Calcule as integrais fazendo a substituição apropriada.
a) \int e^{2 x} d x
b) \int x \cdot\left(2-x^{2}\right)^{3} d x
c) \int \cos 8 x d x
d) \int \frac{d x}{2 x}
e) \int(3 x-1)^{5} d x
f) \int \operatorname{sen} 3 x d x
g) \int \operatorname{sec} 4 x \cdot \operatorname{tg} 4 x d x
h) \int \operatorname{sec} ^{2} 5 x d x
i) \int t \cdot \sqrt{7 t^{2}+12} d t
j) \int \frac{x}{\sqrt{4-5 x^{2}}} d x
k) \int \frac{x^{2}}{\sqrt{x^{3}+1}} d x
m) \int \frac{x}{\left(4 x^{2}+1\right)^{3}} d x
n) \int x \cdot \cos \left(3 x^{2}\right) d x
o) \int e^{\operatorname{sen} x} \cdot \cos x d x
p) \int x^{3} \cdot e^{x^{4}} d x
q) \int x^{2} \cdot e^{-2 x^{3}} d x
r) \int \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}} d x
s) \int \frac{\operatorname{sen}\left(\frac{5}{x}\right)}{x^{2}} d x
t) \int \frac{d x}{e^{x}}
u) \int \frac{e^{\sqrt{y}}}{\sqrt{y}} d y
v) \int \operatorname{sen}{ }^{5} 3 t \cdot \cos 3 t d t
  1. \int \frac{1}{(1-3 x)^{2}} d x
    w) \int \sqrt{e^{y}} d y

Passo 1

O enunciado pede para calcular várias integrais indefinidas usando a técnica de substituição. A ideia geral é identificar uma parte da função cujo diferencial aparece (ou é múltiplo) no integrando, fazer u=g(x), obter du=g'(x)\,dx e reescrever a integral em termos de u. Em seguida, integra-se em u e volta-se a variável original.

Passo 2

Para \int e^{2x}dx faça u=2x, daí du=2\,dx e dx=\tfrac{du}{2}.
\int e^{2x}dx = \int e^{u}\frac{du}{2} = \tfrac12 e^{u} + C = \tfrac12 e^{2x} + C.

Passo 3

Para \int x\,(2-x^{2})^{3}dx use u=2-x^{2}, então du=-2x\,dx e x\,dx=-\tfrac{du}{2}.
\int x\,(2-x^{2})^{3}dx = -\tfrac12 \int u^{3}du = -\tfrac12\cdot\tfrac{u^{4}}{4} + C = -\tfrac{(2-x^{2})^{4}}{8}+C.

Passo 4

Para \int \cos 8x\,dx faça u=8x, logo du=8\,dx, dx=\tfrac{du}{8}.
\int \cos 8x\,dx = \tfrac18\int \cos u\,du = \tfrac18\sin u + C = \tfrac18\sin(8x)+C.

Passo 5

Para \int \frac{dx}{2x} extraímos a constante:
\int \frac{dx}{2x} = \tfrac12\int \frac{dx}{x} = \tfrac12\ln\bigl|x\bigr| + C.

Passo 6

Para \int(3x-1)^{5}dx deixe u=3x-1, então du=3\,dx, dx=\tfrac{du}{3}.
\int(3x-1)^{5}dx = \tfrac13\int u^{5}du = \tfrac{u^{6}}{18} + C = \tfrac{(3x-1)^{6}}{18} + C.

Passo 7

Para \int \operatorname{sen} 3x\,dx tome u=3x, daí du=3\,dx, dx=\tfrac{du}{3}.
\int \operatorname{sen} 3x\,dx = \tfrac13\int \operatorname{sen} u\,du = -\tfrac13\cos u + C = -\tfrac13\cos(3x) + C.

Passo 8

Para \int \operatorname{sec} 4x\;\operatorname{tg} 4x\,dx use u=4x, du=4\,dx, dx=\tfrac{du}{4}.
\int \operatorname{sec} 4x\;\operatorname{tg} 4x\,dx = \tfrac14\int \operatorname{sec} u\,\operatorname{tg} u\,du = \tfrac14\operatorname{sec} u + C = \tfrac14\operatorname{sec}(4x)+C.

