Calcular as seguintes integrais
Calcule as seguintes integrais usando a regra da substituição
\begin{gathered} \int x \cos \left(x^{2}\right) d x, \quad \int x^{2} \mathrm{e}^{x^{3}} d x, \quad \int(4 t-2)^{3} d t, \quad \int_{4}^{9} \frac{d x}{2-3 x}, \quad \int u \sqrt{2+u^{2}} d u, \quad \int e^{x} \operatorname{sen}\left(e^{x}\right) d x \int \frac{1+x}{1-x^{2}} d x, \quad \int x\left(x^{2}-3\right)^{2} d x, \quad \int e^{2+\cos (x)} \operatorname{sen}(x) d x, \quad \int_{0}^{\pi / 2} \cos (x) \operatorname{sen}(x) d x \end{gathered}

Passo 1

O enunciado pede calcular dez integrais diferentes usando a regra da substituição. Todas envolvem reconhecer uma parte da derivada interna da função composta e trocar de variável para simplificar a antiderivada.

Passo 2

Para \displaystyle \int x\cos\bigl(x^{2}\bigr)\,dx faça u=x^{2}, daí du=2x\,dx e x\,dx=\tfrac12\,du.
\int x\cos\bigl(x^{2}\bigr)\,dx =\int \cos(u)\,\frac{du}{2} =\frac12\sin(u)+C =\frac12\sin\bigl(x^{2}\bigr)+C.

Passo 3

Para \displaystyle \int x^{2}e^{x^{3}}\,dx ponha u=x^{3}, logo du=3x^{2}\,dx e x^{2}\,dx=\tfrac{1}{3}du.
\int x^{2}e^{x^{3}}\,dx =\int e^{u}\,\frac{du}{3} =\frac13e^{u}+C =\frac13e^{x^{3}}+C.

Passo 4

Para \displaystyle \int(4t-2)^{3}\,dt use u=4t-2, então du=4\,dt e dt=\tfrac14\,du.
\int(4t-2)^{3}\,dt =\int u^{3}\,\frac{du}{4} =\frac{1}{4}\cdot\frac{u^{4}}{4}+C =\frac{(4t-2)^{4}}{16}+C.

Passo 5

Para \displaystyle \int_{4}^{9}\frac{dx}{2-3x} note que \displaystyle \int\frac{dx}{2-3x}=-\tfrac13\ln\lvert2-3x\rvert+C. Avaliando de 4 a 9:
\Bigl[-\tfrac13\ln\lvert2-3x\rvert\Bigr]_{4}^{9} =-\tfrac13\bigl(\ln|2-27|-\ln|2-12|\bigr) =-\tfrac13(\ln25-\ln10) =\tfrac13\ln\frac{2}{5}.

Passo 6

Para \displaystyle \int u\sqrt{2+u^{2}}\,du tome v=2+u^{2}, assim dv=2u\,du e u\,du=\tfrac12\,dv.
\int u\sqrt{2+u^{2}}\,du =\int \sqrt{v}\,\frac{dv}{2} =\frac12\cdot\frac{v^{3/2}}{3/2}+C =\frac{(2+u^{2})^{3/2}}{3}+C.

Passo 7

Para \displaystyle \int e^{x}\sin\bigl(e^{x}\bigr)\,dx use u=e^{x}, daí du=e^{x}\,dx.
\int e^{x}\sin\bigl(e^{x}\bigr)\,dx =\int \sin(u)\,du =-\cos(u)+C =-\cos\bigl(e^{x}\bigr)+C.

Passo 8

Para \displaystyle \int\frac{1+x}{1-x^{2}}\,dx note que 1-x^{2}=(1-x)(1+x) e faz-se \frac{1+x}{1-x^{2}}=\frac{1}{1-x}. Logo
\int\frac{1+x}{1-x^{2}}\,dx =\int\frac{dx}{1-x} =-\ln\lvert1-x\rvert+C.

Passo 9

Para \displaystyle \int x\bigl(x^{2}-3\bigr)^{2}\,dx ponha u=x^{2}-3, então du=2x\,dx e x\,dx=\tfrac12\,du.
\int x\bigl(x^{2}-3\bigr)^{2}\,dx =\int u^{2}\,\frac{du}{2} =\frac{u^{3}}{6}+C =\frac{(x^{2}-3)^{3}}{6}+C.

Passo 10

Para \displaystyle \int e^{2+\cos x}\sin x\,dx faça u=2+\cos x, logo du=-\sin x\,dx e \sin x\,dx=-du.
\int e^{2+\cos x}\sin x\,dx =-\int e^{u}\,du =-e^{u}+C =-e^{2+\cos x}+C.

Passo 11

Para \displaystyle \int_{0}^{\pi/2}\cos x\,\sin x\,dx use u=\sin x, então du=\cos x\,dx. Os limites viram u(0)=0 e u(\pi/2)=1:
\int_{0}^{\pi/2}\cos x\,\sin x\,dx =\int_{0}^{1}u\,du =\Bigl[\tfrac{u^{2}}{2}\Bigr]_{0}^{1} =\tfrac12.

Resposta

\displaystyle \int x\cos(x^{2})\,dx=\tfrac12\sin(x^{2})+C
\displaystyle \int x^{2}e^{x^{3}}\,dx=\tfrac13e^{x^{3}}+C
\displaystyle \int(4t-2)^{3}\,dt=\tfrac{(4t-2)^{4}}{16}+C
\displaystyle \int_{4}^{9}\frac{dx}{2-3x}=\tfrac13\ln\!\tfrac{2}{5}
\displaystyle \int u\sqrt{2+u^{2}}\,du=\tfrac{(2+u^{2})^{3/2}}{3}+C
\displaystyle \int e^{x}\sin(e^{x})\,dx=-\cos(e^{x})+C
\displaystyle \int\frac{1+x}{1-x^{2}}\,dx=-\ln\lvert1-x\rvert+C
\displaystyle \int x(x^{2}-3)^{2}\,dx=\tfrac{(x^{2}-3)^{3}}{6}+C
\displaystyle \int e^{2+\cos x}\sin x\,dx=-e^{2+\cos x}+C
\displaystyle \int_{0}^{\pi/2}\cos x\,\sin x\,dx=\tfrac12

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