What is the integral of #tan^3xsecxdx#?

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What is the integral of #tan^3xsecxdx#?

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Answer 1 · 2015-03-17T22:10:37

The answer can be written as: #\frac{1}{3}\sec^{3}(x)-\sec(x)+C#.

Use the fact that #\tan^{2}(x)=\sec^{2}(x)-1# to write #\tan^{3}(x)\sec(x)# as #(sec^{2}(x)-1)\tan(x)\sec(x)#. Note now that if #u=\sec(x)# then #du=\sec(x)\tan(x)dx#. Hence

# \int\tan^{3}(x)\sec(x)dx=\int(u^{2}-1)du=u^{3}/3-u+C=\frac{1}{3}\sec^{3}(x)-\sec(x)+C#

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What is the integral of #tan^3xsecxdx#?

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