Question

Calcule as integrais indefinidas usando as substituiçoes dadas:A) f x³(x4-10)² dxB) f x sen (2x²) dxc) f ( x/√3-x²) dx

Answer 1

a)

$\int x^3(x^4-10)^2\ dx\\\\ u=x^4-10\\ \frac{du}{dx}=4x^3\ \to\ \frac{du}{4}=x^3.dx\\\\ \int (x^4-10)^2.x^3\ dx=\int u^2.\frac{1}{4}\ du=\\\\ =\frac{1}{4}\int u^2\ du=\frac{1}{4}.(\frac{u^3}{3})+c=\frac{u^3}{12}+c\\\\ \int x^3(x^4-10)^2\ dx=\frac{(x^4-10)^3}{12}+c$

b)

$\int x.sin(2x^2)\ dx\\\\ u=2x^2\\ \frac{du}{dx}=4x\ \to\ \frac{du}{4}=x.dx\\\\ \int sin(2x^2).x\ dx=\int sin\ u.\frac{1}{4}\ du=\\\\ =\frac{1}{4}\int sin\ u\ du=\frac{1}{4}.(-cos\ u)+c=\frac{-cos\ u}{4}+c\\\\ \int x.sin(2x^2)\ dx=\frac{-cos(2x^2)}{4}+c$

c)

$\int \frac{x}{\sqrt{3-x^2}}\ dx\\\\ u=3-x^2\\ \frac{du}{dx}=-2x\ \to\ \frac{du}{-2}=x.dx\\\\ \int\ \frac{1}{\sqrt{3-x^2}}.x\ dx=\int \frac{1}{\sqrt{u}}.\frac{-1}{2}\ du=\\\\ =\frac{-1}{2}\int u^{\frac{-1}{2}}\ du= \frac{-1}{2}.2u^{\frac{1}{2}}+c=-\sqrt{u}+c\\\\ \int \frac{x}{\sqrt{3-x^2}}\ dx=-\sqrt{3-x^2}+c$