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sin^3x + cos^4x

Mathematics
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what is the problem here?
how to integrate, sorry!
Well, if the power of sine is odd and positive , what you want to do is lop of one of the sine factor, put it to the right of the rest of the expression and convert the remaining (even) sine factors to cosines withe the Pythagorean identity , and then integrate with the substitution method where u=cos(x)

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oh, ok, then, thanks.
hold on let me finish it for you: \[\int sin^3(x)cos^4(x)dx =\int sin^2(x)cos^4(x)sin(x)dx\]
:D
\[\int sin^2(x)cos^4(x)sin(x)dx=\int (1-cos^2(x))cos^4(x)sin(x)dx\]
\[\int (1-cos^2(x))cos^4(x)sin(x)dx=\int (cos^4(x)-cos^6(x))sin(x)dx\]
\[u=cos(x)\rightarrow \frac{du}{dx}=-sin(x)\rightarrow du=-sin(x)dx\rightarrow \frac{-du}{sin(x)}=dx\]
\[\int (cos^4(x)-cos^6(x))sin(x)dx=-\int (u^4-u^6)du\]
\[=-\int (u^4-u^6)du=-\frac{1}{5}u^5+\frac{1}{7}u^7+C\]
\[=\frac{1}{7}cos^7(x)-\frac{1}{5}cos^5(x)+C\]
And thats your answer
so simple. thanks!
well, its simple once you see it

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