Asked by peyton
evaluate the integral I= ∫(0,∏/4) ((5-3e^(-tanx))/(cos^(2)x))dx
Answers
Answered by
Reiny
write it as
5/cos^2 x - 3e^(-tanx)/cos^2x
= 5sec^2x - 3e^(-tanx)/cos^2 x
= 5sec^2x - 3e^(-tanx) (sec^2 x)
now for recognition facts
recall that d(tanx)/dx = sec^2x
so the integral of 5sec^2x is 5tanx
for the integral of the exponential, remember that when we differentiate an e^(anything) function, the original e^anything comes back, times the derivative of "anything"
we have exactly that pattern
so the integral of -3e^(-tanx) (sec^2 x) is
+ 3e^(-tanx)
so ∫((5-3e^(-tanx))/(cos^(2)x))dx
= 5tanx + 3e^(-tanx) + a constant
I will leave the substitution up to you
5/cos^2 x - 3e^(-tanx)/cos^2x
= 5sec^2x - 3e^(-tanx)/cos^2 x
= 5sec^2x - 3e^(-tanx) (sec^2 x)
now for recognition facts
recall that d(tanx)/dx = sec^2x
so the integral of 5sec^2x is 5tanx
for the integral of the exponential, remember that when we differentiate an e^(anything) function, the original e^anything comes back, times the derivative of "anything"
we have exactly that pattern
so the integral of -3e^(-tanx) (sec^2 x) is
+ 3e^(-tanx)
so ∫((5-3e^(-tanx))/(cos^(2)x))dx
= 5tanx + 3e^(-tanx) + a constant
I will leave the substitution up to you
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