Question

O valor da integral ∫ $\frac{3\pi }{2}$ $\frac{\pi }{2}$ cos²(x) é?

a. - π

b. $\frac{-\pi }{2}$ RESPOSTA ERRADA

c. 1

d. 0

e. $\frac{\pi }{2}$

Answer 1

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$\displaystyle\sf{\int_{\frac{3\pi}{2}}^{\frac{\pi}{2}}cos^2(x)~dx=\dfrac{1}{2}\int_{\frac{3\pi}{2}}^{\frac{\pi}{2}}(1+cos(2x))~dx}\\\sf{\dfrac{1}{2}x+\dfrac{1}{4}sen(2x)\Bigg|_{\frac{3\pi}{2}}^{\frac{\pi}{2}}}\\\sf{\dfrac{1}{2}\cdot\left(\dfrac{\pi}{2}\right)+\dfrac{1}{4}sen\left(\diagup\!\!\!2\cdot\dfrac{\pi}{\diagup\!\!\!2}\right)-\left[\dfrac{1}{2}\cdot\left(\dfrac{3\pi}{2}\right)+\dfrac{1}{4}sen\left(\diagup\!\!\!2\cdot\dfrac{3\pi}{\diagup\!\!\!2}\right)\right]}$

$\sf{\dfrac{\pi}{4}+\dfrac{1}{4}sen(\pi)-\dfrac{3\pi}{4}-\dfrac{1}{4}sen(3\pi)}\\\sf{\dfrac{\pi}{4}-\dfrac{3\pi}{4}=-\dfrac{\diagup\!\!\!\!\!\!2^1\pi}{\diagup\!\!\!4_2}=-\dfrac{\pi}{2}}$

$\sf{nota:~acredito~que~os~limites~de~integrac_{\!\!,}\tilde{a}o~estejam~invertidos}\\\sf{por~essa~raz\tilde{a}o~a~resposta~deu~-\dfrac{\pi}{2}}$