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What is #int sec^2(5x) dx #?

1 Answer

Feb 11, 2016

#intsec^2(5x)dx = 1/5tan(5x)+C#

Explanation:

Noting that #d/dx tan(x) = sec^2(x)#

Let #u = 5x => du = 5dx => dx = 1/5du#

#intsec^2(5x)dx = intsec^2(u)/5du#

#=1/5intsec^2(u)du#

#=1/5tan(u)+C#

#=1/5tan(5x)+C#

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