What's the integral of #int tanx / (secx + cosx)dx#?

Question

Christopher Agresta · Accepted Answer

I found: #-arctan[cos(x)]+c# We can try changing everything into sin and cos to get: #int(sin(x)/cos(x))/(1/cos(x)+cos(x))dx=# #=int(sin(x)/cancel(cos(x)))*(cancel(cos(x))/(1+cos^2(x)))dx=# #=intsin(x)/(1+cos^2(x))dx=# consider that #d[cos(x)]=-sin(x)dx# so we have: #=-int1/(1+cos^2(x))d[cos(x)]=# with #cos(x)# as "variable" of integration to get; #=-arctan[cos(x)]+c#

Christopher Agresta · Answer

undefined To solve the integral ∫tan(x) / (sec(x) + cos(x)) dx, perform the following steps:

1. Rewrite tan(x) as sin(x) / cos(x).
2. Rewrite sec(x) as 1 / cos(x).
3. Rewrite the integral as ∫(sin(x) / cos(x)) / (1 / cos(x) + cos(x)) dx.
4. Simplify the denominator to obtain ∫(sin(x) / cos(x)) / (1 + cos^2(x)) dx.
5. Multiply the numerator and denominator by cos(x) to rationalize the expression: ∫sin(x) / (cos(x) + cos^3(x)) dx.
6. Perform a substitution, letting u = cos(x), du = -sin(x) dx.
7. Rewrite the integral in terms of u: ∫-du / (u + u^3) = -∫du / (u(1 + u^2)).
8. Perform partial fraction decomposition on the denominator to obtain: ∫(-1/u) / (1 + u^2) du - ∫(1/(1 + u^2)) du.
9. Integrate each term: -ln|u| - arctan(u) + C.
10. Substitute back u = cos(x): -ln|cos(x)| - arctan(cos(x)) + C.

So, the integral of tan(x) / (sec(x) + cos(x)) dx is -ln|cos(x)| - arctan(cos(x)) + C, where C is the constant of integration.