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paglia
 3 years ago
sin^3x + cos^4x
paglia
 3 years ago
sin^3x + cos^4x

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LagrangeSon678
 3 years ago
Best ResponseYou've already chosen the best response.0what is the problem here?

paglia
 3 years ago
Best ResponseYou've already chosen the best response.1how to integrate, sorry!

LagrangeSon678
 3 years ago
Best ResponseYou've already chosen the best response.0Well, 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)

LagrangeSon678
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1325279411322:dw

LagrangeSon678
 3 years ago
Best ResponseYou've already chosen the best response.0hold on let me finish it for you: \[\int sin^3(x)cos^4(x)dx =\int sin^2(x)cos^4(x)sin(x)dx\]

LagrangeSon678
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int sin^2(x)cos^4(x)sin(x)dx=\int (1cos^2(x))cos^4(x)sin(x)dx\]

LagrangeSon678
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int (1cos^2(x))cos^4(x)sin(x)dx=\int (cos^4(x)cos^6(x))sin(x)dx\]

LagrangeSon678
 3 years ago
Best ResponseYou've already chosen the best response.0\[u=cos(x)\rightarrow \frac{du}{dx}=sin(x)\rightarrow du=sin(x)dx\rightarrow \frac{du}{sin(x)}=dx\]

LagrangeSon678
 3 years ago
Best ResponseYou've already chosen the best response.0\[\int (cos^4(x)cos^6(x))sin(x)dx=\int (u^4u^6)du\]

LagrangeSon678
 3 years ago
Best ResponseYou've already chosen the best response.0\[=\int (u^4u^6)du=\frac{1}{5}u^5+\frac{1}{7}u^7+C\]

LagrangeSon678
 3 years ago
Best ResponseYou've already chosen the best response.0\[=\frac{1}{7}cos^7(x)\frac{1}{5}cos^5(x)+C\]

LagrangeSon678
 3 years ago
Best ResponseYou've already chosen the best response.0And thats your answer

LagrangeSon678
 3 years ago
Best ResponseYou've already chosen the best response.0well, its simple once you see it
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