Question

Evaluate :

`∫_(-pi)^pi (cos ax−sin bx)^2 dx`

Answer 1

 

`∫_(-pi)^pi (cos ax−sin bx)^2 dx`

`=∫_(-pi)^pi(cos^2ax+sin^2bx-2cosaxsinbx)dx`

`=∫_(-pi)^picos^2axdx+∫_(-pi)^pisin^2bxdx-∫_(-pi)^pi2cosaxsinbxdx`

`=2∫_(0)^picos^2axdx+2∫_(0)^pisin^2bxdx-0` [ Since cos²ax and sin²bx are even functions and cosaxsinbx is an odd function.]

`=2∫_(0)^pi(1+cos2ax)/2dx+2∫_(0)^pi(1-cos2bx)/2dx`

`=∫_(0)^pi (1+cos2ax) dx+∫_(0)^pi (1−cos2bx) dx`

`=∫_(0)^pi(1+cos2ax+1−cos2bx)dx`

`=∫_(0)^pi(2+cos2ax−cos2bx)dx`

`=2[x]_0^pi +[(sin2ax)/(2a)]_0^pi−[(sin2bx)/(2b)]_0^pi`

`=2π+(sin2aπ)/(2a)−(sin2bπ)/(2b)`