Encontre às seguintes integrais:
  • \int x^{2} \cos 3 x d x
  • \int x \operatorname{sec} ^{2} 2 x d x
  • \int e^{2 x} \sin 3 x d x
  • \int x \ln (1+x) d x
  • \int \frac{x}{e^{2 x}} d x
  • \int \frac{(\ln x)^{2}}{x^{3}} d x
  • \int \cos \theta \cos ^{3}(\sin \theta) d \theta
  • \int \sin 3 x \cos x d x

Passo 1

Resumindo o enunciado: precisamos calcular as seguintes integrais indefinidas
\displaystyle\int x^{2}\cos(3x)\,dx,
\displaystyle\int x\operatorname{sec}^{2}(2x)\,dx,
\displaystyle\int e^{2x}\sin(3x)\,dx,
\displaystyle\int x\ln(1+x)\,dx,
\displaystyle\int \frac{x}{e^{2x}}\,dx,
\displaystyle\int \frac{(\ln x)^{2}}{x^{3}}\,dx,
\displaystyle\int \cos\theta\,[\cos(\sin\theta)]^{3}\,d\theta,
\displaystyle\int \sin(3x)\cos x\,dx.
Os principais métodos serão: integração por partes, substituição trigonométrica ou exponencial, e as fórmulas de soma de ângulos.

Passo 2

Para \displaystyle\int x^{2}\cos(3x)\,dx usamos duas vezes partes.
Escolhemos u=x^{2}, dv=\cos(3x)\,dx, daí du=2x\,dx, v=\tfrac{\sin(3x)}{3}.
Então
\int x^{2}\cos(3x)\,dx = \frac{x^{2}\sin(3x)}{3} - \frac{2}{3}\int x\sin(3x)\,dx.
Agora em \displaystyle\int x\sin(3x)\,dx usamos u=x, dv=\sin(3x)\,dx, obtendo
\int x\sin(3x)\,dx = -\frac{x\cos(3x)}{3} + \frac{1}{9}\sin(3x).
Substituindo de volta,
\int x^{2}\cos(3x)\,dx = \frac{x^{2}\sin(3x)}{3} + \frac{2x\cos(3x)}{9} - \frac{2\sin(3x)}{27} + C.

Passo 3

Para \displaystyle\int x\operatorname{sec}^{2}(2x)\,dx aplicamos partes com u=x, dv=\operatorname{sec}^{2}(2x)\,dx.
Então du=dx, v=\tfrac{\tan(2x)}{2}, e
\int x\operatorname{sec}^{2}(2x)\,dx = \frac{x\tan(2x)}{2} - \frac{1}{2}\int \tan(2x)\,dx.
Como \displaystyle\int \tan(2x)\,dx=-\tfrac{1}{2}\ln\lvert\cos(2x)\rvert, vem
\int x\operatorname{sec}^{2}(2x)\,dx = \frac{x\tan(2x)}{2} + \frac{1}{4}\ln\bigl\lvert\cos(2x)\bigr\rvert + C.

Passo 4

Para \displaystyle\int e^{2x}\sin(3x)\,dx usamos a fórmula padrão
\int e^{ax}\sin(bx)\,dx = \frac{e^{ax}}{a^{2}+b^{2}}\bigl(a\sin(bx)-b\cos(bx)\bigr)+C.
Aqui a=2, b=3, logo
\int e^{2x}\sin(3x)\,dx = \frac{e^{2x}}{13}\bigl(2\sin(3x)-3\cos(3x)\bigr) + C.

Passo 5

Para \displaystyle\int x\ln(1+x)\,dx usamos partes com u=\ln(1+x), dv=x\,dx.
Então du=\frac{dx}{1+x}, v=\frac{x^{2}}{2},
\int x\ln(1+x)\,dx = \frac{x^{2}}{2}\ln(1+x) - \frac{1}{2}\int \frac{x^{2}}{1+x}\,dx.
Fazendo divisão x^{2}/(1+x)=x-1+\tfrac{1}{1+x} e integrando,
\int \frac{x^{2}}{1+x}\,dx = \frac{x^{2}}{2}-x+\ln(1+x).
Logo
\int x\ln(1+x)\,dx = \frac{x^{2}}{2}\ln(1+x) - \frac{x^{2}}{4} + \frac{x}{2} - \frac{1}{2}\ln(1+x) + C.

