How do you find the integral of #1 / (1 + sin^2 x)#?

Question

Emily Ammerman · Accepted Answer

#arctan(sqrt2tanx)/sqrt2+C#

#I=int1/(1+sin^2x)dx#

Notice that dividing by #cos^2(x)# will get us working with #tanx# and #secx# functions, which are derivatives of themselves:

#I=int(1/cos^2x)/((1+sin^2x)/cos^2x)dx=intsec^2x/(sec^2x+tan^2x)dx#

Recall that #sec^2x=1+tan^2x#:

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

We will use the substitution #tanx=1/sqrt2tantheta#. This implies that #sec^2xdx=1/sqrt2sec^2thetad theta#. Substituting:

#I=int(1/sqrt2sec^2theta)/(1+2(1/sqrt2tantheta)^2)d theta=1/sqrt2int(sec^2theta)/(1+2(1/2tan^2theta))d theta#

#I=1/sqrt2int(sec^2theta)/(1+tan^2theta)d theta#

Using the same trigonometric identity:

#I=1/sqrt2intsec^2theta/sec^2thetad theta=1/sqrt2intd theta=1/sqrt2theta+C#

From #tanx=1/sqrt2tantheta# we see that #sqrt2tanx=tantheta# and #theta=arctan(sqrt2tanx)#. Thus:

#I=arctan(sqrt2tanx)/sqrt2+C#

Ezra Adams · Answer

undefined To find the integral of $ \frac{1}{1 + \sin^2(x)} $, you can use the trigonometric identity $1 + \sin^2(x) = \cos^2(x)$. Then, you can apply a trigonometric substitution. Let $u = \cos(x)$ or $du = -\sin(x)dx$. After substitution, you'll have an integral in terms of $u$, which you can integrate using standard techniques.