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,

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

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

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

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

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

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