$\displaystyle\int (\tan^7 x)(\sec^5 x)\ dx$.

$\tan^2 x = \sec^2 x - 1$

$I\ =\ \displaystyle\int (\tan^7 x)(\sec^5 x)\ dx = \displaystyle\int (\tan x)(\sec x)(\sec^2 x - 1)^3(\sec^4 x)\ dx$

Seja $u = \sec x$, $du = (\tan x)(\sec x) dx$.

$I\ =\ \displaystyle\int (u^2 - 1)^3 \cdot u^4\ du\ =\ \dfrac{u^{11}}{11} - \dfrac{u^9}{3} + \dfrac{3u^7}{7} - \dfrac{u^5}{5} + c$

$\fbox{$\displaystyle\int (\tan^7 x)(\sec^5 x)\ dx\ =\ \dfrac{\sec^{11} x}{11} - \dfrac{\sec^9 x}{3} + \dfrac{3\sec^7 x}{7} - \dfrac{\sec^5 x}{5} + c$}$

Notações. Limites superior e inferior de uma integral.

Seja $f$ uma função descontínua em um conjunto finito de pontos. Sejam $a$ e $b$ elementos de seu domínio.

$\intsup_a^S f(x)\ dx\ \avigual\ b\ \Leftrightarrow\ S = \displaystyle\int_a^b f(x)\ dx$

$\intinf_S^b f(x)\ dx\ \avigual\ a\ \Leftrightarrow\ S = \displaystyle\int_a^b f(x)\ dx$

Observemos que os limites não são únicos, por exemplo $\intsup_{\pi/2}^0 \sin x\ dx$ pode ser $\dfrac{3\pi}{2}$ ou $\dfrac{7\pi}{2}$, razão de não ser utilizada a igualdade "$=$", mas a igualdade conjunta de Antonio Vandré $\{=\}$.

Fórmula da integração por partes.

Pela fórmula do produto para derivadas, $(h \cdot g)'(x) = h'(x)g(x) + h(x)g'(x)$.

Seja $f(x) = h'(x)$ e $F$ a primitiva de $f$.

$\displaystyle\int (F \cdot g)'(x)\ dx\ =\ \displaystyle\int f(x)g(x)\ dx\ +\ \displaystyle\int F(x)g'(x)\ dx\ \Rightarrow$

 

$\Rightarrow\ \fbox{$\displaystyle\int f(x)g(x)\ dx\ =\ F(x)g(x) - \displaystyle\int F(x)g'(x)\ dx$}$

$\displaystyle\int \dfrac{dx}{x^5 + 1}$.

Afim de decompor $\dfrac{1}{x^5 + 1}$ em frações parciais, calculemos as raízes quintas de $-1$, que estão graficamente representadas abaixo:

Logo $\dfrac{1}{x^5 + 1} = \dfrac{Ax + B}{x^2 - \left(2\cos \dfrac{\pi}{5}\right)x + 1} + \dfrac{Cx + D}{x^2 - \left(2\cos \dfrac{3\pi}{5}\right)x + 1} + \dfrac{E}{x + 1}$. ${\large (I)}$

Donde, resolvendo o sistema:

