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

Question

ramya · Answer

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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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

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