Question

O valor da integral ∫_pi/2^3pi/2 (cos^2(x))dx é:
a. -pi
b. 0
c. -pi/2
d. 1
e. pi/2

Answer 1

$\displaystyle \int\limits^{\frac{3\pi}{2}}_{\frac{\pi}{2}}\cos^2(x)\, dx\implies \cos^2(x)=\frac{1+cos(2x)}{2}\implies \int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\frac{1+\cos(2x)}{2}\,dx=\\\\\frac{1}{2}\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}1+cos(2x)\, dx=\frac{1}{2}\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\cos(2x)\,dx+\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}1\,dx=$
$\displaystyle 2x=u\ \ \ \ \frac{du}{dx}=2\implies dx=\frac{du}{2} \\\\ \frac{1}{2}\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\cos(u)\,\frac{du}{2}+\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}1\,dx\implies \frac{1}{2}\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\frac{\cos(u)}{2}\,du+\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}1\,dx\implies\\\\ \int\frac{1}{2}\cos(u)\,du=\frac{1}{2}\sin(u)+C\implies \int dx=x\implies$
$\displaystyle \frac{1}{2}\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\cos(u)\,\frac{du}{2}+\int\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}}1\,dx=\frac{1}{2}(\frac{1}{2}\cdot\sin(2x))\big|\limits^{\frac{3\pi}{2}}_{\frac{\pi}{2}}+x|\limits_{\frac{\pi}{2}}^{\frac{3\pi}{2}})\implies\\\\ \frac{1}{2}(\frac{\sin(2\frac{3\pi}{2})}{2}+\frac{3\pi}{2})-\frac{1}{2}(\frac{\sin(2\frac{\pi}{2})}{2}+\frac{\pi}{2})=\frac{1}{2}(0+\frac{3\pi}{2})-\frac{1}{2}(0+\frac{\pi}{2})=\\\\ \frac{3\pi}{4}-\frac{\pi}{4}=\frac{2\pi}{4}=\boxed{\frac{\pi}{2}}$