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

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Answer 1 · 2018-03-29T16:56:42

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!

Answer 2 · 2018-03-29T16:58:16

#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#

Answer 3 · 2018-03-29T16:59:04

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

Answer 4 · 2018-03-29T17:33:49

#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 #