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Soal dan Pembahasan Integral Trigonometri Dasar

Topik Bahasan
Untuk Rumus Dasar Integral Trigonometri ini, kami rekomendasikan baca di: Rumus Integral Trigonometri Lengkap.

Soal 1. $ \int 2\sin x dx $
$ \int 2\sin x dx = 2 \int \sin x dx = 2(-\cos x ) + c = -2\cos x + c $

Soal 2. $ \int \frac{\sin x + \csc x}{\tan x } dx $
Sederhanakan penyebut dengan rumus trigonometri sudut ganda: $ \sin 2x = 2\sin x \cos x $
$ \begin{align} \int \frac{\tan x + \cot x}{\sin 2 x } dx & = \int \frac{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} }{2 \sin x \cos x } dx \\ & = \int \frac{\frac{\sin x}{\cos x} }{2 \sin x \cos x } + \frac{ \frac{\cos x}{\sin x} }{2 \sin x \cos x } dx \\ & = \int \frac{\sin x}{\cos x . 2 \sin x \cos x} + \frac{\cos x}{\sin x . 2 \sin x \cos x} dx \\ & = \int \frac{1}{\cos x . 2 \cos x} + \frac{1}{\sin x . 2 \sin x } dx \\ & = \int \frac{1}{2} ( \frac{1}{\cos ^2 x } + \frac{1}{\sin ^2 x } ) dx \\ & = \frac{1}{2} \int   \sec ^2 x + \csc ^2 x dx \\ & = \frac{1}{2} (\tan x - \cot x ) + c \end{align} $

Soal 3. $ \int 6\sin (1-3x) dx $
 $ \int 6\cos (1-3x) dx = \frac{6}{-3} \sin (1-3x) + c = -2\sin (1-3x) + c $

Soal 4. $ \int \csc ^2 (-2x + 1) + \sin (2x) dx $
$ = \frac{1}{2} \cot (-2x + 1) - \frac{1}{2} \cos (2x) + c $

Soal 5. $ \int 4\cos 7x \sin 4x dx $
Maka digunakan rumus : $ \, \cos A \sin B = \frac{1}{2} [\sin (A+B) - \sin (A-B)] $
$ \begin{align} \int 4\cos 7x \sin 4x dx & = \int 4 . \frac{1}{2} [ \sin (7x + 4x ) - \sin (7x - 4x)] dx \\ & = \int 2 [ \sin (11x ) - \sin (3x)] dx \\ & = 2 [ - \frac{1}{11} \cos (11x ) - (-\frac{1}{3}\cos (3x)) + c   \\ & = 2 [ - \frac{1}{11} \cos (11x ) + \frac{1}{3}\cos (3x) + c \\ & = - \frac{2}{11} \cos (11x ) + \frac{2}{3}\cos (3x) + c \end{align} $

Soal 6. $ \int 3cos (3x - 1) \cos (2x + 2) dx $
Maka digunakan rumus : $ \, \cos A \cos B = \frac{1}{2} [\cos (A+B) + \cos (A-B)] $
$ \begin{align} & \int 3cos (3x - 1) \cos (2x + 2) dx \\ & = \int 3 . \frac{1}{2} [ \cos ((3x - 1) + (2x + 2)) + \cos ((3x - 1) - (2x + 2))] dx \\ & = \int \frac{3}{2} [ \cos (5x + 1) + \cos (x - 3)] dx \\ & = \frac{3}{2} [ \frac{1}{5} \sin (5x + 1) + \sin (x - 3)] + c \\ & = \frac{3}{10} \sin (5x + 1) + \frac{3}{2} \sin (x - 3) + c \end{align} $
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