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How do you integrate #tanx / (secx + cosx)#?

Answer 1

Ezra Adams

#-arctan(cosx)+C#

We should first try to simplify the integrand.

#tanx/(secx+cosx)=(sinx/cosx)/(1/cosx+cosx)=sinx/(cosx(1/cosx+cosx))=sinx/(1+cos^2x)#

Thus:

#inttanx/(secx+cosx)dx=intsinx/(1+cos^2x)dx#

We can use substitution here. Let #u=cosx#. This implies that #du=-sinxdx#.

#=-int(-sinx)/(1+cos^2x)dx=-int1/(1+u^2)du#

This is the arctangent integral!

#=-arctan(u)+C=-arctan(cosx)+C#

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Answer 2

William Abdalla

To integrate ( \frac{\tan(x)}{\sec(x) + \cos(x)} ), you can use the substitution method. Let ( u = \sec(x) + \cos(x) ). Then, ( du = (\sec(x)\tan(x) - \sin(x))dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \frac{\tan(x)}{\sec(x) + \cos(x)} , dx = \int \frac{1}{u} , du ]

This integral is straightforward to solve:

[ \int \frac{1}{u} , du = \ln|u| + C ]

Substitute ( u = \sec(x) + \cos(x) ) back into the result:

[ \ln|\sec(x) + \cos(x)| + C ]

Therefore, the integral of ( \frac{\tan(x)}{\sec(x) + \cos(x)} ) is ( \ln|\sec(x) + \cos(x)| + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.