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प्रश्न
Evaluate :
`∫_(-pi)^pi (cos ax−sin bx)^2 dx`
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उत्तर
`∫_(-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)`
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