Question

Evalute the integral integration [(Sin2x+cos2x)dx] bw units x=Pi/3 & Pi/2. Answer should be correct to 1 decimal.​

Answer 1

We're asked to evaluate,

$\displaystyle\longrightarrow I=\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}}\left(\sin(2x)+\cos(2x)\right)\ dx$

$\displaystyle\longrightarrow I=\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}}\sin(2x)\ dx+\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}}\cos(2x)\ dx$

Put,

$\longrightarrow u=2x$

$\longrightarrow du=2\ dx$

$\longrightarrow dx=\dfrac{1}{2}\ du$

$\longrightarrow x=\dfrac{\pi}{3}\quad\implies\quad u=\dfrac{2\pi}{3}$

$\longrightarrow x=\dfrac{\pi}{2}\quad\implies\quad u=\pi$

Then the integral becomes,

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int\limits_{\frac{2\pi}{3}}^{\pi}\sin u\ du+\dfrac{1}{2}\int\limits_{\frac{2\pi}{3}}^{\pi}\cos u\ du$

$\displaystyle\longrightarrow I=-\dfrac{1}{2}\Big[\cos u\Big]_{\frac{2\pi}{3}}^{\pi}+\dfrac{1}{2}\Big[\sin u\Big]_{\frac{2\pi}{3}}^{\pi}$

$\displaystyle\longrightarrow I=-\dfrac{1}{2}\left(\cos\pi-\cos\left(\dfrac{2\pi}{3}\right)\right)+\dfrac{1}{2}\left(\sin\pi-\sin\left(\dfrac{2\pi}{3}\right)\right)$

$\displaystyle\longrightarrow I=-\dfrac{1}{2}\left(-1+\dfrac{1}{2}\right)+\dfrac{1}{2}\left(0-\dfrac{\sqrt3}{2}\right)$

$\displaystyle\longrightarrow I=\dfrac{1-\sqrt3}{4}$

$\displaystyle\longrightarrow\underline{\underline{I=-0.2}}$