax + \text{ C}} $$
Integral Cosinus
$$\bbox[lightgreen, 5px, border: 2px solid green] {\int \cos ax \text{ dx} = \dfrac{1}{a} \sin ax + \text{ C}} $$
Integral Tangen
$$\bbox[lightgreen, 5px, border: 2px solid green] {\int \tan ax \text{ dx} = \dfrac{1}{a} \ln |\sec ax| + \text{ C}}$$
$$\bbox[lightgreen, 5px, border: 2px solid green] {\int \tan ax \cdot \sec ax \text{ dx} = \dfrac{1}{a} \sec ax + \text{ C}}$$
Integral Cotangen
$$\bbox[lightgreen, 5px, border: 2px solid green] {\int \cot ax\text{ dx} = \dfrac{1}{a} \ln |\sin ax| + \text{ C}}$$
$$\bbox[lightgreen, 5px, border: 2px solid green] {\int \cot ax \cdot \csc ax \text{ dx} = -\dfrac{1}{a} \csc ax + \text{ C}}$$
Integral Secant
$$\bbox[lightgreen, 5px, border: 2px solid green] {\int \sec ax \text{ dx} = \dfrac{1}{a} \ln |\sec ax + \tan ax| + \text{ C}}$$
$$\bbox[lightgreen, 5px, border: 2px solid green] {\int \sec^2 ax \text{ dx} = \dfrac{1}{a} \tan ax + \text{ C}}$$
Integral Cosecant
$$\bbox[lightgreen, 5px, border: 2px solid green] {\int \csc ax \text{ dx} = \dfrac{1}{a} \ln |\csc ax \:-\: \cot ax| + \text{ C}}$$
$$\bbox[lightgreen, 5px, border: 2px solid green] {\int \csc^2 ax \text{ dx} = -\dfrac{1}{a} \cot ax + \text{ C}}$$
SOAL LATIHAN
Soal 01
\(\color{green} \int \sin^2 x \text{ dx} = \dotso\)
\(\color{blue} \sin^2 x = \dfrac{1}{2}\:-\:\dfrac{1}{2}\cos 2x\)
\(\int \sin^2 x \text{ dx} = \int \dfrac{1}{2}\:-\:\dfrac{1}{2}\cos 2x\text{ dx}\)
\(\int \sin^2 x \text{ dx} = \dfrac{1}{2}\int 1\:-\:\cos 2x\text{ dx}\)
\(\int \sin^2 x \text{ dx} = \dfrac{1}{2}\left(x \:-\:\dfrac{1}{2} \sin 2x \right) + c\)
\(\int \sin^2 x \text{ dx} = \dfrac{1}{2}x \:-\:\dfrac{1}{4} \sin 2x + c\)
Soal 02
\(\color{green}\int \cos^2 x \text{ dx} = \dotso\)
\(\color{blue} \cos^2 x = \dfrac{1}{2} + \dfrac{1}{2}\cos 2x\)
\(\int \cos^2 x \text{ dx} = \int \dfrac{1}{2} + \dfrac{1}{2}\cos 2x\text{ dx}\)
\(\int \cos^2 x \text{ dx} = \dfrac{1}{2}\int 1 + \cos 2x\text{ dx}\)
\(\int \cos^2 x \text{ dx} = \dfrac{1}{2}\left(x + \dfrac{1}{2} \sin 2x \right) + c\)
\(\int \cos^2 x \text{ dx} = \dfrac{1}{2}x + \dfrac{1}{4} \sin 2x + c\)
Soal 03
\(\color{green}\int \sin^3 x \text{ dx} = \dotso\)
\(\int \sin^3 x \text{ dx} = \int \sin^2 x \cdot \sin x \text{ dx}\)
\(\int \sin^3 x \text{ dx} = \int (1\:-\:\cos ^2 x) \cdot \sin x \text{ dx}\)
\(\int \sin^3 x \text{ dx} = \int \sin x \:-\: \cos^2x \cdot \sin x \text{ dx}\)
\(\int \sin^3 x \text{ dx} = -\cos x \:-\: \int \cos^2x \cdot \cancel{\sin x} \dfrac{\text{d}(\cos x)}{- \cancel{\sin x}}\)
\(\int \sin^3 x \text{ dx} = -\cos x + \int \cos^2 x \:\text{d}(\cos x)\)
\(\int \sin^3 x \text{ dx} = -\cos x + \dfrac{1}{3} \cos^3 x + c\)
Soal 04
\(\color{green}\int \cos^3 2x \text{ dx} = \dotso\)
\(\int \cos^3 2x \text{ dx} = \int \cos^2 {2x} \cdot \cos 2x \text{ dx}\)
Gunakan identitas trigonometri:
\(\color{blue} \sin^2 2x + \cos^2 2x =1\)
\(\int \cos^3 2x \text{ dx} = \int (1 \:-\:\sin^2 2x) \cdot \cos 2x \text{ dx}\)
\(\int \cos^3 2x \text{ dx} = \int (1 \:-\:\sin^2 2x) \cdot \cancel{\cos 2x} \cdot \dfrac{ d(\sin 2x)}{2\cancel{\cos 2x}}\)
\(\int \cos^3 2x \text{ dx} = \dfrac{1}{2}\sin 2x \:-\:\dfrac{1}{6} \sin^3 {2x} + c\)
Soal 05
\(\color{green}\int \sin^4 x \text{ dx} = \dotso\)