Passo 9

Para \int \operatorname{sec}^{2}5x\,dx faça u=5x, du=5\,dx, dx=\tfrac{du}{5}.
\int \operatorname{sec}^{2}5x\,dx = \tfrac15\int \operatorname{sec}^{2}u\,du = \tfrac15\tan u + C = \tfrac15\tan(5x)+C.

Passo 10

Para \int t\sqrt{7t^{2}+12}\,dt deixe u=7t^{2}+12, du=14t\,dt, t\,dt=\tfrac{du}{14}.
\int t\sqrt{7t^{2}+12}\,dt = \tfrac1{14}\int u^{\tfrac12}du = \tfrac1{14}\cdot\tfrac{u^{\tfrac32}}{\tfrac32}+C = \tfrac{(7t^{2}+12)^{\tfrac32}}{21}+C.

Passo 11

Para \int \frac{x}{\sqrt{4-5x^{2}}}\,dx faça u=4-5x^{2}, du=-10x\,dx, x\,dx=-\tfrac{du}{10}.
\int \frac{x}{\sqrt{4-5x^{2}}}\,dx = -\tfrac1{10}\int u^{-\tfrac12}du = -\tfrac1{10}\cdot\tfrac{u^{\tfrac12}}{\tfrac12}+C = -\tfrac15\sqrt{4-5x^{2}}+C.

Passo 12

Para \int \frac{x^{2}}{\sqrt{x^{3}+1}}\,dx use u=x^{3}+1, du=3x^{2}dx, x^{2}dx=\tfrac{du}{3}.
\int \frac{x^{2}}{\sqrt{x^{3}+1}}\,dx = \tfrac13\int u^{-\tfrac12}du = \tfrac13\cdot\tfrac{u^{\tfrac12}}{\tfrac12}+C = \tfrac{2}{3}\sqrt{x^{3}+1}+C.

Passo 13

Para \int \frac{x}{(4x^{2}+1)^{3}}\,dx deixe u=4x^{2}+1, du=8x\,dx, x\,dx=\tfrac{du}{8}.
\int \frac{x}{(4x^{2}+1)^{3}}\,dx = \tfrac18\int u^{-3}du = \tfrac18\cdot\tfrac{u^{-2}}{-2}+C = -\tfrac{1}{16}(4x^{2}+1)^{-2}+C.

Passo 14

Para \int x\cos(3x^{2})\,dx faça u=3x^{2}, du=6x\,dx, x\,dx=\tfrac{du}{6}.
\int x\cos(3x^{2})\,dx = \tfrac16\int \cos u\,du = \tfrac16\sin u + C = \tfrac16\sin(3x^{2})+C.

Passo 15

Para \int e^{\operatorname{sen} x}\cos x\,dx use u=\operatorname{sen} x, du=\cos x\,dx.
\int e^{\operatorname{sen} x}\cos x\,dx = \int e^{u}du = e^{u}+C = e^{\operatorname{sen} x}+C.

Passo 16

Para \int x^{3}e^{x^{4}}\,dx deixe u=x^{4}, du=4x^{3}dx, x^{3}dx=\tfrac{du}{4}.
\int x^{3}e^{x^{4}}\,dx = \tfrac14\int e^{u}du = \tfrac14 e^{u}+C = \tfrac14e^{x^{4}}+C.

Passo 17

Para \int x^{2}e^{-2x^{3}}\,dx faça u=-2x^{3}, du=-6x^{2}dx, x^{2}dx=-\tfrac{du}{6}.
\int x^{2}e^{-2x^{3}}\,dx = -\tfrac16\int e^{u}du = -\tfrac16e^{u}+C = -\tfrac16e^{-2x^{3}}+C.

