Question

7) Calcule a seguinte integral definida:

$\int\limits^ \frac{ \pi }{2} _\frac{ \pi }{6} { (sen2x-cosx)} \, dx$

Answer 1

Seja $u = 2x.$
Então, $\dfrac{du}{dx}=2 \implies dx = \dfrac{du}{2}$

$\int(sen \ 2x - cos \ x)dx \\ \\ = \int(sen \ 2x)dx - \int (cos \ x)dx \\ \\ = \int(sen \ u)\dfrac{du}{2} - \int (cos \ x)dx \\ \\ = \dfrac{1}{2}\int(sen \ u)du - \int (cos \ x)dx \\ \\ = \dfrac{1}{2}cos \ u - sen \ x \\ \\ = -\dfrac{1}{2}cos \ 2x - sen \ x + C$

Substituindo x por pi/6 e pi/2 na integral indefinida:
$-\dfrac{1}{2} \ cos\left( 2 \cdot \dfrac{\pi}{6} \right) - sen \ \dfrac{\pi}{6} \\ \\ = -\dfrac{3}{4}$
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$-\dfrac{1}{2} \ cos\left( 2 \cdot \dfrac{\pi}{2} \right) - sen \ \dfrac{\pi}{2} \\ \\ = -\dfrac{1}{2}$

Subtraindo o primeiro resultado do segundo:

$-\dfrac{1}{2}+\dfrac{3}{4}=\boxed{\dfrac{1}{4}}$

Answer 2

$\displaystyle\mathsf{\int\limits_{\frac{\pi}{6}}^{\frac{\pi}{2}}(sen(2x)-cos(x))dx}=\\\mathsf{\left[-\dfrac{1}{2}cos(2x)-sen(x)\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}}$

$\mathsf{-\dfrac{1}{2}cos(2.\frac{\pi}{2})-sen(\dfrac{\pi}{2})} \\ \mathsf {-\left[-\dfrac{1}{2}cos(2\dfrac{\pi}{6})-sen(\dfrac{\pi}{6})\right]}$

$\mathsf{\dfrac{1}{2}-1+\dfrac{1}{4}+\dfrac{1}{2}} = \\ \huge\boxed{\boxed{\mathsf{ \dfrac{1}{4} }}}$