sin^3x + cos^4x

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sin^3x + cos^4x

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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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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|dw:1325279411322:dw|
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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