How do you find #int cot^2x*secxdx#?

Question

Christopher Agresta · Accepted Answer

Notice that #cot^2(x)sec(x) = cot(x)csc(x)# and perform a simple substitution to find that
#intcot^2(x)sec(x)dx = -csc(x) + C# When examining an integral, if a simple substitution does not appear right away, it is frequently possible to manipulate the integrand into a more useful form. Here: #intcot^2(x)sec(x)dx = int(cos^2(x))/(sin^2(x))1/cos(x)dx# #= intcos(x)/sin(x)1/sin(x)dx# #= intcot(x)csc(x)dx# We could go directly to the answer from here, as we know that #csc(x)# is the antiderivative of #-csc(x)cot(x)# but let's do the substitution just to see how it works out. Let #u = csc(x) => du = -csc(x)cot(x)dx# #=> intcot(x)csc(x)dx = int-du# # = -intdu# #= -u+C# #= -csc(x) + C# Thus #intcot^2(x)sec(x)dx = -csc(x)+C#

Cameron Alexander · Answer

undefined To find ∫cot^2(x)sec(x)dx, we can use integration by parts. Let u = cot(x) and dv = sec(x)dx. Then, du = -csc^2(x)dx and v = ln|sec(x) + tan(x)|. Applying the integration by parts formula, we have:

∫cot^2(x)sec(x)dx = cot(x)ln|sec(x) + tan(x)| - ∫ln|sec(x) + tan(x)| * (-csc^2(x))dx

The second integral can be simplified by using a trigonometric identity: csc^2(x) = cot^2(x) + 1. Substituting this identity, we get:

∫cot^2(x)sec(x)dx = cot(x)ln|sec(x) + tan(x)| - ∫ln|sec(x) + tan(x)| * (-cot^2(x) - 1)dx

This integral can be further simplified by distributing the terms inside the logarithm and separating the integrals:

Now, we have two integrals involving ln|sec(x) + tan(x)|. We can use integration by parts again for each of them. After integrating by parts and simplifying the resulting expressions, we can solve for each integral separately.

After calculating both integrals, we can then combine the results to find the final solution for ∫cot^2(x)sec(x)dx.