How do you evaluate #intcos(x)/(9+sin^2x)dx#?

Question

Ezekiel Alexander · Accepted Answer

#1/3tan^-1(sinx/3)+C#

#intcosx/(9+sin^2x)dx#

Let #u=sinx# so #du=cosxcolor(white).dx#:

#=int1/(9+u^2)du=1/9int1/(1+u^2/9)du=1/9int1/(1+(u/3)^2)du#

Now let #v=u/3# so #dv=1/3du#:

#=1/3int(1/3)/(1+(u/3)^2)du=1/3int1/(1+v^2)dv#

Which is a common integral:

#=1/3tan^-1(v)=1/3tan^-1(u/3)=1/3tan^-1(sinx/3)+C#

Isaiah Allen · Answer

undefined To evaluate $ \int \frac{\cos(x)}{9+\sin^2(x)} \, dx $, 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:

$$ \int \frac{\cos(x)}{9+\sin^2(x)} \, dx = \int \frac{1}{9+u^2} \, du $$

This integral can be evaluated using the inverse tangent function. It follows the form $ \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C $.

Applying this, we have:

$$ \int \frac{1}{9+u^2} \, du = \frac{1}{3} \arctan\left(\frac{u}{3}\right) + C $$

Substituting back $ u = \sin(x) $, we get:

$$ \frac{1}{3} \arctan\left(\frac{\sin(x)}{3}\right) + C $$

Therefore, the integral evaluates to $ \frac{1}{3} \arctan\left(\frac{\sin(x)}{3}\right) + C $.