Question #29a24

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Answer 1 · 2016-12-10T17:50:53

#int6sin^2(x)cos^-4(x)dx = 2tan^3(x) + C#

#int6sin^2(x)cos^-4(x)dx #

Use the identities #tan(x) = sin(x)/cos(x) and sec(x) = 1/cos(x)#:

#6inttan^2(x)sec^2(x)dx

Let #u = tan(x)", then, "du = sec^2(x)dx

Then the integral becomes:

#6intu^2du #

This is a trivial integration:

#6/3u^3 + C#

Reverse the substitution:

#2tan^3(x) + C#

The answer is:

#int6sin^2(x)cos^-4(x)dx = 2tan^3(x) + C#