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Question Number 212341 by Spillover last updated on 10/Oct/24

Answered by Ghisom last updated on 10/Oct/24

∫_0 ^(π/4)  ((sin x +cos x)/(9+16sin 2x))dx=       [t=((4(√2))/5)cos (x+(π/4))]  =−(1/(20))∫_0 ^(4/5) (dt/(t^2 −1))=−(1/(40))[ln ∣((t−1)/(t+1))∣]_ =((ln 3)/(20))

π/40sinx+cosx9+16sin2xdx=[t=425cos(x+π4)]=1204/50dtt21=140[lnt1t+1]=ln320

Commented by Spillover last updated on 12/Oct/24

great

great

Answered by MathematicalUser2357 last updated on 13/Oct/24

5.49306144335×10^(−2)

5.49306144335×102

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