Question

calcule as integrais.
a imagem está logo acima dá uma ajuda aí por favor.​

Answer 1

a)

Para se resolver essa integral devemos recorrer à seguinte identidade trigonométrica: $\cos^2x=\frac{1+\cos(2x)}{2}$:

$\int_0^{\pi/2}\cos^2x\;dx=\int_0^{\pi/2}\frac{1+\cos(2x)}{2}\;dx$

$\int_0^{\pi/2}\cos^2x\;dx=\int_0^{\pi/2}\frac{1}{2}+\frac{\cos(2x)}{2}\;dx$

$\int_0^{\pi/2}\cos^2x\;dx=\int_0^{\pi/2}\frac{1}{2}\;dx+\int_0^{\pi/2}\frac{\cos(2x)}{2}\;dx$

$\int_0^{\pi/2}\cos^2x\;dx=\left[\frac{x}{2}\right]_0^{\pi/2}+\frac{1}{2}\int_0^{\pi/2}\cos(2x)\;dx$

Considerando $2x=u$, temos que $\frac{du}{dx}=2\therefore dx=\frac{1}{2}\;du$. Como iremos realizar uma mudança de variáveis, devemos recalcular os limites de integração. Para $x=0$, $u=2\cdot0=0$ e para $x=\pi/2$, $u=2\cdot\pi/2=\pi$. Substituindo:

$\int_0^{\pi/2}\cos^2x\;dx=\frac{\pi}{4}-\frac{0}{2}+\frac{1}{2}\int_0^{\pi}\cos u\cdot\frac{1}{2}\;du$

$\int_0^{\pi/2}\cos^2x\;dx=\frac{\pi}{4}+\frac{1}{4}\int_0^{\pi}\cos u\;du$

$\int_0^{\pi/2}\cos^2x\;dx=\frac{\pi}{4}+\frac{1}{4}\left[\sin u\right]_0^\pi$

$\int_0^{\pi/2}\cos^2x\;dx=\frac{\pi}{4}+\frac{1}{4}\left(\sin \pi-\sin 0\right)$

$\int_0^{\pi/2}\cos^2x\;dx=\frac{\pi}{4}+\frac{1}{4}(0-0)$

$\int_0^{\pi/2}\cos^2x\;dx=\frac{\pi}{4}$

b)

$\int_1^4\frac{1+x}{\sqrt{x}}\;dx=\int_1^4\frac{1}{\sqrt{x}}+\frac{x}{\sqrt{x}}\;dx$

$\int_1^4\frac{1+x}{\sqrt{x}}\;dx=\int_1^4\frac{1}{\sqrt{x}}\;dx+\int_1^4\frac{x}{\sqrt{x}}\;dx$

$\int_1^4\frac{1+x}{\sqrt{x}}\;dx=\int_1^4x^{-1/2}\;dx+\int_1^4x^{1-1/2}\;dx$

$\int_1^4\frac{1+x}{\sqrt{x}}\;dx=\int_1^4x^{-1/2}\;dx+\int_1^4x^{1/2}\;dx$

$\int_1^4\frac{1+x}{\sqrt{x}}\;dx=\left[\frac{x^{-1/2+1}}{-1/2+1}\right]_1^4+\left[\frac{x^{1/2+1}}{1/2+1}\right]_1^4$

$\int_1^4\frac{1+x}{\sqrt{x}}\;dx=\left[\frac{x^{1/2}}{1/2}\right]_1^4+\left[\frac{x^{3/2}}{3/2}\right]_1^4$

$\int_1^4\frac{1+x}{\sqrt{x}}\;dx=2\cdot \sqrt{4}-2\cdot\sqrt{1}+\frac{2}{3}\cdot \sqrt{4^3}-\frac{2}{3}\cdot\sqrt{1^3}$

$\int_1^4\frac{1+x}{\sqrt{x}}\;dx=4-2+\frac{2}{3}\cdot 8-\frac{2}{3}$

$\int_1^4\frac{1+x}{\sqrt{x}}\;dx=\frac{20}{3}$