Double Integral with a trig function

[eddysd]

just wondering if i can still do this, attempted the following:

ʃʃ cos(x+y)dxdy with upper limits of pi/2 and lower limits of 0 for both integrals

My answer came out as 0.

Can anyone confirm this?

 

[Mark44]

just wondering if i can still do this, attempted the following:

ʃʃ cos(x+y)dxdy with upper limits of pi/2 and lower limits of 0 for both integrals

My answer came out as 0.

Can anyone confirm this?

That's what I get.

 

[eddysd]

@Mark44 Thankyou, also what method did you use, I used the identity that
cos(x+y)=cos(x)cos(y)-sin(x)sin(y)

Is there another way of doing it?

 

[Mark44]

Yes. You can integrate cos(x + y) directly. When you integrate cos(x + y) with respect to x (treating y as a constant), you get sin(x + y). Evaluating this at pi/2 and 0 gives sin(pi/2 + y) - sin(y). The sin(pi/2 + y) term can be rewritten as cos(y) using an identity.

Finally, integrate cos(y) - sin(y) with respect to y, and evaluate the antiderivative at pi/2 and 0.