Integral (2) Trigonometri


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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