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प्रश्न
Evaluate :`int Sinx/(sqrt(cos^2 x-2 cos x-3)) dx`
बेरीज
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उत्तर
`"I" = int Sinx/(sqrt(cos^2 x-2 cos x-3)) dx`
Put cos x = t
∴ – sin x dx = dt
sin x dx = – dt
=`int (-dt)/sqrt( t^2-2 t-3)`
=`-int dt/sqrt( t^2-2 t + 1- 1-3)`
=`-int dt /sqrt((t-1)^2-(2)^2` ....(Completing the square)
=`-log|(t-1)+sqrt((t-1)^2-2^2)|+C` ....`{int(dx)/sqrt(x^2-a^2)=log|x+sqrt(x^2-a^2)|+"C"}`
=`-log|t-1+sqrt(t^2-2t-3)|+C`
`=– log|(cosx-1)+sqrt(cos^2x-2cosx-3)|+C`
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