How do you evaluate #int cosx/(1+sin^2x)# from #[0, pi/2]#?

Question

Paisley Johnson · Accepted Answer

#intcosx/(1+sin^2x)dx=pi/4#

First examining without the bounds:

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

Let #sinx=tantheta#. This may look like a wild substitution, but the goal is to get the denominator of the fraction to #1+tan^2theta=sec^2theta#. Note that differentiating #sinx=tantheta# gives #cosxdx=sec^2thetad theta#. It's also helpful that #cosxdx# is already present in the integrand. Substituting these in:

#I=intsec^2theta/(1+tan^2theta)d theta=intd theta=theta+C#

Since #sinx=tantheta#, we see that #theta=arctan(sinx)#:

#I=arctan(sinx)+C#

Adding the bounds:

#I_B=int_0^(pi/2)cosx/(1+sin^2x)dx=[arctan(sinx)]_0^(pi/2)#

#color(white)(I_B)=arctan(sin(pi/2))-arctan(sin(0))#

#color(white)(I_B)=arctan(1)-arctan(0)#

#color(white)(I_B)=pi/4-0#

#color(white)(I_B)=pi/4#

Luke Alvarez · Answer

undefined To evaluate the integral $\int_0^{\pi/2} \frac{\cos(x)}{1 + \sin^2(x)} \, dx$ over the interval $[0, \frac{\pi}{2}]$, we can use a trigonometric substitution.

Let $u = \sin(x)$, then $du = \cos(x) \, dx$.

Substitute $u = \sin(x)$ and $du = \cos(x) \, dx$ into the integral, we get:

$$\int \frac{1}{1 + u^2} \, du$$

This is a standard integral which is equal to $\arctan(u) + C$, where $C$ is the constant of integration.

So, $$\int_0^{\pi/2} \frac{\cos(x)}{1 + \sin^2(x)} \, dx = \left[\arctan(\sin(x))\right]_0^{\pi/2}$$

Evaluating at the upper and lower bounds:

$$\left[\arctan(\sin(\frac{\pi}{2}))\right] - \left[\arctan(\sin(0))\right]$$

$$\arctan(1) - \arctan(0)$$

$$\frac{\pi}{4} - 0 = \frac{\pi}{4}$$

So, $\int_0^{\pi/2} \frac{\cos(x)}{1 + \sin^2(x)} \, dx = \frac{\pi}{4}$.