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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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3 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)} duKnow, 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] duKnow, 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/uHence  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
```
3 years ago
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