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

4 Answers
Mar 29, 2018

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

Explanation:

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!

Mar 29, 2018

#1/3 sin^3x + C#

Explanation:

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#

Mar 29, 2018

Is an inmediate integral. See below

Explanation:

#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

Mar 29, 2018

#1/3sin^3 x + c #

Explanation:

#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 #