[tex]{ {∫}^{2\pi}_{0}} \: \frac{dx}{2 \: - \: sin \: x} \: \: = \: ? [/tex]
ninuninuninuninu...
Hasil dari [tex]\displaystyle{\int\limits^{2\pi}_0 {\frac{dx}{2-sinx}} \, }[/tex] adalah [tex]\displaystyle{\boldsymbol{\frac{2\sqrt{3}}{3}\pi} }[/tex].
PEMBAHASAN
Sifat sifat operasi pada integral :
[tex](i)~\displaystyle{\int\limits {ax^n} \, dx=\frac{a}{n+1}x^{n+1}+C}[/tex]
[tex](ii)~\displaystyle{\int\limits {kf(x)} \, dx=k\int\limits {f(x)} \, dx}[/tex]
[tex](iii)~\displaystyle{\int\limits {\left [ f(x)\pm g(x) \right ]} \, dx=\int\limits {f(x)} \, dx\pm\int\limits {g(x)} \, dx}[/tex]
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DIKETAHUI
[tex]\displaystyle{\int\limits^{2\pi}_0 {\frac{dx}{2-sinx}} \, = }[/tex]
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DITANYA
Tentukan hasilnya.
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PENYELESAIAN
Rumus trigonometri 1/2 sudut:
[tex]\displaystyle{tan\frac{x}{2}=\frac{sinx}{1+cosx}~dan~tan\frac{x}{2}=\frac{1-cosx}{sinx}}[/tex]
[tex]\displaystyle{tan\frac{x}{2}=\frac{1-cosx}{sinx} }[/tex]
[tex]\displaystyle{sinx.tan\frac{x}{2}=1-cosx }[/tex]
[tex]\displaystyle{cosx=1-sinx.tan\frac{x}{2}~~~...(i) }[/tex]
[tex]\displaystyle{tan\frac{x}{2}=\frac{sinx}{1+cosx}~~~...substitusi~pers.(i)}[/tex]
[tex]\displaystyle{tan\frac{x}{2}=\frac{sinx}{1+1-sinx.tan\frac{x}{2}}}[/tex]
[tex]\displaystyle{tan\frac{x}{2}=\frac{sinx}{2-sinx.tan\frac{x}{2}}}[/tex]
[tex]\displaystyle{2tan\frac{x}{2}-sinx.tan\frac{x}{2}=sinx}[/tex]
[tex]\displaystyle{sinx\left ( 1+tan\frac{x}{2} \right )=2tan\frac{x}{2}}[/tex]
[tex]\displaystyle{sinx=\frac{2tan\frac{x}{2}}{1+tan\frac{x}{2}}~~~...(ii) }[/tex]
Substitusi pers.(ii) ke soal.
[tex]\displaystyle{\int\limits^{2\pi}_0 {\frac{dx}{2-sinx}} \, }[/tex]
[tex]\displaystyle{\int\limits^{2\pi}_0 {\frac{1}{2-\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}}} \, dx }[/tex]
[tex]\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 }[/tex]
[tex]\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 }[/tex]
[tex]\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 }[/tex]
[tex]---------------[/tex]
Misal :
[tex]\displaystyle{u=\frac{x}{2}~\to~du=\frac{1}{2}dx}[/tex]
[tex]\displaystyle{Untuk~x=0~\to~u=\frac{0}{2}=0 }[/tex]
[tex]\displaystyle{Untuk~x=2\pi~\to~u=\frac{2\pi}{2}=\pi }[/tex]
[tex]---------------[/tex]
[tex]\displaystyle{=\frac{1}{2}\int\limits^{\pi}_0 {\frac{sec^2u}{tan^2u-tanu+1}} \, (2du) }[/tex]
[tex]\displaystyle{=\int\limits^{\pi}_0 {\frac{sec^2u}{tan^2u-tanu+1}} \, du }[/tex]
[tex]---------------[/tex]
Misal :
