Integrate e^x(sinx-cosx)

CBSE Class 12-science Answered

Integrate e^x(sinx-cosx)

Asked by Anuradha Pradeep | 02 Nov, 2012, 09:00: PM

Expert Answer

Answer :

Using the trigonometric identity
sin(2x) = 2sin(x)cos(x)
giving
sin(x)cos(x) = (1/2)sin(2x)

we have
intg(e^x)sin(x)cos(x) dx = (1/2) intg(e^x)sin(2x) dx

we can now integrate by parts.

Let
u = sin(2x)
then
du/dx = 2cos(2x)
and
dv/dx = e^x
so
v = e^x

Hence
(1/2) intg (e^x) sin(2x) dx
= (1/2)[ (e^x)sin(2x) - 2 intg (e^x)cos(2x) dx]
= (1/2)(e^x)sin(2x) - intg (e^x)cos(2x) dx ............... (a)

then integrate this integral by parts

Let u = cos(2x)
then
du/dx = -2sin(2x)
and
dv/dx = e^x = v

So
intg (e^x)cos(2x) dx
= (e^x)cos(2x) - 2 intg (e^x)(-sin(2x)) dx
= (e^x)cos(2x) + 2 intg (e^x)sin(2x) dx

Substitute this into the original equation (a)

(1/2)intg(e^x)sin(2x) dx = (1/2)(e^x)sin(2x) - (e^x)cos(2x) - 2intg(e^x)sin(2x) dx

Notice that on both sides of the equation we have a intg(e^x)sin(2x) dx term. Take both of these terms to the same side to get:

(1/2)intg(e^x)sin(2x) dx + 2intg(e^x)sin(2x) dx= (1/2)(e^x)sin(2x) - (e^x)cos(2x)

(5/2) intg (e^x)sin(2x) dx = (1/2)(e^x)sin(2x) - (e^x)cos(2x)

divide both sides by 5

(1/2) intg (e^x)sin(2x) dx = (1/5)(e^x)[(1/2)sin(2x) - cos(2x)]

therefore
intg(e^x)sin(x)cos(x) dx = (1/5)(e^x)[(1/2)sin(2x) - cos(2x)] +C Answer

Answered by | 03 Nov, 2012, 11:19: PM