How do you integrate #int e^x/(7+e^(2x))# by trigonometric substitution?

Question

Samantha Adkison · Accepted Answer

#1/sqrt(7) arctan (e^ x/sqrt(7))#

#int e^x/(7+ e^2x)dx = int e^x/(7+ (e^x)^2)dx #

write it in a format that we can look to integrate more easily

Let #u = e^x# #du = e^x dx #

The integral in terms of u is now

#int 1/(7+u^2)*du #

To use a trig substitution on this I used the fact that #int 1/(a^2+x^2)dx = 1/aarctan(x/a)#

Therefore a good trig substitution for this sort of problem would be to let #u = sqrt(7) tan(A)# therefore #u^2 = 7tan^2(A) #

#u = sqrt(7) tan A#

#u = sqrt(7) Sin(A)/Cos(A)#

differentiating using the quotient rule

#du = sqrt(7) ((cos(A)cos(A) - sin(A)(-sin(A)))/cos^2) dA #

#Note cos^2(A) + sin ^2(A) = 1 #

#du = sqrt(7) 1/cos^2(A) dA#

The integral is now of the form

#int 1/7(1+tan^2(A)).sqrt(7) 1/cos^2(A) dA #

#= int sqrt(7)/7(1+sin^2(A)/cos^2(A)).sqrt(7) 1/cos^2(A) dA #

#= int sqrt(7)/7*1/((cos^2(A)+sin^2(A))/cos^2(A)) 1/cos^2(A) dA #

#= int sqrt(7)/7*1 dA# (everything cancels out)

#= sqrt(7)/7A#

Now remember #u = sqrt(7) tan(A)# and #u^2 = 7tan^2(A) #

Therefore # tan(A) = u / sqrt(7)#

A =#arctan (u/sqrt(7))#

So the integral results in = #sqrt(7)/7arctan (u/sqrt(7))#

= #1/sqrt(7)arctan (u/sqrt(7))#

finally we had #u = e^x#

Therefore the answer is #1/sqrt(7) arctan (e^ x/sqrt(7))#

Charles Anctil · Answer

undefined To integrate $ \frac{e^x}{7+e^{2x}} $ by trigonometric substitution, first let $ u = e^x $. Then, $ du = e^x \, dx $. Rewrite the integral in terms of $ u $ to get $ \int \frac{1}{7+u^2} \, du $. This is a standard form for trigonometric substitution. Let $ u = \sqrt{7} \tan{\theta} $. Then, $ du = \sqrt{7} \sec^2{\theta} \, d\theta $. Substitute $ u $ and $ du $ into the integral and simplify. This results in $ \frac{1}{\sqrt{7}} \int \frac{\sec^2{\theta}}{7+7\tan^2{\theta}} \, d\theta $. Now, use the trigonometric identity $ \sec^2{\theta} = 1 + \tan^2{\theta} $ to simplify the expression further. After simplification, the integral becomes $ \frac{1}{\sqrt{7}} \int \frac{1}{14} \, d\theta $. Integrate to get $ \frac{\theta}{\sqrt{7}} + C $. Substitute $ \theta = \arctan{\frac{u}{\sqrt{7}}} $ back into the equation. Thus, the final answer is $ \frac{\arctan{\frac{e^x}{\sqrt{7}}}}{\sqrt{7}} + C $.