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OpenStudy (anonymous):
Find the indefinite integral of (5cosx-2sec^2x)dx
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OpenStudy (amistre64):
who lost it?
OpenStudy (amistre64):
its a sum so we could just find the int of each term
OpenStudy (amistre64):
sec^2 might be tricky tho
OpenStudy (amistre64):
tan goes to sec^2 tho so not all that tricky
OpenStudy (mr.math):
Does "goes to" mean "is the integral of"?
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OpenStudy (amistre64):
derivatives are down to me; and ints are up; tan goes down to sec^2 in my mind
OpenStudy (mr.math):
Hmm interesting :-D
OpenStudy (amistre64):
might be the power thing; x^3 does down a degree ...
OpenStudy (amistre64):
5cos goes up to 5sin
-2sec^2 goes up to -2tan
then we tack on the +C for indefininity
OpenStudy (anonymous):
I can't get it... :/
I know the integral of cosine is sinx+c and the derivative of sec^2x is tanx+c
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OpenStudy (anonymous):
*integral
OpenStudy (amistre64):
then thats it; the +Cs just merge into some uber constant
OpenStudy (amistre64):
\[\int (5cos(x)-2sec^2(x))dx\]
\[\int (5cos(x))dx-\int (2sec^2(x))dx\]
\[5sin(x)+c_1-2tan(x)+c_2\]
\[5sin(x)+-2tan(x)+C\]
OpenStudy (anonymous):
THANKS!
OpenStudy (amistre64):
yw
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