Question Number 101079 by I want to learn more last updated on 30/Jun/20

Commented byI want to learn more last updated on 30/Jun/20

Expecially question (2) please

Commented byDwaipayan Shikari last updated on 01/Jul/20

∫_(−π) ^π (dx/(2−(((1−t^2 )/(1+t^2 ))+((2t)/(1+t^2 ))))) .(((1/2)sec^2 (x/2))/((1/2)sec^2 (x/2)))=2∫_(−π) ^π (dt/((1+t^2 )(((2+2t^2 −1+t^2 −2t)/(1+t^2 )))))  Suppose tanx=t  and (dt/dx)=(1/2)sec^2 x   and   sec^2 (x/2)=1+t^2   =2∫_(−π) ^π (dt/(3t^2 −2t+1))=2∫_(−π) ^π (dt/((3t−(1/3))^2 +(((2(√2))/3))^2 ))=(2/3)∫_(−π) ^π (du/(u^2 +(((2(√2))/3))^2 ))  =(2/3) (3/(2(√2)))tan^(−1) (((3u)/(2(√2))))=(√2)tan^(−1) (((9t−1)/(2(√2))))=[(√2)tan^(−1) (((9tan(x/2)−1)/(2(√2))))]_(−π) ^π                      suppose (3t−(1/3))=u         =(√2)((π/2)+(π/2))=(√2)π★

Answered by abdomathmax last updated on 01/Jul/20

classic method A =∫_(−π) ^π  (dx/(2−(cosx +sinx)))  changement tan((x/2))=t give   A =∫_(−∞) ^(+∞)  ((2dt)/((1+t^2 ){2−(((1−t^2 )/(1+t^2 ))+((2t)/(1+t^2 )))}))  =∫_(−∞) ^(+∞ )  ((2dt)/(2+2t^2 −1+t^2 −2t)) =∫_(−∞) ^(+∞)  ((2dt)/(3t^2 −2t+1))  =(2/3) ∫_(−∞) ^(+∞)   (dt/(t^2 −(2/3)t +(1/3)))  =(2/3) ∫_(−∞) ^(+∞ )   (dt/(t^2 −(2/3)t +(1/9)+(1/3)−(1/9)))  =(2/3) ∫_(−∞) ^(+∞)  (dt/((t−(1/3))^2  +(2/9)))  =_(t−(1/3)=((√2)/3)u)   (2/3)×(9/2) ∫_(−∞) ^(+∞)  (1/(u^2  +1))×((√2)/3) du  =(√2)∫_(−∞) ^(+∞)  (du/(1+u^2 )) =(√2)[ arctanu]_(−∞) ^(+∞)   =(√2){(π/2)−(−(π/2))} =π(√2)  ★A =π(√2)★

Commented byI want to learn more last updated on 02/Jul/20

Thanks sir

Answered by mr W last updated on 30/Jun/20

∫_(−π) ^π (dx/(2−(cos x+sin x)))  =(1/(√2))∫_(−π) ^π (dx/((√2)−cos (x−(π/4))))  =(1/(√2))∫_(−((5π)/4)) ^((3π)/4) (dt/((√2)−cos t))  =(√2)[tan^(−1) {((√2)+1)tan (t/2)}]_(−((5π)/4)) ^(−π)        +(√2)[tan^(−1) {((√2)+1)tan (t/2)}]_(−π) ^((3π)/4)   =(√2)[(π/2)−tan^(−1) ((√2)+1)^2 +tan^(−1) ((√2)+1)^2 +(π/2)]  =(√2)π

Commented byI want to learn more last updated on 30/Jun/20

Thanks sir. I appreciate

Commented bymr W last updated on 30/Jun/20