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Question Number 133107 by mnjuly1970 last updated on 18/Feb/21
...nicecalculus...find::ϕ=???∫01(sin(x)+sin(1x))dxx
Answered by Dwaipayan Shikari last updated on 18/Feb/21
Φ=∫01(sin(x)+sin(1x))dxx1x=u⇒−1x2=dudx=∫1∞sin(1u)u+sinuudu=∫0∞sin(1u)u+sinuudu−Φ2Φ=∫0∞sinuu+1u.sin(1u)duΦ=π4+12∫0∞sin(t)tdu1u=t⇒−1u2=dtdu=π4+π4=π2
Commented by mnjuly1970 last updated on 19/Feb/21
gratefulmrpayan...
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