integration of sin square tita into cos cube tita into d tita
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∫sin²θ cosθ dθ
put sinθ=t
diffrentiate
cosθ dθ =dt
∫t² dt= t³/3 + C
=(sin³θ)/3 +C
hope it helps mark as brainliest if you think
put sinθ=t
diffrentiate
cosθ dθ =dt
∫t² dt= t³/3 + C
=(sin³θ)/3 +C
hope it helps mark as brainliest if you think
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