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Evaluate int(cosx)/((sin^(2)x+4sinx+5))d...

Evaluate `int(cosx)/((sin^(2)x+4sinx+5))dx`.

Text Solution

Verified by Experts

Putting `sinx=t " and "cosx dx=dt` we get
`int(cosx)/((sin^(2)x+4sinx+5))dx=int(dt)/((t^(2)+4t+5))=int(dt)/({(t^(2)+4t+4)+1}}`
`=int(dt)/({(t+2)^(2)+1^(2)}}=int(du)/((u^(2)+1)), "where "u=(t+2)`
`=tan^(-1)u+C=tan^(-1)(t+2)+C`
`=tan^(-1)(sinx+2)+C`.
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