What is the integral of #int sin^3x/cos^2x dx#?

Question

Aurora Alvarado · Accepted Answer

I found: #1/cos(x)+cos(x)+c# Let us try writing it as: #int(sin(x)*sin^2(x))/(cos^2(x))dx=# #=int(sin(x)(1-cos^2(x)))/(cos^2(x))dx=# #=int(sin(x))/(cos^2(x))dx-intsin(x)(cancel(cos^2(x)))/cancel((cos^2(x)))dx=# let us use a little manipulation into the first integral: #=-int(d[cos(x)])/cos^2(x)-intsin(x)dx=# #=-int(cos(x))^-2d[cos(x)]-intsin(x)dx=# #=(cos(x))^-1+cos(x)+c=1/cos(x)+cos(x)+c#

Luke Alvarez · Answer

undefined To solve the integral ∫(sin^3(x) / cos^2(x)) dx, we can use the substitution method. Let u = sin(x) and du = cos(x) dx. Then the integral becomes ∫(u^3 / cos(x) du). Using trigonometric identity cos^2(x) = 1 - sin^2(x), we can replace cos(x) with √(1 - u^2). The integral then becomes ∫(u^3 / √(1 - u^2)) du. This integral can be solved using a trigonometric substitution. Let u = sin(θ), then du = cos(θ) dθ and √(1 - u^2) = cos(θ). Substituting these into the integral gives us ∫(sin^3(θ)) dθ. Using trigonometric identity sin^3(θ) = (3sin(θ) - sin(3θ))/4, the integral becomes ∫((3sin(θ) - sin(3θ))/4) dθ. Integrating term by term yields (-3cos(θ) + cos(3θ)/12 + C), where C is the constant of integration. Finally, substitute back u = sin(x) and θ = arcsin(u) to get the integral in terms of x: (-3cos(arcsin(u)) + cos(3arcsin(u))/12 + C). Simplifying further, we get (-3√(1 - u^2) + u(9 - 12u^2)/12 + C), which is the integral of the given function.