Question

resolva
 ∫$sen^{5} x cos^{2} dx$
sabendo que abre-se a potência impar e substitui pela identidade cos²x+sen²=1$sen^{5} x cos^{2} dx$

Answer 1

∫sen⁵xcos²xdx = ∫sen⁴xcos²xsenxdex = ∫(1 - cos²x)²cos²x senx dx =
Seja cosx = u => -senx dx = du => senxdx = -du
∫(1- u)² u²(-du) = -∫u²(1 - 2u + u²) du = -∫(u⁴-2u³ + u²)du = 
-(1/5u⁵ - 2/4u⁴ + 1/3u³) = -(1/5cos⁵x -1/2 cos⁴x + 1/3cos³x) + C

Answer 2

          $\displaystyle I=\int \sin^{5} x \cos^{2}x \;dx\\ \\ \\ I=\int \sin^{4} x \cos^{2}x \cdot\sin x\;dx\\ \\ \\ I=\int (\sin^{2} x)^2 \cos^{2}x \cdot\;d(-\cos x)\\ \\ \\ I=-\int (1-\cos^{2} x)^2 \cos^{2}x \cdot\;d(\cos x)\\ \\ \\ I=-\int (1-2\cos^{2} x+\cos^4x) \cos^{2}x \cdot\;d(\cos x)\\ \\ \\ I=-\int \cos^2x-2\cos^{4} x+\cos^6x\cdot\;d(\cos x)\\ \\ \\ I=-\int \cos^2x\;d(\cos x)+2\int\cos^{4} x\;d(\cos x)-\int\cos^6x\;d(\cos x)$
          
           $\boxed{\boxed{I=-\dfrac{1}{3}\cos^3x+\dfrac{2}{5}\cos^5x-\dfrac{1}{7}\cos^7 x+C}}$