这题怎么做?求不定积分
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∫ dx/(3sinx+4cosx +5)
=(1/5)∫ dx/[(3/5)sinx+(4/5)cosx +1 ]
=(1/5)∫ dx/[cos(x-φ) +1]
where cosφ = 4/5 , sinφ = 3/5
=(1/5)∫ dx/ { 2(cos[(x-φ)/2] )^2}
=(1/10)∫ { sec[(x-φ)/2] }^2 dx
=(1/5) tan[(x-φ)/2] + C
=(1/5)∫ dx/[(3/5)sinx+(4/5)cosx +1 ]
=(1/5)∫ dx/[cos(x-φ) +1]
where cosφ = 4/5 , sinφ = 3/5
=(1/5)∫ dx/ { 2(cos[(x-φ)/2] )^2}
=(1/10)∫ { sec[(x-φ)/2] }^2 dx
=(1/5) tan[(x-φ)/2] + C
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