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प्रश्न
Evaluate `int_0^(pi/4) (sinx + cosx)/(16 + 9sin2x) dx`
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उत्तर
Let I = `int_0^(pi/4) (sinx + cosx)/(16 + 9sin2x) dx`
Put -cosx + sin x = t .....(1)
Then
(sin x + cos x) dx= dt
As x → 0, t → -1
Also `x = pi/4`, t → 0
Squaring (1) both sides we get
`cos^2x + sin^2x - 2cosxsinx = t^2`
`=> 1 - sin2x = t^2`
`=> sin 2x =1 - t^2`
Substituting these values, we get
`I = int_(-1)^0 dt/(16+9(1-t^2))`
`= int_(-1)^0 dt/(25 - 9t^2)`
`= 1/9 int_(-1)^0 dt/((5/3)^2 - t^2)`
`= 1/9[1/(2a) log |(a+t)/(a-t)|]_(-1)^0` where a = 5/3
`= 1/9 [3/(2(5)) "" log |(5/3+t)/(5/3-t)|]_(-1)^0`
`= 1/9 [3/10 {log 1 - log 1/4}]^(-1)`
`= 3/90 (-log 1/4) = 1/30 log 4`
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