anonymous
  • anonymous
sin^3x + cos^4x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
what is the problem here?
anonymous
  • anonymous
how to integrate, sorry!
anonymous
  • anonymous
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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anonymous
  • anonymous
|dw:1325279411322:dw|
anonymous
  • anonymous
oh, ok, then, thanks.
anonymous
  • anonymous
hold on let me finish it for you: \[\int sin^3(x)cos^4(x)dx =\int sin^2(x)cos^4(x)sin(x)dx\]
anonymous
  • anonymous
:D
anonymous
  • anonymous
\[\int sin^2(x)cos^4(x)sin(x)dx=\int (1-cos^2(x))cos^4(x)sin(x)dx\]
anonymous
  • anonymous
\[\int (1-cos^2(x))cos^4(x)sin(x)dx=\int (cos^4(x)-cos^6(x))sin(x)dx\]
anonymous
  • anonymous
\[u=cos(x)\rightarrow \frac{du}{dx}=-sin(x)\rightarrow du=-sin(x)dx\rightarrow \frac{-du}{sin(x)}=dx\]
anonymous
  • anonymous
\[\int (cos^4(x)-cos^6(x))sin(x)dx=-\int (u^4-u^6)du\]
anonymous
  • anonymous
\[=-\int (u^4-u^6)du=-\frac{1}{5}u^5+\frac{1}{7}u^7+C\]
anonymous
  • anonymous
\[=\frac{1}{7}cos^7(x)-\frac{1}{5}cos^5(x)+C\]
anonymous
  • anonymous
And thats your answer
anonymous
  • anonymous
so simple. thanks!
anonymous
  • anonymous
well, its simple once you see it

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