[tex]v=tanu~\to~dv=sec^2udu[/tex]
[tex]---------------[/tex]
[tex]\displaystyle{=\int\limits^{\pi}_0 {\frac{sec^2u}{v^2-v+1}} \, \frac{dv}{sec^2u} }[/tex]
[tex]\displaystyle{=\int\limits^{\pi}_0 {\frac{1}{(v^2-v+\frac{1}{4})+(1-\frac{1}{4})}} \, dv }[/tex]
[tex]\displaystyle{=\int\limits^{\pi}_0 {\frac{1}{(v-\frac{1}{2})^2+\frac{3}{4}}} \, dv }[/tex]
[tex]---------------[/tex]
Misal :
[tex]\displaystyle{w=v-\frac{1}{2}~\to~dw=dv }[/tex]
[tex]---------------[/tex]
[tex]\displaystyle{=\int\limits^{\pi}_0 {\frac{1}{w^2+\frac{3}{4}}} \, dw }[/tex]
[tex]\displaystyle{=\int\limits^{\pi}_0 {\frac{1}{\frac{3}{4}(\frac{4}{3}w^2+1)}} \, dw }[/tex]
[tex]\displaystyle{=\frac{4}{3}\int\limits^{\pi}_0 {\frac{1}{(\frac{2}{\sqrt{3}}w)^2+1}} \, dw }[/tex]
[tex]---------------[/tex]
Misal :
[tex]\displaystyle{y=\frac{2}{\sqrt{3}}w~\to~dy=\frac{2}{\sqrt{3}}dw }[/tex]
[tex]---------------[/tex]
[tex]\displaystyle{=\frac{4}{3}\int\limits^{\pi}_0 {\frac{1}{y^2+1}} \, \frac{\sqrt{3}dy}{2} }[/tex]
[tex]\displaystyle{=\frac{2\sqrt{3}}{3}\int\limits^{\pi}_0 {\frac{1}{y^2+1}} \, dy }[/tex]
[tex]\displaystyle{=\frac{2\sqrt{3}}{3}arctany\Bigr|^{\pi}_0 }[/tex]
[tex]\displaystyle{=\frac{2\sqrt{3}}{3}arctan\left ( \frac{2}{\sqrt{3}}w \right )\Bigr|^{\pi}_0 }[/tex]
[tex]\displaystyle{=\frac{2\sqrt{3}}{3}arctan\left [ \frac{2}{\sqrt{3}}\left ( v-\frac{1}{2} \right ) \right ]\Bigr|^{\pi}_0 }[/tex]
[tex]\displaystyle{=\frac{2\sqrt{3}}{3}arctan\left [ \frac{2}{\sqrt{3}}\left ( tanu-\frac{1}{2} \right ) \right ]\Bigr|^{\pi}_0 }[/tex]
[tex]\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 \} }[/tex]
[tex]\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 \} }[/tex]
[tex]\displaystyle{=\frac{2\sqrt{3}}{3}\left \{ arctan\left [ -\frac{1}{\sqrt{3}} \right ]-arctan\left [ -\frac{1}{\sqrt{3}} \right ] \right \} }[/tex]
[tex]\displaystyle{=\frac{2\sqrt{3}}{3}\left ( \frac{11}{6}\pi-\frac{5}{6}\pi \right ) }[/tex]
[tex]\displaystyle{=\frac{2\sqrt{3}}{3}\pi }[/tex]
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KESIMPULAN
Hasil dari [tex]\displaystyle{\int\limits^{2\pi}_0 {\frac{dx}{2-sinx}} \, }[/tex] adalah [tex]\displaystyle{\boldsymbol{\frac{2\sqrt{3}}{3}\pi} }[/tex].
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PELAJARI LEBIH LANJUT
- Integral : https://brainly.co.id/tugas/30176534
- Luas kurva : https://brainly.co.id/tugas/30113906
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DETAIL JAWABAN
Kelas : 11
Mapel: Matematika
Bab : Integral
Kode Kategorisasi: 11.2.10
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