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Question Number 144282 by SOMEDAVONG last updated on 24/Jun/21
I=∫_(π/6) ^(π/3) ((sin^(2021) x)/(sin^(2021) x+cos^(2021) x))dx=?
I=π6π3sin2021xsin2021x+cos2021xdx=?
Answered by som(math1967) last updated on 24/Jun/21
I=∫_(π/6) ^(π/3) ((sin^(2021) ((π/3)+(π/6)−x))/(sin^(2021) ((π/3)+(π/6)−x)+cos((π/3)+(π/6)−x)))dx  I=∫_(π/6) ^(π/3) ((cos^(2021) x)/(sin^(2021) x+cos^(2021) x))dx  ∴2I=∫_(π/6) ^(π/3) ((cos^(2021) x+sin^(2021) x)/(sin^(2021) x+cos^(2021) x))dx  2I=∫_(π/6) ^(π/3) dx  I=[(x/2)]_(π/6) ^(π/3) =(π/(12)) ans
I=π6π3sin2021(π3+π6x)sin2021(π3+π6x)+cos(π3+π6x)dxI=π6π3cos2021xsin2021x+cos2021xdx2I=π6π3cos2021x+sin2021xsin2021x+cos2021xdx2I=π6π3dxI=[x2]π6π3=π12ans
Commented by SOMEDAVONG last updated on 24/Jun/21
Thanks sir!
Thankssir!

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