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Question Number 52197 by maxmathsup by imad last updated on 04/Jan/19
calculate ∫_(π/4) ^(π/3)  ((cosx −sinx)/(2 +sin(2x)))dx
calculateπ4π3cosxsinx2+sin(2x)dx
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Jan/19
2+sin2x  1+sin^2 x+cos^2 x+2sinxcosx  1+(sinx+cosx)^2   ∫_(π/4) ^(π/3)  ((d(sinx+cosx))/(1+(sinx+cosx)^2 ))  ∣tan^(−1) (sinx+cosx)∣_(π/4) ^(π/3)   tan^(−1) (((√3)/2)+(1/2))−tan^(−1) ((1/( (√2)))+(1/( (√2))))  tan^(−1) (((1+(√3))/2))−tan^(−1) ((√2) )  tan^(−1) (((((1+(√3))/2)−(√2))/(1+((1+(√3))/( (√2))))))  =tan^(−1) ((((1+(√3) −2(√2))/2)/(((√2) +1+(√3))/( (√2)))))  =tan^(−1) (((1+(√3) −2(√2))/( (√2) ((√2) +1+(√3))))  tan^(−1) (((1+(√3) −2(√2))/(2+(√2) +(√6))))
2+sin2x1+sin2x+cos2x+2sinxcosx1+(sinx+cosx)2π4π3d(sinx+cosx)1+(sinx+cosx)2tan1(sinx+cosx)π4π3tan1(32+12)tan1(12+12)tan1(1+32)tan1(2)tan1(1+3221+1+32)=tan1(1+32222+1+32)=tan1(1+3222(2+1+3)tan1(1+3222+2+6)
Commented by Abdo msup. last updated on 05/Jan/19
thank you sir Tanmay.
thankyousirTanmay.

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