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∫ xe x cosx dx= f(x)+c, then find f(x).

∫ xex cosx dx= f(x)+c, then find f(x).


1 Answers

Rathod Shankar AskiitiansExpert-IITB
69 Points
10 years ago


integral by parts take  x and ex cosx as two parts

\int f g\, dx = f \int g\, dx - \int \left ( f' \int g\,dx \right )\, dx.\!

take f=x   and g=ex cosx
here iam giving method for integration for ex cosx ...i think u can solve now right. do practice so that u can get more concept

\int e^{x} \cos (x) \, dx\!
u = \cos(x) \Rightarrow du = -\sin(x)\,dx
dv = e^x \, dx \Rightarrow v = \int e^x \, dx = e^x
\int e^{x} \cos (x) \, dx = e^{x} \cos (x) + \int e^{x} \sin (x) \, dx.\!
u = \sin(x) \Rightarrow du = \cos(x)\, dx
dv = e^x \, dx \Rightarrow v = \int e^x \, dx = e^x
\int e^x \sin (x) \, dx = e^x \sin (x) - \int e^x \cos (x) \,dx

Putting these together,

\int e^x \cos (x) \,dx = e^x \cos (x) + e^x \sin (x) - \int e^x \cos (x) \, dx.
2 \int e^{x} \cos (x) \, dx = e^{x} ( \sin (x) + \cos (x) ) + C\!
\int e^x \cos (x) \,dx = {e^x ( \sin (x) + \cos (x) ) \over 2} + C'\!

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