Calcule
(a) \int\left(e^{-x}+\operatorname{sen}(6 x)+\frac{2}{x^{5}}\right) d x
(b) \int\left(\sqrt[7]{x^{3}}+\operatorname{sen} \frac{x}{5}\right) d x
(c) \int_{0}^{x} \frac{1}{\sqrt{1-t^{2}}} d t
(d) \int 13^{x} \pi^{x} d x
(e) \int_{0}^{2} x^{3} \sqrt{2+3 x^{4}} d x
(f) \int_{-\pi / 3}^{\pi / 4}(7+\operatorname{sen} 5 x)^{2} d x
(g!) \int_{0}^{2 \pi} \sqrt{1+\cos x} d x
(h) \int \operatorname{sec} x d x
(i) \int_{-\pi / 2}^{\pi / 2} \operatorname{sen} 6 x \cos 3 x d x
(j) \int \cos 4 x \cos 2 x d x

Passo 1

O enunciado pede para calcular várias integrais, algumas indefinidas e outras definidas. Vamos usar as regras básicas de integração: linearidade, regra da potência, substituição simples, identidades trigonométricas, e o Teorema Fundamental do Cálculo.

Passo 2

Calcular \displaystyle\int\bigl(e^{-x}+\sin(6x)+\tfrac{2}{x^{5}}\bigr)\,dx.
A integral distribui-se em três termos:
\int e^{-x}\,dx = -e^{-x},\quad \int \sin(6x)\,dx = -\frac{1}{6}\cos(6x),\quad \int 2x^{-5}\,dx = 2\cdot\frac{x^{-4}}{-4} = -\tfrac{1}{2}x^{-4}.
Logo,
\int\bigl(e^{-x}+\sin(6x)+\tfrac{2}{x^{5}}\bigr)\,dx = -e^{-x} - \frac{1}{6}\cos(6x) - \frac{1}{2x^{4}} + C.

Passo 3

Calcular \displaystyle\int\bigl(\sqrt[7]{x^{3}} + \sin\tfrac{x}{5}\bigr)\,dx.
Note que \sqrt[7]{x^{3}} = x^{3/7}.
\int x^{3/7}\,dx = \frac{x^{10/7}}{10/7} = \frac{7}{10}x^{10/7},\quad \int \sin\!\bigl(\tfrac{x}{5}\bigr)\,dx \;\text{fazendo }u=\tfrac{x}{5},\,dx=5\,du, \;=\;5\!\int\sin u\,du = -5\cos u = -5\cos\!\bigl(\tfrac{x}{5}\bigr).
Assim,
\int\bigl(\sqrt[7]{x^{3}} + \sin\tfrac{x}{5}\bigr)\,dx = \frac{7}{10}x^{10/7} - 5\cos\!\bigl(\tfrac{x}{5}\bigr) + C.

Passo 4

Calcular \displaystyle\int_{0}^{x}\frac{1}{\sqrt{1-t^{2}}}\,dt.
Pelo Teorema Fundamental,
\int_{0}^{x}\frac{1}{\sqrt{1-t^{2}}}\,dt = \arcsin(t)\Big|_{0}^{x} = \arcsin(x) - \arcsin(0) = \arcsin(x).

Passo 5

Calcular \displaystyle\int 13^{x}\,\pi^{x}\,dx.
Note que 13^{x}\pi^{x}=(13\pi)^{x} e
\int a^{x}\,dx = \frac{a^{x}}{\ln(a)} + C.
Aqui a=13\pi, logo
\int 13^{x}\pi^{x}\,dx = \int(13\pi)^{x}\,dx = \frac{(13\pi)^{x}}{\ln(13\pi)} + C.

Passo 6

Calcular \displaystyle\int_{0}^{2} x^{3}\sqrt{2+3x^{4}}\,dx.
Use u=2+3x^{4}\implies du=12x^{3}dx\implies x^{3}dx=\tfrac{du}{12}.
Quando x=0, u=2; quando x=2, u=2+3\cdot16=50.
\int_{0}^{2} x^{3}\sqrt{2+3x^{4}}\,dx = \int_{2}^{50}\sqrt{u}\,\frac{du}{12} = \frac{1}{12}\int_{2}^{50}u^{1/2}\,du = \frac{1}{12}\cdot\frac{2}{3}u^{3/2}\Big|_{2}^{50} = \frac{1}{18}\bigl(50^{3/2}-2^{3/2}\bigr).
Como 50^{3/2}=50\sqrt{50}=250\sqrt{2} e 2^{3/2}=2\sqrt{2},
\int_{0}^{2} x^{3}\sqrt{2+3x^{4}}\,dx = \frac{1}{18}\bigl(250\sqrt{2}-2\sqrt{2}\bigr) = \frac{248\sqrt{2}}{18} = \frac{124\sqrt{2}}{9}.

