Calcular $f(x) = \dfrac{d}{dx}\displaystyle\int_1^{\sin x} 3t^2\ dt$.

$f(x) = \dfrac{d}{dx} \left.t^3\right|_1^{\sin x} = \dfrac{d}{dx} (\sin^3 x - \sin 1) = \fbox{$3(\sin^2 x)(\cos x)$}$

Determine a curva $y = f(x)$ no plano $xy$ que passa pelo ponto $(9, 4)$ e cujo coeficiente angular em cada ponto é $3\sqrt{x}$.

$\dfrac{d}{dx}f(x) = 3\sqrt{x}\ \Rightarrow\ f(x) = 2\sqrt{x^3} + c$

$f(9) = 4\ \Rightarrow\ c = -50\ \Rightarrow\ \fbox{$f(x) = 2\sqrt{x^3} - 50$}$

Calcular $I\ =\ \displaystyle\int \dfrac{x}{x^2 + 4}\ dx$.

Seja $x = 2\tan u$, $dx = 2\sec^2 u\ du$.

$I\ =\ \displaystyle\int \tan u\ du\ =\ \displaystyle\int \dfrac{\sin u}{\cos u} du$

Seja $v = \cos u$, $dv = -\sin u\ du$.

$I\ =\ -\displaystyle\int \dfrac{dv}{v} = -(\log |v|) + c = -(\log |\cos u|) + c = \fbox{$-\left(\log \cos \arctan \dfrac{x}{2}\right) + c$}$

Determinar a primitiva de $f(x) = \dfrac{2}{3}\sec^2 \dfrac{x}{3}$.

Seja $u = \dfrac{x}{3}$, $du = \dfrac{dx}{3}$.

$\displaystyle\int \dfrac{2}{3}\sec^2 \dfrac{x}{3}\ dx\ =\ 2\displaystyle\int \sec^2 u\ du\ =\ 2\tan u + c = \fbox{$2\left(\tan \dfrac{x}{3}\right) + c$}$

Determinar a primitiva de $f(x) = -\pi \sin (\pi x)$.

Seja $u = \pi x$, $du = \pi dx$.

$\displaystyle\int -\pi \sin (\pi x)\ dx\ =\ -\displaystyle\int \sin u\ du\ =\ (\cos u) + c = \fbox{$\cos (\pi x) + c$}$

Calcular $I\ =\ \displaystyle\int (x + 1)(3x - 2)\ dx$.

$I = \displaystyle\int 3x^2 + x - 2\ dx = \fbox{$x^3 + \dfrac{x^2}{2} - 2x + c$}$

Calcular $I\ =\ \displaystyle\int \dfrac{x^2 + x + 1}{\sqrt{x}}\ dx$.

$I\ =\ \displaystyle\int \dfrac{x^2}{\sqrt{x}}\ dx\ +\ \displaystyle\int \dfrac{x}{\sqrt{x}}\ dx\ +\ \displaystyle\int \dfrac{dx}{\sqrt{x}}$

$\fbox{$I = \dfrac{2}{5}\sqrt{x^5} + \dfrac{2}{3}\sqrt{x^3} + 2\sqrt{x} + c$}$

Calcular $I\ =\ \displaystyle\int \dfrac{t\sqrt{t} + \sqrt{t}}{t^2}\ dt$.

$I = \displaystyle\int \dfrac{dt}{\sqrt{t}} + \displaystyle\int \dfrac{dt}{\sqrt{t^3}} = 2\sqrt{t} - \dfrac{2}{\sqrt{t}} + c = \fbox{$\dfrac{2t - 2}{\sqrt{t}} + c$}$

Calcular $I\ =\ \displaystyle\int \theta \sqrt[4]{1 - \theta^2}\ d\theta$.

Seja $u = 1 - \theta^2$, $du = -2\theta d\theta$.

$I\ =\ -\dfrac{1}{2}\displaystyle\int \sqrt[4]{u}\ du\ =\ -\dfrac{2}{5}\sqrt[4]{u^5} + c = \fbox{$-\dfrac{2\sqrt[4]{(1 - \theta^2)^5}}{5} + c$}$

Calcular $\displaystyle\int \sqrt{3 - 2s}\ ds$.

Seja $u = 3 - 2s$, $du = -2ds$.

$\displaystyle\int \sqrt{3 - 2s}\ ds\ =\ -\dfrac{1}{2}\displaystyle\int \sqrt{u}\ du\ =\ -\dfrac{1}{3}\sqrt{u^3} + c = \fbox{$-\dfrac{\sqrt{(3 - 2s)^3}}{3} + c$}$

Exercício: queda livre na Lua.

Na Lua, a aceleração da gravidade é $1,6\ m/s^2$. Uma pedra é solta de um penhasco na Lua e atinge sua superfície 20 segundos depois. Quão fundo ela caiu? Qual era a velocidade no instante do impacto?

$\dfrac{dv}{dt} = 1,6$

$\displaystyle\int_0^{20} 1,6\ dt = \left.1,6t\right|_0^{20} = 32$

$\displaystyle\int_0^{20} v\ dt\ =\ \dfrac{5}{8}\displaystyle\int_0^{32} v\ dv\ =\ \dfrac{5}{8}\left.\dfrac{v^2}{2}\right|_0^{32} = 320$

$\fbox{$320$ metros, e $32$ m/s}$.

Obter a primitiva de $f(x) = \left(\sec \dfrac{\pi x}{2}\right)\left(\tan \dfrac{\pi x}{2}\right)$.

