How can I integrate #sin^2 xcosx#??

Question

How can I integrate  #sin^2 xcosx#??

William Abdalla · Accepted Answer

#1/3sin^3 x + c # #int sin^2 x cosx dx # Make an appropriate u sub: # u = sinx # #du = cosx dx # #=> int u^2 du # #=> 1/3u^3 +c # #=> 1/3sin^3 x + c #

Charles Arocha · Answer

The integral is equal to #1/3sin^3x + C#

Let #u = sinx#. Then #du = cosx dx# and #dx = (du)/cosx#.

#I = int u^2 cosx * (du)/cosx#

#I = int u^2 du#

#I = 1/3u^3 + C#

#I = 1/3sin^3x + C#

Hopefully this helps!

Charles Anctil · Answer

#1/3 sin^3x + C# Given: #int sin^2x cos x dx# Use #u#-substitution. Let #u = sin x; " "du = cos x dx; " "dx = (du)/(cos x)# #int sin^2x cos x dx = int u^2 cancel(cos x) (du)/(cancel(cos x)) = int u^2 du = 1/3 u^3 +C# #int sin^2x cos x dx = 1/3 sin^3x + C#

Christopher Agresta · Answer

Is an inmediate integral. See below

#intsin^2xcosxdx=1/3sin^3x+C# because the derivative of #sin^3x# is #3sin^2xcosx#. To remove #3# we need to insert #1/3# before

Emilia Aguilar · Answer

undefined To integrate $ \sin^2(x)\cos(x) $, you can use the substitution method. Here's how:

1. Let $ u = \sin(x) $, then $ du = \cos(x) dx $.
2. Rewrite the integral in terms of $ u $: $ \int u^2 du $.
3. Integrate $ u^2 $ with respect to $ u $.
4. Replace $ u $ with $ \sin(x) $ in the final result.

The integral of $ u^2 $ with respect to $ u $ is $ \frac{1}{3}u^3 + C $, where $ C $ is the constant of integration.

So, the integral of $ \sin^2(x)\cos(x) $ is $ \frac{1}{3}\sin^3(x) + C $, where $ C $ is the constant of integration.