ILMU PELAJARAN DAN PENDIDIKAN SEKOLAH: INTEGRAL 2 TRIGONOMETRI dalam MATEMATIKA

Sabtu, 07 Agustus 2010

INTEGRAL 2 TRIGONOMETRI dalam MATEMATIKA

Berikut ini identitas trigonometri yang sering digunakan untuk menyelesaikan soal-soal integral trigonometri

1. $\sin^{2}x +\cos^{2}x = 1$

2. $1+\tan^{2}x = \sec^{2}x$

3. $1+\cot^{2}x = \csc^{2}x$

4. $\sin^{2}x = \frac{1}{2}(1-\cos 2x)$

5. $\cos^{2}x = \frac{1}{2}(1+\cos 2x)$

6. $\sin x \cos x =\frac{1}{2}\sin 2x$

7. $\sin x \cos y =\frac{1}{2}[\sin(x+y)+\sin(x-y)]$

8. $\sin x \sin y =-\frac{1}{2}[\cos(x+y)-\cos(x-y)]$

9. $\cos x \cos y =\frac{1}{2}[\cos(x+y)+\cos(x-y)]$

10. $1-\cos x = 2 \sin^{2}(\frac{1}{2}x)$

11. $1+\cos x = 2 \cos^{2}(\frac{1}{2}x)$

12. $1\pm\sin x = 1\pm \cos(\frac{1}{2}\pi - x)$

Contoh :

1. $\int_{}^{}\sin^{2}x dx =\int_{}^{}\frac{1}{2}(1-\cos 2x) dx$

$= \int_{}^{}(\frac{1}{2}-\frac{1}{2}\cos 2x) dx$

$= \frac{1}{2}x -\frac{1}{2}.\frac{1}{2}\sin 2x + C$

$= \frac{1}{2}x -\frac{1}{4}\sin 2x + C$

Jadi $\int_{}^{}\sin^{2}x dx = \frac{1}{2}x -\frac{1}{4}\sin 2x + C$

2. $\int_{}^{}\cos^{2}(3x) dx$

$= \int_{}^{}\frac{1}{2}[1+\cos(6x)] dx$

$= \int_{}^{}[\frac{1}{2}+\frac{1}{2}\cos(6x)] dx$

$= \frac{1}{2}x+\frac{1}{2}.\frac{1}{6}\sin(6x)+ C$

$= \frac{1}{2}x+\frac{1}{12}\sin(6x)+ C$

Jadi $\int_{}^{}\cos^{2}(3x) dx = \frac{1}{2}x+\frac{1}{12}\sin(6x)+ C$

3. $\int_{}^{}\sin^{3}x dx$

$= \int_{}^{}\sin^{2}x \sin x dx$

$= \int_{}^{}[1-\cos^{2}x] \sin x dx$

$= \int_{}^{}[\sin x-\cos^{2}x \sin x] dx$

$= -\cos x+\frac{1}{3}\cos^{3}x +C$

Jadi $\int_{}^{}\sin^{3}x dx = -\cos x+\frac{1}{3}\cos^{3}x +C$

4. $\int_{}^{}\sin ^{4}x dx$

$\int_{}^{}\sin ^{4}x dx = \int_{}^{}(\sin ^{2}x)^{2} dx$

$= \int_{}^{}(\frac{1}{2}[1-\cos 2x])^{2} dx$

$= \int_{}^{}\frac{1}{4}[1-\cos 2x]^{2} dx$

$= \int_{}^{}\frac{1}{4}[1-2\cos 2x +\cos^{2} 2x] dx$

$= \int_{}^{}\frac{1}{4}dx-\int_{}^{}\frac{1}{2}\cos 2xdx +\int_{}^{}\frac{1}{4}\cos^{2} 2x dx$

$= \int_{}^{}\frac{1}{4}dx-\int_{}^{}\frac{1}{2}\cos 2xdx +\int_{}^{}\frac{1}{4}.\frac{1}{2}[1+\cos 4x] dx$

$= \int_{}^{}\frac{1}{4}dx-\int_{}^{}\frac{1}{2}\cos 2xdx +\int_{}^{}\frac{1}{8}[1+\cos 4x] dx$

$= \frac{1}{4}x-\frac{1}{2}.\frac{1}{2}\sin 2x +\frac{1}{8}x+\frac{1}{8}.\frac{1}{4}.\sin 4x + C$

$= \frac{1}{4}x-\frac{1}{4}\sin 2x +\frac{1}{8}x+\frac{1}{32}\sin 4x + C$

$= \frac{3}{8}x-\frac{1}{4}\sin 2x +\frac{1}{32}\sin 4x + C$

Jadi $\int_{}^{}\sin ^{4}x dx =\frac{3}{8}x-\frac{1}{4}\sin 2x +\frac{1}{32}\sin 4x + C$

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