\(\int \sin^4 x \text{ dx} = \int \left(\dfrac{1\:-\:\cos 2x}{2} \right)^2 \text{ dx}\)
\(\int \sin^4 x \text{ dx} = \dfrac{1}{4}\int (1 \:-\:\cos 2x)^2 \text{ dx}\)
\(\int \sin^4 x \text{ dx} = \dfrac{1}{4}\int 1 \:-\:2\cos 2x + \cos^2 2x \text{ dx}\)
\(\int \sin^4 x \text{ dx} = \dfrac{1}{4} \left [x \:-\:\cancel{2}\cdot \dfrac{1}{\cancel{2}} \cdot \sin 2x + \int \cos^2 2x \text{ dx} \right]\)
\(\int \sin^4 x \text{ dx} = \dfrac{1}{4} \:-\:\dfrac{1}{4}\sin 2x + \dfrac{1}{4}\cdot \int \left(\dfrac{1}{2} + \dfrac{1}{2} \cos 4x \right) \text{ dx}\)
\(\int \sin^4 x \text{ dx} = \dfrac{1}{4}x \:-\:\dfrac{1}{4}\sin 2x + \dfrac{1}{8}\left(x + \dfrac{1}{4}\cdot \sin 4x \right) + c\)
\(\int \sin^4 x \text{ dx} = \dfrac{3}{8}x \:-\:\dfrac{1}{4}\sin 2x + \dfrac{1}{32}\sin 4x + c\)
Soal 06
\(\color{green}\int \cos^4 x \text{ dx} = \dotso\)
\(\int \cos^4 x \text{ dx} = \int \left(\dfrac{1 + \cos 2x}{2} \right)^2 \text{ dx}\)
\(\int \cos^4 x \text{ dx} = \dfrac{1}{4}\int 1 + 2\cos 2x + \cos^2 2x \text{ dx}\)
\(\int \cos^4 x \text{ dx} = \dfrac{1}{4}\left (x + 2 \cdot \dfrac{1}{2} \sin 2x + \int \dfrac{1}{2} + \dfrac{1}{2} \cos 4x \text{ dx} \right)\)
\(\int \cos^4 x \text{ dx} = \dfrac{1}{4}x + \dfrac{1}{4} \sin 2x + \dfrac{1}{4} \dfrac {1}{2} (x + \dfrac{1}{4}\sin 4x) + c\)
\(\int \cos^4 x \text{ dx} = \dfrac{3}{8}x + \dfrac{1}{4}\sin 2x + \dfrac{1}{32} \sin 4x + c\)
Soal 07
\(\color{green}\int \tan x \text{ dx} = \dotso\)
\(\int \tan x \text{ dx} = \int \dfrac{\cancel{\sin x}}{\cos x} \cdot \dfrac{\text{d} (\cos x)}{-\cancel{\sin x}}\)
\(\int \tan x \text{ dx} = – \int \dfrac{1}{\cos x} \text{d}(\cos x)\)
\(\int \tan x \text{ dx} = – \ln |\cos x| + c\)
Soal 08
\(\color{green}\int \tan^2 x \text{ dx} = \dotso\)
\(\int \tan^2 x \text{ dx} = \int (\sec^2 x \:-\: 1) \text{ dx}\)
Note: \(\color{blue} 1 + \tan^2 x = \sec^2 x\)
\(\int \tan^2 x \text{ dx} = \int \sec^2 x \text{ dx} \:-\: \int \text{ dx}\)
\(\int \tan^2 x \text{ dx} = \tan x \:-\: x + c\)
Soal 09
\(\color{green}\int \tan^3 (\frac{1}{2} x) \text{ dx} = \dotso\)
\(\int \tan^2 (\frac{1}{2} x)\cdot \tan (\frac{1}{2} x) \text{ dx}\)
\(\int (\sec^2 \frac{1}{2}x \:-\: 1) \cdot \tan (\frac{1}{2} x) \text{ dx}\)
\(\int \sec^2 \frac{1}{2} x \cdot \tan \frac{1}{2} x \text{ dx} \:-\: \int \tan \frac{1}{2} x \text{ dx}\)
\(\int \cancel{\sec^2 \frac{1}{2}x} \cdot \tan \frac{1}{2}x \cdot \dfrac{\text{d } (\tan \frac{1}{2}x)}{\frac{1}{2} \cdot \cancel{\sec^2 \frac{1}{2} x}} \:-\: \left(- \dfrac{1}{\frac{1}{2}} \ln | \cos \frac{1}{2} x|\right)\)
\(2 \cdot \frac{1}{2} \tan^2 \frac{1}{2} x + 2 \ln |\cos \frac{1}{2}x| + c\)
\(\tan^2 \frac{1}{2} x + 2 \ln |\cos \frac{1}{2}x| + c\)
Soal 10
\(\color{green}\int \sin^2 x \cdot \cos^2 x \text{ dx} = \dotso\)
Note:
\(\color{blue} \sin^2 x = \dfrac{1\:-\:\cos 2x}{2}\)
\(\color{blue} \cos^2 x = \dfrac{1 + \cos 2x}{2}\)
\(\int \dfrac{1\:-\:\cos 2x}{2} \cdot \dfrac{1 + \cos 2x}{2} \text{ dx}\)
\(\dfrac{1}{4}\int (1 \:-\: \cos^2 2x \text{ dx}\)
\(\dfrac{1}{4} \left(x \:-\: \int \cos^2 2x \text{ dx} \right)\)
\(\dfrac{1}{4}x \:-\: \dfrac{1}{4} \int \dfrac{1 + \cos 4x}{2} \text{ dx}\)
\(\dfrac{1}{8}x \:-\: \dfrac{1}{32} \sin 4x + c\)