Passo 7

Calcular \displaystyle\int_{-\pi/3}^{\pi/4}\bigl(7+\sin(5x)\bigr)^{2}\,dx.
Expanda o quadrado:
(7+\sin5x)^{2} = 49 + 14\sin5x + \sin^{2}5x.
Então
\int(49 + 14\sin5x + \sin^{2}5x)\,dx = 49x - \frac{14}{5}\cos5x + \int\sin^{2}5x\,dx.
Use \sin^{2}5x=\tfrac{1-\cos10x}{2}:
\int\sin^{2}5x\,dx = \frac{1}{2}\int(1-\cos10x)\,dx = \frac{x}{2} - \frac{1}{20}\sin10x.
Somando,
F(x)=49x - \frac{14}{5}\cos5x + \frac{x}{2} - \frac{1}{20}\sin10x = \frac{99}{2}x - \frac{14}{5}\cos5x - \frac{1}{20}\sin10x.
Avalie de -\pi/3 a \pi/4.
Após cálculo dos valores nos limites, obtém-se
\int_{-\pi/3}^{\pi/4}(7+\sin5x)^{2}dx = \frac{231\pi}{8} \;+\;\frac{56\sqrt{2}+\sqrt{3}+54}{40}.

Passo 8

Calcular \displaystyle\int_{0}^{2\pi}\sqrt{1+\cos x}\,dx.
Use 1+\cos x=2\cos^{2}(x/2), então
\sqrt{1+\cos x}=\sqrt{2}\,\bigl|\cos(x/2)\bigr|.
Sobre [0,\pi] \cos(x/2)\ge0, e sobre [\pi,2\pi] muda sinal. Fazendo a divisão e integrando,
\int_{0}^{2\pi}\sqrt{1+\cos x}\,dx =4\sqrt{2}.

Passo 9

Calcular \displaystyle\int \operatorname{sec} x\,dx.
Fórmula padrão:
\int \operatorname{sec} x\,dx = \ln\bigl|\operatorname{sec} x + \tan x\bigr| + C.

Passo 10

Calcular \displaystyle\int_{-\pi/2}^{\pi/2}\sin(6x)\cos(3x)\,dx.
Use produto-soma:
\sin6x\cos3x = \tfrac12\bigl(\sin9x + \sin3x\bigr).
Como \sin(kx) é ímpar, a integral simétrica é zero:
\int_{-\pi/2}^{\pi/2}\sin(6x)\cos(3x)\,dx = 0.

Passo 11

Calcular \displaystyle\int \cos(4x)\cos(2x)\,dx.
Produto-soma:
\cos4x\cos2x = \tfrac12\bigl(\cos6x + \cos2x\bigr).
Logo
\int\cos4x\cos2x\,dx = \frac12\int(\cos6x+\cos2x)\,dx = \frac{1}{12}\sin6x + \frac{1}{4}\sin2x + C.

Resposta

(a) \,-e^{-x} - \tfrac{1}{6}\cos(6x) - \tfrac{1}{2x^{4}} + C.
(b) \,\tfrac{7}{10}x^{10/7} - 5\cos\bigl(\tfrac{x}{5}\bigr) + C.
(c) \,\arcsin(x).
(d) \,\tfrac{(13\pi)^{x}}{\ln(13\pi)} + C.
(e) \,\tfrac{124\sqrt{2}}{9}.
(f) \,\tfrac{231\pi}{8} + \tfrac{56\sqrt{2}+\sqrt{3}+54}{40}.
(g) \,4\sqrt{2}.
(h) \,\ln\bigl|\operatorname{sec} x + \tan x\bigr| + C.
(i) \,0.
(j) \,\tfrac{1}{12}\sin(6x) + \tfrac{1}{4}\sin(2x) + C.

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