Question

Ajuda com essa integral indefinida ∫x^1/2(sen^2(x^(3/2)-1))dx

Answer 1

Olá

$\displaystyle \int (x^ \frac{1}{2}(sen^2(x^ \frac{3}{2}-1 )) )dx \\ \\ \\ u=x^ \frac{3}{2}-1 \\ \\ du= \frac{3}{2}x^ \frac{1}{2} dx \\ \\ \frac{2}{3}du= x^ \frac{1}{2}dx \\ \\ \\ \\ Substituindo~ na~ integral \\ \\ \\ \displaystyle \int sen^2(u)\cdot \frac{2}{3}du \\ \\ \\ \\ \frac{2}{3} \displaystyle \int sen^2(u)\cdot du \\ \\ \\ Pelas~identidades~trigonometrica~temos~que: \\ \\ sen^2(x) ~=~ \frac{1-cos(2x)}{2} \\ \\ Substituindo$

$\frac{2}{3} \displaystyle \int \frac{1-cos(2u)}{2} du \\ \\ \\ \frac{\diagup\!\!\!\!2}{3} \cdot \frac{1}{\diagup\!\!\!\!2} \displaystyle \int 1-cos(2u) du \\ \\ \\ \frac{1}{3} \displaystyle \int 1-cos(2u) du \\ \\ \\\ \frac{1}{3} \displaystyle \int 1du~~-~~ \frac{1}{3} \displaystyle \int cos(2u)du \\ \\ \\\text{ Antes de resolvermos, temos que fazer a substituicao no cos(2u)} \\ \\ w=2u \\ dw=2du \\ \\ \frac{1}{2} dw=du \\ \\ \\ \text{Substituindo na integral}$

$\frac{1}{3} \displaystyle \int 1du~~-~~ \frac{1}{3} \displaystyle \int cos(w)\cdot \frac{1}{2} dw \\ \\ \\ \frac{1}{3} \displaystyle \int 1du~~-~~ \frac{1}{3} \cdot \frac{1}{2}\displaystyle \int cos(w)\cdot dw \\ \\ \\ \frac{1}{3} \displaystyle \int 1du~~-~~ \frac{1}{6} }\displaystyle \int cos(w)\cdot dw \\ \\ \\ ( \frac{1}{3}u ) ~-~( \frac{1}{6} sen(w)) \\ \\ \\ \text{colocando de volta os valores de "u" e "w"} \\ \\ \frac{ 1 }{3}(x^ \frac{3}{2} -1) ~-~ \frac{1}{6}(2u)$

$\frac{ 1 }{3}(x^ \frac{3}{2} -1) ~-~ \frac{1}{6}(2(x^ \frac{3}{2} -1)) \\ \\ \\ \boxe{\boxed{\frac{ 1 }{3}(x^ \frac{3}{2} -1) ~-~ \frac{1}{6}(2x^ \frac{3}{2} -2) + C}}$







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