Passo 6

Para \displaystyle\int \frac{x}{e^{2x}}\,dx = \int x e^{-2x}\,dx usamos partes com u=x, dv=e^{-2x}\,dx.
du=dx, v=-\tfrac{e^{-2x}}{2},
\int x e^{-2x}\,dx = -\frac{x e^{-2x}}{2} + \frac{1}{2}\int e^{-2x}\,dx = -\frac{x e^{-2x}}{2} - \frac{e^{-2x}}{4} + C = -\frac{2x+1}{4}\,e^{-2x} + C.

Passo 7

Para \displaystyle\int \frac{(\ln x)^{2}}{x^{3}}\,dx fazemos t=\ln x, dt=\frac{dx}{x}, e x^{-3}dx=x^{-2}\cdot\frac{dx}{x}=e^{-2t}dt.
Assim reduz-se a \displaystyle\int t^{2}e^{-2t}\,dt, que por partes duas vezes fornece
\int t^{2}e^{-2t}\,dt = -\frac{2t^{2}+2t+1}{4}\,e^{-2t} + C.
Voltando a x, e^{-2t}=x^{-2} e t=\ln x,
\int \frac{(\ln x)^{2}}{x^{3}}\,dx = -\,\frac{2(\ln x)^{2}+2\ln x+1}{4\,x^{2}} + C.

Passo 8

Para \displaystyle\int \cos\theta\,[\cos(\sin\theta)]^{3}\,d\theta fazemos u=\sin\theta, du=\cos\theta\,d\theta.
A integral vira \displaystyle\int \cos^{3}(u)\,du=\int(1-\sin^{2}u)\cos u\,du.
Com w=\sin u, dw=\cos u\,du, obtemos
\int \cos^{3}(u)\,du = \int (1-w^{2})\,dw = w-\frac{w^{3}}{3} = \sin u-\frac{\sin^{3}u}{3}.
Retornando u=\sin\theta,
\int \cos\theta\,[\cos(\sin\theta)]^{3}\,d\theta = \sin(\sin\theta) - \frac{\sin^{3}(\sin\theta)}{3} + C.

Passo 9

Para \displaystyle\int \sin(3x)\cos x\,dx usamos a identidade de produto para soma:
\sin(3x)\cos x = \tfrac{1}{2}\bigl[\sin(4x)+\sin(2x)\bigr].
Então
\int \sin(3x)\cos x\,dx = \frac{1}{2}\int\sin(4x)\,dx + \frac{1}{2}\int\sin(2x)\,dx = -\frac{\cos(4x)}{8} - \frac{\cos(2x)}{4} + C.

Resposta

\begin{aligned} \int x^{2}\cos(3x)\,dx &= \frac{x^{2}\sin(3x)}{3} + \frac{2x\cos(3x)}{9} - \frac{2\sin(3x)}{27} + C,\\\ \int x\operatorname{sec}^{2}(2x)\,dx &= \frac{x\tan(2x)}{2} + \frac{1}{4}\ln\bigl\lvert\cos(2x)\bigr\rvert + C,\\\ \int e^{2x}\sin(3x)\,dx &= \frac{e^{2x}}{13}\bigl(2\sin(3x)-3\cos(3x)\bigr) + C,\\\ \int x\ln(1+x)\,dx &= \frac{x^{2}}{2}\ln(1+x) - \frac{x^{2}}{4} + \frac{x}{2} - \frac{1}{2}\ln(1+x) + C,\\\ \int \frac{x}{e^{2x}}\,dx &= -\frac{2x+1}{4}\,e^{-2x} + C,\\\ \int \frac{(\ln x)^{2}}{x^{3}}\,dx &= -\,\frac{2(\ln x)^{2}+2\ln x+1}{4\,x^{2}} + C,\\\ \int \cos\theta\,[\cos(\sin\theta)]^{3}\,d\theta &= \sin(\sin\theta) - \frac{\sin^{3}(\sin\theta)}{3} + C,\\\ \int \sin(3x)\cos x\,dx &= -\frac{\cos(4x)}{8} - \frac{\cos(2x)}{4} + C. \end{aligned}

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