${\tiny \begin{cases}B + D + E = 1\\ \left[4 - 4\cos \left(\dfrac{3\pi}{5}\right)\right]A + \left[4 - 4\cos \left(\dfrac{3\pi}{5}\right)\right]B + \left[4 - 4\cos \left(\dfrac{\pi}{5}\right)\right]C + \left[4 - 4\cos \left(\dfrac{\pi}{5}\right)\right]D + \left[4 - 4\cos \left(\dfrac{3\pi}{5}\right)\right]\left[4 - 4\cos \left(\dfrac{\pi}{5}\right)\right]E = 1\\ \left[30 - 24\cos \left(\dfrac{3\pi}{5}\right)\right]A + \left[15 - 12\cos \left(\dfrac{3\pi}{5}\right)\right]B + \left[30 - 24\cos \left(\dfrac{\pi}{5}\right)\right]C + \left[15 - 12\cos \left(\dfrac{\pi}{5}\right)\right]D + \left[5 - 4\cos \left(\dfrac{3\pi}{5}\right)\right]\left[5 - 4\cos \left(\dfrac{\pi}{5}\right)\right]E = 1\\ \left[120 - 72\cos \left(\dfrac{3\pi}{5}\right)\right]A + \left[40 - 24\cos \left(\dfrac{3\pi}{5}\right)\right]B + \left[120 - 72\cos \left(\dfrac{\pi}{5}\right)\right]C + \left[40 - 24\cos \left(\dfrac{\pi}{5}\right)\right]D + \left[10 - 6\cos \left(\dfrac{3\pi}{5}\right)\right]\left[10 - 6\cos \left(\dfrac{\pi}{5}\right)\right]E = 1\\ \left[10 + 8\cos \left(\dfrac{3\pi}{5}\right)\right]A - \left[5 + 4\cos \left(\dfrac{3\pi}{5}\right)\right]B + \left[10 + 8\cos \left(\dfrac{\pi}{5}\right)\right]C - \left[5 + 4\cos \left(\dfrac{\pi}{5}\right)\right]D + \left[5 + 4\cos \left(\dfrac{3\pi}{5}\right)\right]\left[5 + 4\cos \left(\dfrac{\pi}{5}\right)\right]E = 1\end{cases}}$,

obtemos:

${\tiny \begin{cases}A = \dfrac{\left( 24 {\cos^2 { \dfrac{\pi }{5} }}+92 \cos{ \dfrac{\pi }{5} }+200\right) \cos{ \dfrac{3 \pi }{5} }+32 {\cos^2 { \dfrac{\pi }{5} }}+170 \cos{ \dfrac{\pi }{5} }+285}{\left( 288 \cos{ \dfrac{\pi }{5} }+480\right) {\cos^2 { \dfrac{3 \pi }{5} }}+\left( 720-288 {\cos^2 { \dfrac{\pi }{5} }}\right) \cos{ \dfrac{3 \pi }{5} }-480 {\cos^2 { \dfrac{\pi }{5} }}-720 \cos{ \dfrac{\pi }{5} }}\\ B = -\dfrac{\left( 288 {\cos^2 { \dfrac{\pi }{5} }}+420 \cos{ \dfrac{\pi }{5} }-200\right) \cos{ \dfrac{3 \pi }{5} }+480 {\cos^2 { \dfrac{\pi }{5} }}+592 \cos{ \dfrac{\pi }{5} }-285}{\left( 288 \cos{ \dfrac{\pi }{5} }+480\right) {\cos^2 { \dfrac{3 \pi }{5} }}+\left( 720-288 {\cos^2 { \dfrac{\pi }{5} }}\right) \cos{ \dfrac{3 \pi }{5} }-480 {\cos^2 { \dfrac{\pi }{5} }}-720 \cos{ \dfrac{\pi }{5} }}\\ C = -\dfrac{\left( 24 \cos{ \dfrac{\pi }{5} }+32\right) {\cos^2 { \dfrac{3 \pi }{5} }}+\left( 92 \cos{ \dfrac{\pi }{5} }+170\right) \cos{ \dfrac{3 \pi }{5} }+200 \cos{ \dfrac{\pi }{5} }+285}{\left( 288 \cos{ \dfrac{\pi }{5} }+480\right) {\cos^2 { \dfrac{3 \pi }{5} }}+\left( 720-288 {\cos^2 { \dfrac{\pi }{5} }}\right) \cos{ \dfrac{3 \pi }{5} }-480 {\cos^2 { \dfrac{\pi }{5} }}-720 \cos{ \dfrac{\pi }{5} }}\\ D = \dfrac{\left( 288 \cos{ \dfrac{\pi }{5} }+480\right) {\cos^2 { \dfrac{3 \pi }{5} }}+\left( 420 \cos{ \dfrac{\pi }{5} }+592\right) \cos{ \dfrac{3 \pi }{5} }-200 \cos{ \dfrac{\pi }{5} }-285}{\left( 288 \cos{ \dfrac{\pi }{5} }+480\right) {\cos^2 { \dfrac{3 \pi }{5} }}+\left( 720-288 {\cos^2 { \dfrac{\pi }{5} }}\right) \cos{ \dfrac{3 \pi }{5} }-480 {\cos^2 { \dfrac{\pi }{5} }}-720 \cos{ \dfrac{\pi }{5} }}\\ E = -\dfrac{3}{\left( 12 \cos{ \dfrac{\pi }{5} }+20\right) \cos{ \dfrac{3 \pi }{5} }+20 \cos{ \dfrac{\pi }{5} }+30}\end{cases}}$. ${\large (II)}$

