INT[min(sinx, cosx)dx] from 0 to pi equals to ??

Apr 2009
28
0
Dear Members,

Any clue how to solve the following integrals?

Thanks in advance!!

1. INT[min(sinx, cosx)dx] from 0 to pi equals to
(A) 1 ? 2squrt(2)
(B) 1
(C) 0
(D) 1 ? squrt(2).

What’s the meaning of min(sinx, cosx), because no other criteria is given?


2. INT[sinxcosx/(x+1)^2 dx] from 0 to 4 == ?

Tried to solve the above integral by 'Integration by Parts'. Since (x+1)^2 in the denominator it's power is increasing.
 

mathman

Forum Staff
May 2007
6,888
759
min(sinx,cosx) simply means take the smaller one for a given x. In the range 0 to pi, sin is smaller 0<x<pi/4, while cos is smaller pi/4 < x < pi. You just integrate each part separately and add the two integrals.

As for the second integral, you can integrate by parts to go the other way, but it looks somewhat formidable.
 
Apr 2009
28
0
mathman said:
min(sinx,cosx) simply means take the smaller one for a given x. In the range 0 to pi, sin is smaller 0<x<pi/4, while cos is smaller pi/4 < x < pi. You just integrate each part separately and add the two integrals.

As for the second integral, you can integrate by parts to go the other way, but it looks somewhat formidable.


knowledgegain said:
Thank you very much for your time and help, mathman.
I am able to solve the first integral.

But for the 2nd integral, i think it could be solved by proper trigonometric substitution, because upper limit is given as 4 and the terms sinx and cosx are there in the integral.