hasil integral 0 sampai π/3 4 sin (2x-π/2) dx adalah hal 24

Penyelesaian:

[tex] \int\limits^ \frac{ \pi }{3} _0 {4.sin(2x - \frac{ \pi }{2}) } \, dx [/tex]

d([tex]2x - \frac{ \pi }{2} [/tex]) = 2 dx
d([tex]2x - \frac{ \pi }{2} [/tex]) / 2 = dx

[tex]\int\limits^ \frac{ \pi }{3} _0 {4.sin(2x - \frac{ \pi }{2}) } \, d(2x - \frac{ \pi }{2}) . \frac{1}{2} [/tex]
= [tex]2 \int\limits^ \frac{ \pi }{3} _0 {sin(2x - \frac{ \pi }{2}) } \, d(2x - \frac{ \pi }{2}) [/tex]

∫ sin x dx = -cos x

[tex]2 \int\limits^ \frac{ \pi }{3} _0 {sin(2x - \frac{ \pi }{2}) } \, d(2x - \frac{ \pi }{2}) [/tex]
= [tex]-2.cos(2x - \frac{ \pi }{2}) ] 0-\ \textgreater \ \frac{ \pi }{3} [/tex]
= [tex](-2.cos(2( \frac{ \pi }{3}) - \frac{ \pi }{2}) - (-2.cos(2(0) - \frac{ \pi }{2}) [/tex]
= [tex]-2.cos( \frac{ \pi }{6}) - 2.cos( \frac{ \pi }{2}) [/tex]
= -2.cos30° - 2.cos90°
= [tex]-2. \frac{1}{2} \sqrt{3} - 0 [/tex]
= [tex] -\sqrt{3} [/tex]

Integral Tentu Fungsi Trigonometri.

[tex]\displaystyle \int_{0}^{\pi /3}4\sin \left ( 2x-\frac{\pi }{2} \right )dx=-4\int_{0}^{\pi /3}\cos 2x~dx\\ =-4\left [ \frac{1}{2}\sin 2x \right ]_{0}^{\pi /3}[/tex]
[tex]\displaystyle =\left [ -2\sin 2\left ( \frac{\pi}{3} \right ) \right ]-\left [ -2\sin 2\left ( 0\right ) \right ]\\ =-\sqrt{3}-0\\=-\sqrt{3}[/tex]