De ${\large (I)}$ obtemos:

${\tiny \dfrac{1}{x^5 + 1} = \dfrac{A}{2} \cdot \dfrac{2x - 2\cos \dfrac{\pi}{5}}{x^2 - \left(2\cos \dfrac{\pi}{5}\right)x + 1} + \dfrac{B + A\cos \dfrac{\pi}{5}}{\sin^2 \dfrac{\pi}{5}} \cdot \dfrac{1}{\left(\dfrac{x - \cos \dfrac{\pi}{5}}{\sin \dfrac{\pi}{5}}\right)^2 + 1} + \dfrac{C}{2} \cdot \dfrac{2x - 2\cos \dfrac{3\pi}{5}}{x^2 - \left(2\cos \dfrac{3\pi}{5}\right)x + 1} + \dfrac{D + C\cos \dfrac{3\pi}{5}}{\sin^2 \dfrac{3\pi}{5}} \cdot \dfrac{1}{\left(\dfrac{x - \cos \dfrac{3\pi}{5}}{\sin \dfrac{3\pi}{5}}\right)^2 + 1} + \dfrac{E}{x + 1}}$. ${\large (III)}$
 
Substituindo ${\large (II)}$ em ${\large (III)}$:

$\fbox{$\begin{array}{l}{\tiny \displaystyle\int \dfrac{dx}{x^5 + 1} \overset{x\ \neq\ \cos \dfrac{\pi}{5}}{=} \dfrac{\left( 24 {\cos^2 { \dfrac{\pi }{5} }}+92 \cos{ \dfrac{\pi }{5} }+200\right) \cos{ \dfrac{3 \pi }{5} }+32 {\cos^2 { \dfrac{\pi }{5} }}+170 \cos{ \dfrac{\pi }{5} }+285}{2 \left[ \left( 288 \cos{ \dfrac{\pi }{5} }+480\right) {\cos^2 { \dfrac{3 \pi }{5} }}+\left( 720-288 {\cos^2 { \dfrac{\pi }{5} }}\right) \cos{ \dfrac{3 \pi }{5} }-480 {\cos^2 { \dfrac{\pi }{5} }}-720 \cos{ \dfrac{\pi }{5} }\right] } \log \left|x^2 - \left(2\cos \dfrac{\pi}{5}\right)x + 1\right| +}\\ \\ {\tiny \dfrac{\dfrac{\cos{ \dfrac{\pi }{5} } \left[ \left( 24 {\cos^2 { \dfrac{\pi }{5} }}+92 \cos{ \dfrac{\pi }{5} }+200\right) \cos{ \dfrac{3 \pi }{5} }+32 {\cos^2 { \dfrac{\pi }{5} }}+170 \cos{ \dfrac{\pi }{5} }+285\right] }{\left( 288 \cos{ \dfrac{\pi }{5} }+480\right) {\cos^2 { \dfrac{3 \pi }{5} }}+\left( 720-288 {\cos^2 { \dfrac{\pi }{5} }}\right) \cos{ \dfrac{3 \pi }{5} }-480 {\cos^2 { \dfrac{\pi }{5} }}-720 \cos{ \dfrac{\pi }{5} }} - \dfrac{\left( 288 {\cos^2 { \dfrac{\pi }{5} }}+420 \cos{ \dfrac{\pi }{5} }-200\right) \cos{ \dfrac{3 \pi }{5} }+480 {\cos^2 { \dfrac{\pi }{5} }}+592 \cos{ \dfrac{\pi }{5} }-285}{\left( 288 \cos{ \dfrac{\pi }{5} }+480\right) {\cos^2 { \dfrac{3 \pi }{5} }}+\left( 720-288 {\cos^2 { \dfrac{\pi }{5} }}\right) \cos{ \dfrac{3 \pi }{5} }-480 {\cos^2 { \dfrac{\pi }{5} }}-720 \cos{ \dfrac{\pi }{5} }}}{\left(\sin \dfrac{\pi}{5}\right) ⋅ \arctan^{-1} \dfrac{x - \cos \dfrac{\pi}{5}}{\sin \dfrac{\pi}{5}}} -} \\ \\ {\tiny -\dfrac{\left( 24 \cos{ \dfrac{\pi }{5} }+32\right) {\cos^2 { \dfrac{3 \pi }{5} }}+\left( 92 \cos{ \dfrac{\pi }{5} }+170\right) \cos{ \dfrac{3 \pi }{5} }+200 \cos{ \dfrac{\pi }{5} }+285}{2 \left[ \left( 288 \cos{ \dfrac{\pi }{5} }+480\right) {\cos^2 { \dfrac{3 \pi }{5} }}+\left( 720-288 {\cos^2 { \dfrac{\pi }{5} }}\right] \cos{ \dfrac{3 \pi }{5} }-480 {\cos^2 { \dfrac{\pi }{5} }}-720 \cos{ \dfrac{\pi }{5} }\right) } \log \left|x^2 - \left(2\cos \dfrac{3\pi}{2}\right)x + 1\right| +}\\ \\ {\tiny +\dfrac{\dfrac{\left( 288 \cos{ \dfrac{\pi }{5} }+480\right) {\cos^2 { \dfrac{3 \pi }{5} }}+\left( 420 \cos{ \dfrac{\pi }{5} }+592\right) \cos{ \dfrac{3 \pi }{5} }-200 \cos{ \dfrac{\pi }{5} }-285}{\left( 288 \cos{ \dfrac{\pi }{5} }+480\right) {\cos^2 { \dfrac{3 \pi }{5} }}+\left( 720-288 {\cos^2 { \dfrac{\pi }{5} }}\right) \cos{ \dfrac{3 \pi }{5} }-480 {\cos^2 { \dfrac{\pi }{5} }}-720 \cos{ \dfrac{\pi }{5} - \dfrac{\cos{ \dfrac{3 \pi }{5} } \left( \left( 24 \cos{ \dfrac{\pi }{5} }+32\right) {\cos^2 { \dfrac{3 \pi }{5} }}+\left( 92 \cos{ \dfrac{\pi }{5} }+170\right) \cos{ \dfrac{3 \pi }{5} }+200 \cos{ \dfrac{\pi }{5} }+285\right) }{\left( 288 \cos{ \dfrac{\pi }{5} }+480\right) {\cos^2 { \dfrac{3 \pi }{5} }}+\left( 720-288 {\cos^2 { \dfrac{\pi }{5} }}\right) \cos{ \dfrac{3 \pi }{5} }-480 {\cos^2 { \dfrac{\pi }{5} }}-720 \cos{ \dfrac{\pi }{5} }}}}}{\left(\sin \dfrac{3\pi}{5}\right) ⋅ \arctan^{-1} \dfrac{x - \cos \dfrac{3\pi}{5}}{\sin \dfrac{3\pi}{5}}} -}\\ \\ {\tiny -\dfrac{3 \log \left|x + 1\right|}{\left( 12 \cos{ \dfrac{\pi }{5} }+20\right) \cos{ \dfrac{3 \pi }{5} }+20 \cos{ \dfrac{\pi }{5} }+30} + c}\\ \\ {\tiny c\ \in\ \mathbb{R}}\end{array}$}$.

Calcular $I\ =\ \displaystyle\int_0^1 \sqrt{x + 3}\ dx$.

Seja $u = x + 3$, $du = dx$.

