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Integration of [(cosx)^3+(cosx)^5]/[(sinx)^2+(sinx)^4] with respect to x

Integration of [(cosx)^3+(cosx)^5]/[(sinx)^2+(sinx)^4] with respect to x

Grade:12

1 Answers

ramya
20 Points
6 years ago
int = integration
 
= int {[cos^3(x)][(cos^2(x)] + 1}/{[sin^2(x)][sin^2(x)+1]} dx
 
use cos^2(x) as 1 - sin^2(x)
= int cos(x) * {[sin^2(x)][sin^2(x)-1}/{[sin^2(x)][sin^2(x)+1} dx
                    substitute u = sinx
                            dx = 1/[cos(x)] du
= int {[u^2 - 2][u^2 - 1]}/{u^2[u^2 + 1]} du
= int {[2 - 4u^2]/[u^2(u^2 + 1)} + 1 du
= int 1 du - 2 int {[2u^2 - 1]/u^2(u^2 + 1)} du
Know, int {[2u^2 - 1]/u^2(u^2 + 1)} du
      = int {[3/(u^2 + 1)] - [1/u^2]} du
      = 3 int [1/(u^2 + 1)] du - int [1/u^2] du
Know, int [1/(u^2 + 1)] du
      = tan^-1(u)
                   and int [1/u^2] du
                        = -1/u
                             int 1 du = u
 Therefore, 3 int [1/(u^2 + 1)] du - int [1/u^2] du
            = 3 tan^-1(u) + 1/u
Hence  int 1 du - 2 int {[2u^2 - 1]/u^2(u^2 + 1)} du 
      = u - 6 tan^-1(u) - 2/u
    => sin(x) - 6 tan^-1(sin(x)) - 2/sin(x) + C
    => sin(x) - 6 tan^-1(sin(x)) - 2 cos(x) + C

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