[tex]\begin{aligned}\vphantom{\Bigg|}&\frac{1}{2}\int_{\pi/6}^{\pi/3}\left(\frac{8\tan(x)-\sin(2x)}{\sin^2(x)+2\cos(x)-1}\right)dx\\\vphantom{\Bigg|}&=\boxed{\vphantom{\bigg|}\ln\left(e^{\frac{1}{3}\left(12-4\sqrt{3}\right)}\sqrt{3}\right)}\quad{\sf(opsi\ C.)}\end{aligned}[/tex]
Integral
Karena saya bukan ahli integral dan trigonometri, maka saya tentukan integral tak tentunya terlebih dahulu.
[tex]\begin{aligned}&\frac{1}{2}\int\left(\frac{8\tan(x)-\sin(2x)}{\sin^2(x)+2\cos(x)-1}\right)dx\\&{=\ }\frac{1}{2}\int\left(\frac{8\sin(x)\sec(x)-2\sin(x)\cos(x)}{1-\cos^2(x)+2\cos(x)-1}\right)dx\\&{=\ }\frac{1}{2}\int\left(\frac{2\sin(x)\left[4\sec(x)-\cos(x)\right]}{2\cos(x)-\cos^2(x)}\right)dx\\&{=\ }\frac{1}{2}\int\left(\frac{2\sin(x)\left[4\sec(x)-\cos(x)\right]}{\cos(x)\left[2-\cos(x)\right]}\right)dx\end{aligned}[/tex][tex]\begin{aligned}&{=\ }\frac{1}{2}\int\left(2\tan(x)\cdot\frac{4\sec(x)-\cos(x)}{2-\cos(x)}\right)dx\\&{=\ }\frac{1}{2}\int\left(2\tan(x)\cdot\frac{\left(\dfrac{4-\cos^2(x)}{\cos(x)}\right)}{2-\cos(x)}\right)dx\\&{=\ }\frac{1}{2}\int\left(2\tan(x)\cdot\frac{4-\cos^2(x)}{\cos(x)\left(2-\cos(x)\right)}\right)dx\\&{=\ }\frac{1}{2}\int\left(2\tan(x)\cdot\frac{2+\cos(x)}{\cos(x)}\right)dx\\&{=\ }\frac{1}{2}\int\left(2\tan(x)\sec(x)\left[2+\cos(x)\right]\right)dx\end{aligned}[/tex][tex]\begin{aligned}&{=\ }\frac{1}{2}\int\left(4+2\cos(x)\right)\tan(x)\sec(x)\,dx\\&{=\ }\frac{1}{2}\int\left(4+\frac{2}{\sec(x)}\right)\tan(x)\sec(x)\,dx\\&\quad...\ {\sf Ambil\ }u=\sec(x)\\&\qquad\Rightarrow du=\tan(x)\sec(x)\,dx\\&{=\ }\frac{1}{2}\int\left(4+\frac{2}{u}\right)du\\&{=\ }\frac{1}{2}\int4\,du\:+\:\frac{1}{2}\int\frac{2}{u}\,du\end{aligned}[/tex][tex]\begin{aligned}&{=\ }2\int\,du\:+\:\int\frac{1}{u}\,du\\&{=\ }2u+\ln(u)\\&\quad...\ \textsf{Substitusi kembali }u.\\&{=\ }2\sec(x)+\ln\left(\sec(x)\right)+C\end{aligned}[/tex]
Maka:
[tex]\begin{aligned}&\frac{1}{2}\int_{\pi/6}^{\pi/3}\left(\frac{8\tan(x)-\sin(2x)}{\sin^2(x)+2\cos(x)-1}\right)dx\\&{=\ }\Bigl[2\sec(x)+\ln\left(\sec(x)\right)\Bigr]_{\pi/6}^{\pi/3}\\&{=\ }2\sec\left(\frac{\pi}{3}\right)+\ln\left(\sec\left(\frac{\pi}{3}\right)\right)-\left[2\sec\left(\frac{\pi}{6}\right)+\ln\left(\sec\left(\frac{\pi}{6}\right)\right)\right]\\&\quad...\ \sec\left(\frac{\pi}{3}\right)=\frac{1}{\frac{1}{2}}=2\end{aligned}[/tex][tex]\begin{aligned}&\quad...\ \sec\left(\frac{\pi}{6}\right)=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\&{=\ }2\cdot2+\ln\left(2\right)-\left[2\cdot\frac{2\sqrt{3}}{3}+\ln\left(\frac{2\sqrt{3}}{3}\right)\right]\\&{=\ }4-\frac{4\sqrt{3}}{3}+\ln(2)-\ln\left(\frac{2\sqrt{3}}{3}\right)\\&{=\ }\frac{1}{3}\left(12-4\sqrt{3}\right)+\ln\left(\frac{\cancel{2}}{\frac{\cancel{2}\sqrt{3}}{3}}\right)\end{aligned}[/tex][tex]\begin{aligned}&{=\ }\frac{1}{3}\left(12-4\sqrt{3}\right)+\ln\left(\frac{3}{\sqrt{3}}\right)\\&{=\ }\boxed{\frac{1}{3}\left(12-4\sqrt{3}\right)+\ln\left(\sqrt{3}\right)}\\&{=\ }\ln\left(e^{\frac{1}{3}\left(12-4\sqrt{3}\right)}\right)+\ln\left(\sqrt{3}\right)\\&{=\ }\boxed{\vphantom{\bigg|}\ln\left(e^{\frac{1}{3}\left(12-4\sqrt{3}\right)}\sqrt{3}\right)}\quad{\sf(opsi\ C.)}\end{aligned}[/tex]