 

$I\ =\ \displaystyle\int_3^4 \sqrt{u}\ du\ =\ \left.\dfrac{2\sqrt{u^3}}{3}\right|_3^4 = \fbox{$\dfrac{16}{3} - 2\sqrt{3}$}$

Considerações sobre o comprimento da senoide.

O comprimento da senoide é dado por $S = 4\displaystyle\int_0^{\pi / 2} \sqrt{1 + \cos^2 x}\ dx$.

Notemos que $0 \le \cos^2 x \le 1$, logo $4\displaystyle\int_0^{\pi / 2} \sqrt{1}\ dx\ \le\ S\ \le\ 4\displaystyle\int_0^{\pi / 2} \sqrt{1 + 1}\ dx\ \Rightarrow$

 

$\Rightarrow\ \fbox{$2\pi \le S \le 2\sqrt{2}\pi$}$.

Mostre que $\displaystyle\int_1^{+\infty} \dfrac{x}{x^4 + 1}\ dx$ é convergente.

$x^4 < x^4 + 1\ \Rightarrow\ \dfrac{1}{x^4 + 1} < \dfrac{1}{x^4}\ \overset{x \ge 1}{\Rightarrow}\ 0 < \dfrac{x}{x^4 + 1} < \dfrac{1}{x^3}$

Como $\displaystyle\int_1^{+\infty} \dfrac{dx}{x^3}$ converge, pelo critério da comparação, $\displaystyle\int_1^{+\infty} \dfrac{x}{x^4 + 1}\ dx$ é convergente.

 

Quod Erat Demonstrandum.

Calcular $I\ =\ \displaystyle\int \sec(x)\csc(x)\ dx$.

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

 

Seja $u = 2x$, $du = 2dx$.

 

$I\ =\ \displaystyle\int \csc(u)\ du\ =\ -\log |\cot(u) + \csc(u)| + c = \fbox{$-\log |\cot(2x) + \csc(2x)| + c$}$

Calcular $I\ =\ \displaystyle\int_0^{\sqrt{3}} \dfrac{4x}{\sqrt{x^2 + 1}}\ dx$.

Seja $u = x^2 + 1$, $du = 2x\ dx$.

 

$I\ =\ 2\displaystyle\int_1^4 \dfrac{du}{\sqrt{u}} = 4\left.\sqrt{u}\right|_1^4 = \fbox{$4$}$

Calcular $I\ =\ \displaystyle\int_0^\pi \sin^2\left(\dfrac{x}{4}\right) \cos\left(\dfrac{x}{4}\right)\ dx$.

Seja $u = \sin\left(\dfrac{x}{4}\right)$, $du\ =\ \dfrac{\cos\left(\dfrac{x}{4}\right)}{4}\ dx$.

 

$I\ =\ 4\displaystyle\int_0^{\sqrt{2}/2} u^2\ du\ = \fbox{$\dfrac{\sqrt{2}}{3}$}$

Calcular $I\ =\ \displaystyle\int_0^\pi \sin^2 \left(1 + \dfrac{\theta}{2}\right)\ d\theta$.

$I\ =\ \displaystyle\int_0^\pi \cos^2 \left(\dfrac{\pi}{2} - 1 - \dfrac{\theta}{2}\right)\ d\theta\ =\ \displaystyle\int_0^\pi \dfrac{\cos (\pi - 2 - \theta) + 1}{2}\ d\theta$

 

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

 

$I\ =\ \displaystyle\int_{-2}^{\pi - 2} \dfrac{1 + \cos u}{2}\ du\ =\ \left.\dfrac{u}{2}\right|_{-2}^{\pi - 2} + \left.\dfrac{\sin(u)}{2}\right|_{-2}^{\pi - 2} =$

 

$= \dfrac{\pi - 2}{2} + 1 + \dfrac{\sin(\pi - 2)}{2} + \dfrac{\sin 2}{2} = \fbox{$\dfrac{\pi}{2} + \sin 2$}$

Calcular $I\ =\ \displaystyle\int_4^9 2x\sqrt{x}\ dx$.

