(1) \(\color{green}\int \cos 4x \text{ dx} = \dfrac{1}{4} \sin 4x + \text{ c}\)
(3) \(\color{green}\int \csc^2 4x \text{ dx} = \dfrac{1}{4} \cot 4x + \text{ c}\)
(4) \(\color{green}\int \sec 4x \cdot \tan 4x \text{ dx} = \dfrac{1}{4} \sec 4x + \text{ c}\)
Answer: D
(1) \(\color{blue}\int \cos 4x \text{ dx} = \dfrac{1}{4} \sin 4x + \text{ c}\)
(2) \(\color{red}\int \sin 4x \text{ dx} = -\dfrac{1}{4} \cos 4x + \text{ c}\)
(3) \(\color{red}\int \csc^2 4x \text{ dx} = -\dfrac{1}{4} \cot 4x + \text{ c}\)
(4) \(\color{blue}\int \sec 4x \cdot \tan 4x \text{ dx} = \dfrac{1}{4} \sec 4x + \text{ c}\)
\(\color{green}\int \left(10\sin 3x \:-\: \sec^{2} \dfrac{1}{4}x \right) \text{ dx} = \dotso\)
\(\int \left(10\sin 3x \:-\: \sec^{2} \dfrac{1}{4}x \right) \text{ dx}\)
\(10 \cdot \left(-\dfrac{1}{3} \cos 3x\right) \:-\: \dfrac{1}{\frac{1}{4}} \tan \dfrac{1}{4}x + \text{ c}\)
\(-\dfrac{10}{3} \cos 3x \:-\: 4\tan \dfrac{1}{4}x + \text{ c}\)
\(\color{blue}\int \left(\csc ax \cdot \cot ax \right) \text{ dx} = -\dfrac{1}{a} \csc ax + \text{ c}\)
\(\int \left(6\csc 5x \cdot \cot 5x \right) \text{ dx}\)
\(6 \cdot \left(-\dfrac{1}{5} \csc 5x \right) + \text{ c}\)
\(-\dfrac{6}{5} \csc 5x + \text{ c}\)
Gunakan rumus perkalian fungsi trigonometri
\(\color{blue} 2 \cos x \cos y = \cos (x + y) + \cos (x\:-\:y)\)
\(\int \left(6\cos 4x \cdot \cos 2x \right) \text{ dx}\)
\(3\int \left(2\cos 4x \cdot \cos 2x \right) \text{ dx}\)
\(3\int \left(\cos (4x + 2x) + \cos (4x\:-\:2x) \right) \text{ dx}\)
\(3\int \left(\cos 6x + \cos 2x \right) \text{ dx}\)
\(3\left(\dfrac{1}{6} \sin 6x + \dfrac{1}{2} \sin 2x \right) + \text{ c}\)
\(\dfrac{1}{2} \sin 6x + \dfrac{3}{2} \sin 2x + \text{ c}\)
Gunakan rumus perkalian fungsi trigonometri
\(\color{blue} -2 \sin x \sin y = \cos (x + y) \:-\: \cos (x\:-\:y)\)
\(-3\int \left(-2\sin 8x \cdot \sin2x \right) \text{ dx}\)
\(-3\int \left(\cos (8x + 2x) \:-\: \cos (8x \:-\:2x)\right) \text{ dx}\)
\(-3\int \left(\cos 10x \:-\: \cos 6x\right) \text{ dx}\)
\(-3\left(\dfrac{1}{10} \sin 10x \:-\: \dfrac{1}{6} \sin 6x \right) + \text{ c}\)
\(-\dfrac{3}{10} \sin 10x + \dfrac{1}{2} \sin 6x + \text{ c}\)
Gunakan rumus sudut rangkap cosinus
\(\color{blue} \cos^2 x = \dfrac{1}{2} + \dfrac{1}{2} \cos 2x \)
\(\int 5\cos^2 4x \text{ dx}\)
\(5\int \left(\dfrac{1}{2} + \dfrac{1}{2} \cos 8x\right) \text{ dx}\)
\(5 \left(\dfrac{1}{2}x + \dfrac{1}{2} \cdot \dfrac{1}{8} \sin 8x \right) + \text{ c}\)
\(\dfrac{5}{2}x + \dfrac{5}{16} \sin 8x + \text{ c}\)
Jabarkan \(\color{green}\int (\sec 3x \:-\: \tan 3x)^2 \text{ dx}\)
\(\int \left(\sec^2 3x \:-\:2 \sec 3x \tan 3x + \tan^2 3x \right) \text{ dx}\)
Dengan menggunakan identitas trigonometri, \(\color{blue} \tan^2 3x = \sec^2 3x \:-\:1\)
\(\int \left(\sec^2 3x \:-\:2 \sec 3x \tan 3x + \sec^2 3x \:-\:1 \right) \text{ dx}\)
\(\int \left(2\sec^2 3x \:-\:2 \sec 3x \tan 3x +\:-\:1 \right) \text{ dx}\)
\(2 \cdot \dfrac{1}{3} \tan 3x \:-\:2 \cdot \dfrac{1}{3} \sec 3x \:-\: x + \text{ c}\)
\(\dfrac{2}{3} \tan 3x \:-\: \dfrac{2}{3} \sec 3x \:-\: x + \text{ c}\)
\(\color{green}\int \left(\dfrac{8\cos x \:-\: \sin x}{2 \cos x + 3 \sin x}\right)\text{ dx} = \dotso\)
\(\int \left(\dfrac{8\cos x \:-\: \sin x}{2 \cos x + 3 \sin x}\right)\text{ dx}\)
\(\int \left(\dfrac{\color{blue} 2 \cos x + 3 \sin x \color{black}+ 6\cos x \:-\:4 \sin x}{2 \cos x + 3 \sin x}\right)\text{ dx}\)
\(\int \left(1 + \dfrac{6\cos x \:-\:4 \sin x}{2 \cos x + 3 \sin x} \right)\text{ dx}\)
\(x + \int \left( \dfrac{6\cos x \:-\:4 \sin x}{2 \cos x + 3 \sin x} \right)\text{ dx}\)
Misal:
\(\color{blue} u = 2 \cos x + 3 \sin x\)
\(\color{blue} \dfrac{du}{dx} =-2\sin x + 3\cos x\)
\(x + \int \left(\dfrac{6\cos x \:-\:4 \sin x}{u}\right)\dfrac{du}{-2\sin x + 3\cos x}\)
\(x + \int \left(\dfrac{2\cancel{(-2\sin x + 3\cos x)}}{u}\right)\dfrac{du}{\cancel{-2\sin x + 3\cos x}}\)
\(x + 2\int \dfrac{1}{u} \text{ du}\)
\(x + 2 \ln |u| + \text{ c}\)
\(x + 2 \ln |2 \cos x + 3 \sin x| + \text{ c}\)