Jawab:
C
Penjelasan dengan langkah-langkah:
[tex]\begin{aligned}&\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{8\tan x-\sin 2x}{\sin^2 x+2\cos x-1}~dx\\&=\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\frac{8\sin x}{\cos x}-2\sin x\cos x}{1-\cos^2 x+2\cos x-1}~dx\\&=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\frac{4\sin x}{\cos x}-\sin x\cos x}{(2-\cos x)\cos x}~dx\\&=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{4\sin x-\sin x\cos^2 x}{(2-\cos x)\cos^2 x}~dx\\\end{aligned}[/tex]
[tex]\begin{aligned}&=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{(4-\cos^2 x)\sin x}{(2-\cos x)\cos^2 x}~dx\\&=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{(2-\cos x)(2+\cos x)\sin x}{(2-\cos x)\cos^2 x}~dx\\&=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{(2+u)\sin x}{u^2}~\frac{du}{-\sin x}\\&=-\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left ( \frac{1}{u}+\frac{2}{u^2} \right )du\end{aligned}[/tex]
[tex]\begin{aligned}&=-\left [ \ln|u|-\frac{2}{u} \right ]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\\&=\left [ \frac{2}{\cos x}-\ln|\cos x| \right ]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\\&=\left [2 \sec x+\ln|\cos x|^{-1} \right ]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\\&=\left [ 2\sec x+\ln\left | \frac{1}{\cos x} \right | \right ]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\\&=\left [2 \sec x+\ln|\sec x| \right ]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\end{aligned}[/tex]
[tex]\begin{aligned}&=2\sec\frac{\pi}{3}+\ln\left | \sec\frac{\pi}{3} \right |-2\sec\frac{\pi}{6}-\ln\left | \sec\frac{\pi}{3} \right |\\&=4-\frac{4\sqrt{3}}{3}+\ln|2|-\ln\left | \frac{2\sqrt{3}}{3} \right |\\&=\frac{12-4\sqrt{3}}{3}+\ln\left | \frac{2}{\frac{2\sqrt{3}}{3}} \right |\\&=\ln e^{\frac{1}{3}\left ( 12-4\sqrt{3} \right )}+\ln\left | \sqrt{3} \right |\\&=\ln\left [ e^{\frac{1}{3}\left ( 12-4\sqrt{3} \right )}\sqrt{3} \right ]\end{aligned}[/tex]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
[tex]\begin{aligned}\vphantom{\Bigg|}&\frac{1}{2}\int_{\pi/6}^{\pi/3}\left(\frac{8\tan(x)-\sin(2x)}{\sin^2(x)+2\cos(x)-1}\right)dx\\\vphantom{\Bigg|}&=\boxed{\vphantom{\bigg|}\ln\left(e^{\frac{1}{3}\left(12-4\sqrt{3}\right)}\sqrt{3}\right)}\quad{\sf(opsi\ C.)}\end{aligned}[/tex]
Penjelasan dengan langkah-langkah:
Integral
Karena saya bukan ahli integral dan trigonometri, maka saya tentukan integral tak tentunya terlebih dahulu.
