Question

integrate 0 to pie by 2(cosx-sinx)dx​

Answer 1

Answer:

∫

0

2

π

(cosx−sinx)e

x

dx

=∫e

x

(cosx−sinx)dx

It is in the form of

∫e

x

(f(x)+f

1

(x))dx=e

x

f(x)+c

Thus ∫e

x

(cosx−sinx)dx=e

x

cosx+c

∫

0

2

π

e

x

(cosx−sinx)dx=(e

x

cosx+c)

0

π/2

=(e

2

π

⋅cos

2

π

+c)−(e

0

cos0+c)

=−1

Thus ∫

0

2

π

e

x

(cosx−sinx)dx=−1