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sin-2021-x-sin-2023-x-dx-

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Question Number 163580 by mkam last updated on 08/Jan/22
∫ sin^(2021) (x) . sin(2023 x ) dx
∫sin2021(x).sin(2023x)dx
Commented by mkam last updated on 08/Jan/22
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Answered by abdullahhhhh last updated on 08/Jan/22
I=∫sin(2022x+x) sin^(2021) x dx ∫sin^(2021) x [sin(2022x)cosx(x)+cosx(2022x) sinx] I=∫[cosx sin^(2021) x sin(2022x)] dx+∫sin^(2022) x cos(2022x) dx I=I_1 (By parts)+I_2 I_1 =∫cosx sin^(2021) x sin(2022x) dx u=sin(2022x) dv =cosx sin^(2021) x du=2022 cos(2022x) v=((sin^(2022) x)/(2022)) (1/(2022)) sin(2022x) sin^(2022) x−∫cos(2022) sin^(2022) x dx I_1 =((sin(2022x) sin^(2022) x)/(2022))−I_2 I=((sin(2022x) sin^(2022) x)/(2022))+C
I=∫sin(2022x+x)sin2021xdx∫sin2021x[sin(2022x)cosx(x)+cosx(2022x)sinx]I=∫[cosxsin2021xsin(2022x)]dx+∫sin2022xcos(2022x)dxI=I1(Byparts)+I2I1=∫cosxsin2021xsin(2022x)dxu=sin(2022x)dv=cosxsin2021xdu=2022cos(2022x)v=sin2022x202212022sin(2022x)sin2022x−∫cos(2022)sin2022xdxI1=sin(2022x)sin2022x2022−I2I=sin(2022x)sin2022x2022+C

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