[tex]\begin{aligned}&\frac{1}{2}\int\left(\frac{8\tan(x)-\sin(2x)}{\sin^2(x)+2\cos(x)-1}\right)dx\\&{=\ }\frac{1}{2}\int\left(\frac{8\sin(x)\sec(x)-2\sin(x)\cos(x)}{1-\cos^2(x)+2\cos(x)-1}\right)dx\\&{=\ }\frac{1}{2}\int\left(\frac{2\sin(x)\left[4\sec(x)-\cos(x)\right]}{2\cos(x)-\cos^2(x)}\right)dx\\&{=\ }\frac{1}{2}\int\left(\frac{2\sin(x)\left[4\sec(x)-\cos(x)\right]}{\cos(x)\left[2-\cos(x)\right]}\right)dx\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\frac{1}{2}\int\left(2\tan(x)\cdot\frac{4\sec(x)-\cos(x)}{2-\cos(x)}\right)dx\\&{=\ }\frac{1}{2}\int\left(2\tan(x)\cdot\frac{\left(\dfrac{4-\cos^2(x)}{\cos(x)}\right)}{2-\cos(x)}\right)dx\\&{=\ }\frac{1}{2}\int\left(2\tan(x)\cdot\frac{4-\cos^2(x)}{\cos(x)\left(2-\cos(x)\right)}\right)dx\\&{=\ }\frac{1}{2}\int\left(2\tan(x)\cdot\frac{2+\cos(x)}{\cos(x)}\right)dx\\&{=\ }\frac{1}{2}\int\left(2\tan(x)\sec(x)\left[2+\cos(x)\right]\right)dx\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\frac{1}{2}\int\left(4+2\cos(x)\right)\tan(x)\sec(x)\,dx\\&{=\ }\frac{1}{2}\int\left(4+\frac{2}{\sec(x)}\right)\tan(x)\sec(x)\,dx\\&\quad...\ {\sf Ambil\ }u=\sec(x)\\&\qquad\Rightarrow du=\tan(x)\sec(x)\,dx\\&{=\ }\frac{1}{2}\int\left(4+\frac{2}{u}\right)du\\&{=\ }\frac{1}{2}\int4\,du\:+\:\frac{1}{2}\int\frac{2}{u}\,du\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }2\int\,du\:+\:\int\frac{1}{u}\,du\\&{=\ }2u+\ln(u)\\&\quad...\ \textsf{Substitusi kembali }u.\\&{=\ }2\sec(x)+\ln\left(\sec(x)\right)+C\end{aligned}[/tex]
Maka:
[tex]\begin{aligned}&\frac{1}{2}\int_{\pi/6}^{\pi/3}\left(\frac{8\tan(x)-\sin(2x)}{\sin^2(x)+2\cos(x)-1}\right)dx\\&{=\ }\Bigl[2\sec(x)+\ln\left(\sec(x)\right)\Bigr]_{\pi/6}^{\pi/3}\\&{=\ }2\sec\left(\frac{\pi}{3}\right)+\ln\left(\sec\left(\frac{\pi}{3}\right)\right)-\left[2\sec\left(\frac{\pi}{6}\right)+\ln\left(\sec\left(\frac{\pi}{6}\right)\right)\right]\\&\quad...\ \sec\left(\frac{\pi}{3}\right)=\frac{1}{\frac{1}{2}}=2\end{aligned}[/tex]
[tex]\begin{aligned}&\quad...\ \sec\left(\frac{\pi}{6}\right)=\frac{1}{\frac{\sqrt{3}}{2}}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\\&{=\ }2\cdot2+\ln\left(2\right)-\left[2\cdot\frac{2\sqrt{3}}{3}+\ln\left(\frac{2\sqrt{3}}{3}\right)\right]\\&{=\ }4-\frac{4\sqrt{3}}{3}+\ln(2)-\ln\left(\frac{2\sqrt{3}}{3}\right)\\&{=\ }\frac{1}{3}\left(12-4\sqrt{3}\right)+\ln\left(\frac{\cancel{2}}{\frac{\cancel{2}\sqrt{3}}{3}}\right)\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\frac{1}{3}\left(12-4\sqrt{3}\right)+\ln\left(\frac{3}{\sqrt{3}}\right)\\&{=\ }\boxed{\frac{1}{3}\left(12-4\sqrt{3}\right)+\ln\left(\sqrt{3}\right)}\\&{=\ }\ln\left(e^{\frac{1}{3}\left(12-4\sqrt{3}\right)}\right)+\ln\left(\sqrt{3}\right)\\&{=\ }\boxed{\vphantom{\bigg|}\ln\left(e^{\frac{1}{3}\left(12-4\sqrt{3}\right)}\sqrt{3}\right)}\quad{\sf(opsi\ C.)}\end{aligned}[/tex]
Verified answer
Jawab:
C
Penjelasan dengan langkah-langkah:
[tex]\begin{aligned}&\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{8\tan x-\sin 2x}{\sin^2 x+2\cos x-1}~dx\\&=\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\frac{8\sin x}{\cos x}-2\sin x\cos x}{1-\cos^2 x+2\cos x-1}~dx\\&=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{\frac{4\sin x}{\cos x}-\sin x\cos x}{(2-\cos x)\cos x}~dx\\&=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{4\sin x-\sin x\cos^2 x}{(2-\cos x)\cos^2 x}~dx\\\end{aligned}[/tex]
[tex]\begin{aligned}&=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{(4-\cos^2 x)\sin x}{(2-\cos x)\cos^2 x}~dx\\&=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{(2-\cos x)(2+\cos x)\sin x}{(2-\cos x)\cos^2 x}~dx\\&=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{(2+u)\sin x}{u^2}~\frac{du}{-\sin x}\\&=-\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left ( \frac{1}{u}+\frac{2}{u^2} \right )du\end{aligned}[/tex]
[tex]\begin{aligned}&=-\left [ \ln|u|-\frac{2}{u} \right ]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\\&=\left [ \frac{2}{\cos x}-\ln|\cos x| \right ]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\\&=\left [2 \sec x+\ln|\cos x|^{-1} \right ]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\\&=\left [ 2\sec x+\ln\left | \frac{1}{\cos x} \right | \right ]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\\&=\left [2 \sec x+\ln|\sec x| \right ]_{\frac{\pi}{6}}^{\frac{\pi}{3}}\end{aligned}[/tex]
[tex]\begin{aligned}&=2\sec\frac{\pi}{3}+\ln\left | \sec\frac{\pi}{3} \right |-2\sec\frac{\pi}{6}-\ln\left | \sec\frac{\pi}{3} \right |\\&=4-\frac{4\sqrt{3}}{3}+\ln|2|-\ln\left | \frac{2\sqrt{3}}{3} \right |\\&=\frac{12-4\sqrt{3}}{3}+\ln\left | \frac{2}{\frac{2\sqrt{3}}{3}} \right |\\&=\ln e^{\frac{1}{3}\left ( 12-4\sqrt{3} \right )}+\ln\left | \sqrt{3} \right |\\&=\ln\left [ e^{\frac{1}{3}\left ( 12-4\sqrt{3} \right )}\sqrt{3} \right ]\end{aligned}[/tex]