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Integral of : cosx dx. = ——————— [cosx/2 +sinx/2]^5

Integral of :      cosx dx.    = 
                     ———————
                      [cosx/2 +sinx/2]^5

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Grade:11

1 Answers

Myra
12 Points
3 years ago
Let cosx= (cosx/2-sinx/2)(cosx/2+sinx/2)
After substituting 
∫(cosx/2-sinx/2)  dx
——————————
(cosx/2+sinx/2)^4
Let (cosx/2+sinx/2) = t
By differentiating
[cosx/2 - sinx/2] dx= 2dt
After substitution
2∫dt
———     =
    t^4
After integration 
-2/3( t^-3) + C     
-2/3 [ sinx/2 +cosx/2]^-3 +C
t^4

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