$f(x) = \dfrac{\sin \dfrac{\pi x}{2}}{\cos^2 \dfrac{\pi x}{2}}$

Seja $u = \cos \dfrac{\pi x}{2}$, $du = -\dfrac{\pi}{2}\sin \dfrac{\pi x}{2}\ dx$.

$\displaystyle\int f(x)\ dx\ =\ -\dfrac{2}{\pi} \displaystyle\int \dfrac{du}{u^2}\ =\ \dfrac{2}{\pi u} + c = \fbox{$\dfrac{2\sec \dfrac{\pi x}{2}}{\pi} + c$}$

Calcular $\displaystyle\int x(\sin x)\ dx$.

$\displaystyle\int x(\sin x)\ dx\ =\ -x(\cos x) + \displaystyle\int \cos x\ dx\ =\ \fbox{$-x(\cos x) + \sin x + c$}$

Encontrar $\displaystyle\int \sin^3 x\ dx$.

$I\ =\ \displaystyle\int \sin^3 x\ dx\ =\ \displaystyle\int (\sin x)(1 - \cos^2 x) x\ dx\ =$

$=\ -(\cos x)(1 - \cos^2 x) + 2\displaystyle\int (\cos^2 x)(\sin x)\ dx\ =$

$=\ -\cos x + \cos^3 x + 2\displaystyle\int (1 - \sin^2 x)(\sin x)\ dx\ =$

$=\ -\cos x + \cos^3 x + 2\displaystyle\int \sin x\ dx - 2\underset{I}{\underbrace{\displaystyle\int\sin^3 x\ dx}}$

$\fbox{$\displaystyle\int \sin^3 x\ dx\ =\ \dfrac{\cos^3 x}{3} - \cos x + c$}$

Duas formas de encontrar $I = \displaystyle\int (\sin x)(\cos x)\ dx$.

Primeira:

$I = \dfrac{1}{2}\displaystyle\int \sin (2x)\ dx = -\dfrac{\cos(2x)}{4} + c$

Segunda:

Seja $u = \sin x$, $du = \cos x\ dx$.

$I = \displaystyle\int u\ du = \dfrac{u^2}{2} + C = \dfrac{\sin^2 x}{2} + C$

Observemos que $-\dfrac{\cos(2x)}{4} - \dfrac{\sin^2 x}{2} = -\dfrac{1}{4}$, que é constante. Logo as duas respostas estão corretas, pois tratam-se de integrais indefinidas.

Calcular $I\ =\ \displaystyle\int \sin (\sqrt{x})\ dx$.

$I\ =\ x\sin (\sqrt{x}) - \displaystyle\int \dfrac{x\cos (\sqrt{x})}{2\sqrt{x}}\ dx$

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

$I\ =\ x\sin (\sqrt{x}) - \displaystyle\int u^2\cos u\ du$

$I\ =\ x\sin (\sqrt{x}) - u^2\sin u + 2\displaystyle\int u\sin u\ du$

$I\ =\ x\sin (\sqrt{x}) - u^2\sin u - 2u\cos u + 2\displaystyle\int \cos u\ du$

$I\ =\ \cancel{x\sin (\sqrt{x})} - \cancel{x\sin (\sqrt{x})} - 2\sqrt{x}\cos (\sqrt{x}) + 2\sin (\sqrt{x}) + c$

$\fbox{$I = -2\sqrt{x}\cos (\sqrt{x}) + 2\sin (\sqrt{x}) + c$}$

Calcular $I\ =\ \displaystyle\int_0^{2\pi} cos^3 x\ dx$.

 $I\ =\ \cancelto{0}{\left.[(\sin x)(\cos^2 x)]\right|_0^{2\pi}} + \cancelto{0}{2\displaystyle\int_0^{2\pi} (\sin^2 x)(\cos x)\ dx} = \fbox{$0$}$

Integral da cotangente.

Seja $u = \sin x$, $du\ =\ \cos x\ dx$.

$\displaystyle\int \cot x\ dx = \displaystyle\int \dfrac{\cos x}{\sin x}\ dx\ = \displaystyle\int \dfrac{du}{u} = \fbox{$\log |\sin x| + c$}$

Integral da tangente.

Seja $u = \cos x$, $du\ =\ -\sin x\ dx$.

$\displaystyle\int \tan x\ dx = \displaystyle\int \dfrac{\sin x}{\cos x}\ dx\ =\ -\displaystyle\int \dfrac{du}{u}\ = \fbox{$-\log |\cos x| + c$}$

Calcular $I\ =\ \displaystyle\int \dfrac{\sqrt{x^2 + 4}}{4}\ dx$.

Seja $x = 2\tan \theta$, $-\dfrac{\pi}{2} < \theta < \dfrac{\pi}{2}$. $dx = 2\sec^2 \theta\ d\theta$

$I\ =\ \displaystyle\int \sec^3 \theta\ d\theta\ =\ (\sec \theta)(\tan \theta) - \displaystyle\int (\sec \theta)(\tan^2 \theta)\ d\theta\ =$

$=\ (\sec \theta)(\tan \theta) - \displaystyle\int \sec^3 \theta\ d\theta + \log |\sec \theta + \tan \theta|\ \Rightarrow$

$\Rightarrow\ I = \dfrac{(\sec \theta)(\tan \theta) + \log |\sec \theta + \tan \theta|}{2} + c\ =\ \fbox{$\dfrac{x\sqrt{4 + x^2}}{8} + \dfrac{\log \left|\sqrt{4 + x^2} + x\right|}{2} + c$}$