[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}

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DIKETAHUI

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

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DITANYA

Tentukan hasilnya.

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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 }

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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 }

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\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 }

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Misal :

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

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\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 }

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Misal :

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

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\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 }

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Misal :

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

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\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 \} }

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

\displaystyle{=\frac{2\sqrt{3}}{3}\left ( \frac{11}{6}\pi-\frac{5}{6}\pi \right ) }

\displaystyle{=\frac{2\sqrt{3}}{3}\pi }

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KESIMPULAN

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

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PELAJARI LEBIH LANJUT

  1. Integral : yomemimo.com/tugas/30176534
  2. Luas kurva : yomemimo.com/tugas/30113906

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DETAIL JAWABAN

Kelas : 11

Mapel: Matematika

Bab : Integral

Kode Kategorisasi: 11.2.10

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Last Update: Sat, 11 Jun 22