Question

como calcular a integral seno ao quadradr de x cosseno ao cubo dx​

Answer 1

$\displaystyle\int {sin^2x\,.\,cos^3x}~dx~=\\\\\\=~\displaystyle\int {sin^2x\,.\,cos^2x\,.\,cos\,x}~dx\\\\\\Utilizando~a~identidade~trigonometrica~~cos^2x~=~1-sin^2x\\\\\\=~\displaystyle\int {sin^2x\,.\,\left(1-sin^2x\right)\,.\,cos\,x}~dx$

$=~\displaystyle\int {\left(sin^2x-sin^4x\right)\,.\,cos\,x}~dx\\\\\\Utilizando~a~substituicao:\\\\\left \{ {u~=~sin\,x~~} \atop {du~=~cos\,x\,dx}} \right.\\\\\\=~\displaystyle\int \left(u^2-u^4\right)\,.\,du\\\\\\=~\frac{u^3}{3}~-~\frac{u^5}{5}~+~K\\\\\\Voltando~a~Substituicao\\\\\\=~\boxed{\frac{sin^3x}{3}~-~\frac{sin^5x}{5}~+~K}$