Passo 18

Para \int \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\,dx use u=e^{x}-e^{-x}, du=(e^{x}+e^{-x})dx.
\int \frac{e^{x}+e^{-x}}{e^{x}-e^{-x}}\,dx = \int \frac{du}{u} = \ln\bigl|u\bigr|+C = \ln\bigl|e^{x}-e^{-x}\bigr|+C.

Passo 19

Para \int \frac{\operatorname{sen}\bigl(\tfrac5x\bigr)}{x^{2}}\,dx tome u=\tfrac5x, du=-\tfrac{5}{x^{2}}dx, \tfrac{dx}{x^{2}}=-\tfrac{du}{5}.
\int \frac{\operatorname{sen}(\tfrac5x)}{x^{2}}\,dx = -\tfrac15\int \operatorname{sen} u\,du = \tfrac15\cos u + C = \tfrac15\cos\bigl(\tfrac5x\bigr)+C.

Passo 20

Para \int \frac{dx}{e^{x}} note que isso é \int e^{-x}dx.
\int e^{-x}dx = -e^{-x}+C.

Passo 21

Para \int \frac{e^{\sqrt{y}}}{\sqrt{y}}\,dy faça u=\sqrt{y}, du=\tfrac{1}{2\sqrt{y}}dy, dy=2u\,du.
\int \frac{e^{\sqrt{y}}}{\sqrt{y}}\,dy = \int \frac{e^{u}}{u}\,2u\,du = 2\int e^{u}du = 2e^{u}+C = 2e^{\sqrt{y}}+C.

Passo 22

Para \int \operatorname{sen}^{5}(3t)\cos(3t)\,dt use u=\operatorname{sen}(3t), du=3\cos(3t)\,dt, \cos(3t)\,dt=\tfrac{du}{3}.
\int \operatorname{sen}^{5}(3t)\cos(3t)\,dt = \tfrac13\int u^{5}du = \tfrac{u^{6}}{18}+C = \tfrac{\operatorname{sen}^{6}(3t)}{18}+C.

Passo 23

Para \int \frac{1}{(1-3x)^{2}}\,dx deixe u=1-3x, du=-3\,dx, dx=-\tfrac{du}{3}.
\int \frac{1}{(1-3x)^{2}}\,dx = -\tfrac13\int u^{-2}du = -\tfrac13\cdot\bigl(\tfrac{u^{-1}}{-1}\bigr)+C = \tfrac{1}{3(1-3x)}+C.

Passo 24

Para \int \sqrt{e^{y}}\,dy note que \sqrt{e^{y}}=e^{y/2}. Então
\int e^{y/2}dy.
Use u=\tfrac{y}{2}, du=\tfrac12dy, dy=2du:
\int e^{y/2}dy = 2\int e^{u}du = 2e^{u}+C = 2e^{y/2}+C.

Resposta

a) \tfrac12e^{2x}+C
b) -\tfrac{(2-x^{2})^{4}}{8}+C
c) \tfrac18\sin(8x)+C
d) \tfrac12\ln\lvert x\rvert +C
e) \tfrac{(3x-1)^{6}}{18}+C
f) -\tfrac13\cos(3x)+C
g) \tfrac14\operatorname{sec}(4x)+C
h) \tfrac15\tan(5x)+C
i) \tfrac{(7t^{2}+12)^{3/2}}{21}+C
j) -\tfrac15\sqrt{4-5x^{2}}+C
k) \tfrac{2}{3}\sqrt{x^{3}+1}+C
m) -\tfrac{1}{16}(4x^{2}+1)^{-2}+C
n) \tfrac16\sin(3x^{2})+C
o) e^{\operatorname{sen} x}+C
p) \tfrac14e^{x^{4}}+C
q) -\tfrac16e^{-2x^{3}}+C
r) \ln\bigl|e^{x}-e^{-x}\bigr|+C
s) \tfrac15\cos\bigl(\tfrac5x\bigr)+C
t) -e^{-x}+C
u) 2e^{\sqrt{y}}+C
v) \tfrac{\operatorname{sen}^{6}(3t)}{18}+C
  1. \tfrac{1}{3(1-3x)}+C
w) 2e^{y/2}+C

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