$I = \left.\dfrac{4}{5}\sqrt{x^5}\right|_4^9 = \dfrac{972}{5} - \dfrac{128}{5} = \fbox{$\dfrac{844}{5}$}$

Calcular a integral definida $I\ =\ \displaystyle\int_0^1 \dfrac{dx}{\sqrt{3x + 1}}$.

Seja $u = 3x + 1$, $du = 3dx$.

 

$I = \dfrac{1}{3}\displaystyle\int_1^4 \dfrac{du}{\sqrt{u}} = \dfrac{1}{3} \cdot \left.2\sqrt{u}\right|_1^4 = \dfrac{4}{3} - \dfrac{2}{3} = \fbox{$\dfrac{2}{3}$}$

Calcular a integral definida $I\ =\ \displaystyle\int_{\pi/6}^{\pi/2} \left(x + \dfrac{2}{\sin^2 x}\right)\ dx$.

$I = \left.\dfrac{x^2}{2}\right|_{\pi/6}^{\pi/2} - 2\left.\cot x\right|_{\pi/6}^{\pi/2} = \fbox{$\dfrac{\pi^2}{9} + 2\sqrt{3}$}$

Calcular a integral definida $I\ =\ \displaystyle\int_0^1 \dfrac{y^2}{\sqrt{4 - 3y}}\ dy$.

Seja $4 - 3y = x$, $y = \dfrac{4 - x}{3}$ e $dx = -3dy$.

 

$I\ =\ -\dfrac{1}{27}\displaystyle\int_4^1 \dfrac{16 - 8x + x^2}{\sqrt{x}}\ dx\ =\ -\dfrac{1}{27}\left.\left(32\sqrt{x} - \dfrac{16}{3}\sqrt{x^3} + \dfrac{2}{5}\sqrt{x^5}\right)\right|_4^1 =$

 

$=\ -\dfrac{32 - 64 - \dfrac{16}{3} + \dfrac{128}{3} + \dfrac{2}{5} - \dfrac{64}{5}}{27} = \fbox{$\dfrac{106}{405}$}$

Calcular a integral definida $I = \displaystyle\int_0^e \dfrac{dx}{x + e}$.

Seja $u = x + e$, $du = dx$.

 

$I = \displaystyle\int_e^{2e} \dfrac{du}{u} = \left.\log |u|\right|_e^{2e} = \log 2e - \log e = \fbox{$\log 2$}$

Calcular $I = \displaystyle\int \dfrac{9r^2\ dr}{\sqrt{1 - r^3}}$.

Seja $u = 1 - r^3,\ du\ =\ -3r^2\ dr$.

 

$I\ =\ -3\displaystyle\int \dfrac{du}{\sqrt{u}} = -6\sqrt{u} + c = \fbox{$-6\sqrt{1 - r^3} + c$}$

Calcular $I\ =\ \displaystyle\int_0^1 (1 - 2x)^3\ dx$.

Seja $u = 1 - 2x,\ du = -2dx$.

$I\ =\ -\dfrac{1}{2}\displaystyle\int_1^{-1} u^3\ du\ =\ -\dfrac{1}{2} \cdot \left.\dfrac{u^4}{4}\right|_1^{-1} = -\dfrac{1}{2}\left(\cancel{\dfrac{1}{4}} - \cancel{\dfrac{1}{4}}\right) = \fbox{$0$}$

Calcular $I\ =\ \displaystyle\int x^2 \sec(x^3)\ dx$.

Seja $u = x^3,\ du = 3x^2\ dx$.

$I\ =\ \dfrac{1}{3}\displaystyle\int \sec{u}\ du\ =\ \dfrac{\log |\sec u + \tan u|}{3} + c = \fbox{$\log \left|\sqrt[3]{\sec(x^3) + \tan(x^3)}\right| + c$}$

Calcular $I\ =\ \displaystyle\int e^{-x} + 4^x\ dx$.

$I = -e^{-x} +\ \displaystyle\int e^{x\log 4}\ dx\ + c = \fbox{$-e^{-x} + \dfrac{e^{x\log 4}}{\log 4} + c$}$