How do you integrate #int cos^3xsin^4xdx#?

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Answer 1 · 2017-03-05T12:47:14

The answer is #==(sin^5x)/5-(sin^7x)/7+C#

We need

#cos^2x=1-sin^2x#

#intx^ndx=x^(n+1)/(n+1)+C(n!=-1)#

Therefore,

#cos^3xsin^4x=cosxsin^4x(1-sin^2x)#

We perform a substitution

Let #u=sinx#, #=>#, #du=dxcosx#

Therefore,

#intcos^3xsin^4xdx=intcosxsin^4x(1-sin^2x)dx#

#=intu^4(1-u^2)du#

#=int(u^4-u^6)du#

#=u^5/5-u^7/7+C#

#=(sin^5x)/5-(sin^7x)/7+C#