OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

So yes I agree so far.

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

sin0=0 I thought? So it should be -cos(1)+cos(0) ?

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

|dw:1344817963593:dw|

5 years ago
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