Question Number 116998 by TANMAY PANACEA last updated on 08/Oct/20

Commented by Dwaipayan Shikari last updated on 08/Oct/20

Commented by TANMAY PANACEA last updated on 08/Oct/20

Answered by TANMAY PANACEA last updated on 08/Oct/20

Answered by mathmax by abdo last updated on 08/Oct/20
![let∫_(−(π/2)) ^(π/2) ((sin^2 x)/(1+2^x ))dx =I changement x=−t give I =∫_(−(π/2)) ^(π/2) ((sin^2 t)/(1+2^(−t) ))dt ⇒2I =∫_(−(π/2)) ^(π/2) ((sin^2 x)/(1+2^x )) dx +∫_(−(π/2)) ^(π/2) ((sin^2 x)/(1+2^(−x) ))dx =∫_(−(π/2)) ^(π/2) ((1/(1+2^x ))+(1/(1+2^(−x) )))sin^2 x dx =∫_(−(π/2)) ^(π/2) (((2+2^x +2^(−x) )/(1+2^(−x) +2^x +1)))sin^2 x dx =∫_(−(π/2)) ^(π/2) ((1−cos(2x))/2)dx =(π/2)−(1/4)[sin(2x)]_(−(π/2)) ^(π/2) =(π/2) ⇒2I =(π/2) ⇒★I =(π/4) ★](https://www.tinkutara.com/question/Q117015.png)
Commented by TANMAY PANACEA last updated on 08/Oct/20

Commented by Bird last updated on 09/Oct/20
