[tex]{ {∫}^{2\pi}_{0}} \: \frac{dx}{2 \: -

Berikut ini adalah pertanyaan dari unknown pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Pertama

${ {∫}^{2\pi}_{0}} \: \frac{dx}{2 \: - \: sin \: x} \: \: = \: ?$ ninuninuninuninu...

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

Hasil dari $\displaystyle{\int\limits^{2\pi}_0 {\frac{dx}{2-sinx}} \, }$ adalah $\displaystyle{\boldsymbol{\frac{2\sqrt{3}}{3}\pi} }$ .

PEMBAHASAN

Sifat sifat operasi pada integral :

$(i)~\displaystyle{\int\limits {ax^n} \, dx=\frac{a}{n+1}x^{n+1}+C}$

$(ii)~\displaystyle{\int\limits {kf(x)} \, dx=k\int\limits {f(x)} \, dx}$

$(iii)~\displaystyle{\int\limits {\left [ f(x)\pm g(x) \right ]} \, dx=\int\limits {f(x)} \, dx\pm\int\limits {g(x)} \, dx}$

DIKETAHUI

$\displaystyle{\int\limits^{2\pi}_0 {\frac{dx}{2-sinx}} \, = }$

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DITANYA

Tentukan hasilnya.

PENYELESAIAN

Rumus trigonometri 1/2 sudut:

$\displaystyle{tan\frac{x}{2}=\frac{sinx}{1+cosx}~dan~tan\frac{x}{2}=\frac{1-cosx}{sinx}}$

$\displaystyle{tan\frac{x}{2}=\frac{1-cosx}{sinx} }$

$\displaystyle{sinx.tan\frac{x}{2}=1-cosx }$

$\displaystyle{cosx=1-sinx.tan\frac{x}{2}~~~...(i) }$

$\displaystyle{tan\frac{x}{2}=\frac{sinx}{1+cosx}~~~...substitusi~pers.(i)}$

$\displaystyle{tan\frac{x}{2}=\frac{sinx}{1+1-sinx.tan\frac{x}{2}}}$

$\displaystyle{tan\frac{x}{2}=\frac{sinx}{2-sinx.tan\frac{x}{2}}}$

$\displaystyle{2tan\frac{x}{2}-sinx.tan\frac{x}{2}=sinx}$

$\displaystyle{sinx\left ( 1+tan\frac{x}{2} \right )=2tan\frac{x}{2}}$

$\displaystyle{sinx=\frac{2tan\frac{x}{2}}{1+tan\frac{x}{2}}~~~...(ii) }$

Substitusi pers.(ii) ke soal.

$\displaystyle{\int\limits^{2\pi}_0 {\frac{dx}{2-sinx}} \, }$

$\displaystyle{\int\limits^{2\pi}_0 {\frac{1}{2-\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}} \, dx }$

$\displaystyle{=\int\limits^{2\pi}_0 {\frac{1}{\frac{2+2tan^2\frac{x}{2}-2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}} \, dx }$

$\displaystyle{=\int\limits^{2\pi}_0 {\frac{1+tan^2\frac{x}{2}}{2(tan^2\frac{x}{2}-tan\frac{x}{2}+1)}} \, dx }$

$\displaystyle{=\frac{1}{2}\int\limits^{2\pi}_0 {\frac{sec^2\frac{x}{2}}{tan^2\frac{x}{2}-tan\frac{x}{2}+1}} \, dx }$

$---------------$

Misal :

$\displaystyle{u=\frac{x}{2}~\to~du=\frac{1}{2}dx}$

$\displaystyle{Untuk~x=0~\to~u=\frac{0}{2}=0 }$

$\displaystyle{Untuk~x=2\pi~\to~u=\frac{2\pi}{2}=\pi }$

$---------------$

$\displaystyle{=\frac{1}{2}\int\limits^{\pi}_0 {\frac{sec^2u}{tan^2u-tanu+1}} \, (2du) }$

$\displaystyle{=\int\limits^{\pi}_0 {\frac{sec^2u}{tan^2u-tanu+1}} \, du }$

$---------------$

Misal :

$v=tanu~\to~dv=sec^2udu$

$---------------$

$\displaystyle{=\int\limits^{\pi}_0 {\frac{sec^2u}{v^2-v+1}} \, \frac{dv}{sec^2u} }$

$\displaystyle{=\int\limits^{\pi}_0 {\frac{1}{(v^2-v+\frac{1}{4})+(1-\frac{1}{4})}} \, dv }$

$\displaystyle{=\int\limits^{\pi}_0 {\frac{1}{(v-\frac{1}{2})^2+\frac{3}{4}}} \, dv }$

$---------------$

Misal :

$\displaystyle{w=v-\frac{1}{2}~\to~dw=dv }$

$---------------$

$\displaystyle{=\int\limits^{\pi}_0 {\frac{1}{w^2+\frac{3}{4}}} \, dw }$

$\displaystyle{=\int\limits^{\pi}_0 {\frac{1}{\frac{3}{4}(\frac{4}{3}w^2+1)}} \, dw }$

$\displaystyle{=\frac{4}{3}\int\limits^{\pi}_0 {\frac{1}{(\frac{2}{\sqrt{3}}w)^2+1}} \, dw }$

$---------------$

Misal :

$\displaystyle{y=\frac{2}{\sqrt{3}}w~\to~dy=\frac{2}{\sqrt{3}}dw }$

$---------------$

$\displaystyle{=\frac{4}{3}\int\limits^{\pi}_0 {\frac{1}{y^2+1}} \, \frac{\sqrt{3}dy}{2} }$

$\displaystyle{=\frac{2\sqrt{3}}{3}\int\limits^{\pi}_0 {\frac{1}{y^2+1}} \, dy }$

$\displaystyle{=\frac{2\sqrt{3}}{3}arctany\Bigr|^{\pi}_0 }$

$\displaystyle{=\frac{2\sqrt{3}}{3}arctan\left ( \frac{2}{\sqrt{3}}w \right )\Bigr|^{\pi}_0 }$

$\displaystyle{=\frac{2\sqrt{3}}{3}arctan\left [ \frac{2}{\sqrt{3}}\left ( v-\frac{1}{2} \right ) \right ]\Bigr|^{\pi}_0 }$

$\displaystyle{=\frac{2\sqrt{3}}{3}arctan\left [ \frac{2}{\sqrt{3}}\left ( tanu-\frac{1}{2} \right ) \right ]\Bigr|^{\pi}_0 }$

$\displaystyle{=\frac{2\sqrt{3}}{3}\left \{ arctan\left [ \frac{2}{\sqrt{3}}\left ( tan\pi-\frac{1}{2} \right ) \right ]-arctan\left [ \frac{2}{\sqrt{3}}\left ( tan0-\frac{1}{2} \right ) \right ] \right \} }$

$\displaystyle{=\frac{2\sqrt{3}}{3}\left \{ arctan\left [ \frac{2}{\sqrt{3}}\left ( 0-\frac{1}{2} \right ) \right ]-arctan\left [ \frac{2}{\sqrt{3}}\left ( 0-\frac{1}{2} \right ) \right ] \right \} }$