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INTEGRAL OF THE DAY - QUIZ #20Tentukan hasil dari :[tex]

Berikut ini adalah pertanyaan dari e18ht1nFinity pada mata pelajaran Matematika untuk jenjang Sekolah Menengah Pertama

INTEGRAL OF THE DAY - QUIZ #20Tentukan hasil dari :
\displaystyle \int^{ \frac{ \pi}{3}}_{ \frac{\pi}{12}} \frac{ \cos(x) +(2 + \sqrt{3} ) \sin(x) }{2 + \sqrt{3} } \cdot \sin(x) \: \sf {dx}

golden hint : (2 + √3) = tan (π/3 + π/12)
​

Jawaban dan Penjelasan

Berikut ini adalah pilihan jawaban terbaik dari pertanyaan diatas.

\int \cos~x.\sin~x~dx

=\int \left(\frac{1}{2}\times 2.\cos~x.\sin~x\right)~dx

=\frac{1}{2}~\int \sin~2x~dx

\boxed{\boxed{\int \cos~x.\sin~x~dx=-\frac{1}{4}\cos~2x+c}}

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

\int \sin^2~x~dx

=\int \left(\frac{1}{2}\times (1-\cos~2x)\right)~dx

\boxed{\boxed{\int \sin^2~x~dx=\frac{1}{2}\left(x-\frac{1}{2}.\sin~2x\right)+c}}

••••••••••••••••••••••••••••••••••••••••••••••••••

\int \limits_\frac{\pi}{12}^\frac{\pi}{3} \left(\frac{\cos~x~+~\left(2~+~\sqrt{3}\right).\sin~x}{2~+~\sqrt{3}}.\sin~x\right)~dx

\begin{array}{ll}=&\frac{1}{2~+~\sqrt{3}}.\int \limits_\frac{\pi}{12}^\frac{\pi}{3} (\cos~x.\sin~x)~dx\\&+~\int \limits_\frac{\pi}{12}^\frac{\pi}{3} \sin^2~x~dx\\~\\=&\frac{1}{2~+~\sqrt{3}}.\left[-\frac{1}{4}\cos~2x\right]_\frac{\pi}{12}^\frac{\pi}{3}\\&+~\left[\frac{1}{2}\left(x-\frac{1}{2}.\sin~2x\right)\right]_\frac{\pi}{12}^\frac{\pi}{3}\\~\\=&\frac{1}{2~+~\sqrt{3}}.\left[-\frac{1}{4}.\cos~\frac{2\pi}{3}-\left(-\frac{1}{4}.\cos~\frac{\pi}{6}\right)\right]\\&+\left[\frac{1}{2}\left(\frac{\pi}{3}-\frac{1}{2}.\sin~\frac{2\pi}{3}\right)\right.\\&\left.-\left(\frac{1}{2}\left(\frac{\pi}{12}-\frac{1}{2}.\sin~\frac{\pi}{6}\right)\right)\right]\\~\\=&\frac{1}{2~+~\sqrt{3}}.\left[\left(-\frac{1}{4}\right).\left(-\frac{1}{2}\right)-\left(-\frac{1}{4}\right).\left(\frac{1}{2}\sqrt{3}\right)\right]\\&+\left[\frac{1}{2}\left(\frac{\pi}{3}-\frac{1}{2}.\left(\frac{1}{2}\sqrt{3}\right)\right)\right.\\&\left.-\left(\frac{1}{2}\left(\frac{\pi}{12}-\frac{1}{2}.\left(\frac{1}{2}\right)\right)\right)\right]\\~\\=&\frac{1}{2~+~\sqrt{3}}.\left[\frac{1}{8}+\frac{1}{8}\sqrt{3}\right]\\&+\left[\frac{1}{2}\left(\frac{\pi}{3}-\frac{1}{4}\sqrt{3}\right)\right.\\&\left.-\left(\frac{1}{2}\left(\frac{\pi}{12}-\frac{1}{4}\right)\right)\right]\\~\\=&\frac{1+\sqrt{3}}{8\left(2+\sqrt{3}\right)}+\frac{\pi}{6}-\frac{1}{8}\sqrt{3}\\&-\frac{\pi}{24}+\frac{1}{8}\\~\\=&\frac{1+\sqrt{3}}{8\left(2+\sqrt{3}\right)}+\frac{1}{8}\left(\pi+1-\sqrt{3}\right)\\~\\=&\frac{1+\sqrt{3}+\left(2+\sqrt{3}\right)\left(\pi+1-\sqrt{3}\right)}{8\left(2+\sqrt{3}\right)}\\~\\=&\frac{\left(2+\sqrt{3}\right)\pi}{8\left(2+\sqrt{3}\right)}~+~\frac{1+\sqrt{3}+\left(2+\sqrt{3}\right)\left(1-\sqrt{3}\right)}{8\left(2+\sqrt{3}\right)}\end{array}

\boxed{\boxed{\red{\begin{array}{c}\int \limits_\frac{\pi}{12}^\frac{\pi}{3} \left(\frac{\cos~x~+~\left(2~+~\sqrt{3}\right).\sin~x}{2~+~\sqrt{3}}.\sin~x\right)~dx\\~\\\huge{=\frac{\pi}{8}}\end{array}}}}

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Last Update: Fri, 27 Aug 21
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