Question

Solve the definite integral from 0 to pi/2 of cosx*sin(sinx)dx

Answer 1

u=sinx
du=cosx

so then it's -cos(sin(pi/2))+ cos((sin0))

which is cos(0)-cos(1)

and now I'm lost

Answer 2

\[\Large \int \frac{du}{dx} \cdot \sin(u)dx \]

Answer 3

\[\Large \int \sin(u)du \]

Answer 4

\[ \Large =-\cos(u)=-\cos(\sin x) \]

Answer 5

So yes I agree so far.

Answer 6

\[ \Large \left. - \cos(\sin x)\right|_0^{\frac{\pi}{2}} \]

Answer 7

\[\Large - \cos(1)- (-1) = 1- \cos(1) \]

Answer 8

sin0=0 I thought?

So it should be

-cos(1)+cos(0) ?

Answer 9

\[ \sin(0)=0 \]
But you are taking
\[-\cos(\sin0) = -\cos(0) = -1 \]