How do you find #int sec^2x/(1-sin^2x) dx #?

Question

How do you find #int sec^2x/(1-sin^2x)  dx #?

Christopher Allers · Accepted Answer

Shown below

#sin^2 x + cos^2 x = 1 => cos^2x = 1 - sin^2 x # #=> int sec^2x / ( cos^2 x ) dx # #=> int sec^2 x * 1/cos^2 x dx # #=> int sec^4 x dx # #=> int sec^2 x * sec^2 x dx # #=> int ( 1 + tan^2x) * sec^2 x dx # # u = tanx # #du = sec^2 x dx # #=> int 1 + u ^2 du # #=> u + 1/3 u^3 + c # # = tanx + 1/3 tan^3 x + c #

Christopher Allers · Answer

#tan(x) + frac(1)(3) tan^(3)(x) + C#

We have: #int# #frac(sec^(2)(x))(1 - sin^(2)(x))# #dx#

#= int# #frac(sec^(2)(x))(cos^(2)(x))# #dx#

#= int# #frac(sec^(2)(x))(frac(1)(sec^(2)(x)))# #dx#

#= int# #sec^(2)(x) cdot sec^(2)(x)# #dx#

Then, the Pythagorean identity is #cos^(2)(x) + sin^(2)(x) = 1#.

We can divide through by #cos^(2)(x)# it to get:

#Rightarrow 1 + tan^(2)(x) = sec^(2)(x)#

Now let's apply our integral to this rearranged identity:

#= int# #(1 + tan^(2)(x)) cdot sec^(2)(x)# #dx#

Now, let's use #u#-substitution, where #u = tan(x) Rightarrow du = sec^(2)(x)# #dx#:

#= int# #(1 + u^(2))# #du#

#= int# #1# #du + int# #u^(2)# #du#

#= u + frac(1)(3) u^(3) + C#

Finally, we can substitute #tan(x)# in place of #u#:

#= tan(x) + frac(1)(3) tan^(3)(x) + C#

Aurora Alvarado · Answer

undefined Integrate sec^2(x) / (1 - sin^2(x)) dx by first simplifying the expression in the integrand. Recall that sec^2(x) = 1 / cos^2(x) and 1 - sin^2(x) = cos^2(x). Therefore, the expression becomes 1 / cos^2(x) / cos^2(x) = 1 / cos^4(x). Now, rewrite cos^4(x) as (cos^2(x))^2 and use the substitution u = cos(x), du = -sin(x) dx:

∫ 1 / cos^4(x) dx
= ∫ 1 / (cos^2(x))^2 dx
= ∫ 1 / u^2 du
= ∫ u^(-2) du
= -u^(-1) + C
= -1 / cos(x) + C
= -sec(x) + C