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

Question

Charles Arocha · Accepted Answer

#int\ 1/sin^2(x)+cos(2x)\ dx=1/2sin(2x)-cot(x)+C#

First, I will split up the integral into two parts: #int\ 1/sin^2(x)+cos(2x)\ dx=int\ 1/sin^2(x)\ dx + int\ cos(2x)\ dx#

I will call the left one Integral 1 and the right one Integral 2

Integral 1 We can use the following trigonometric identity: #1/sin(theta)=csc(theta)#

This gives: #int\ 1/sin^2(x)\ dx=int\ csc^2(x)\ dx#

The derivative of #cot(x)# is #-csc^2(x)#, so we can deduce that the answer to this integral must be #-cot(x)#: #int\ csc^2(x)\ dx=-cot(x)+C#

Integral 2 Here I will do a u-substitution with #u=2x#. The derivative of #u# is #2#, so we divide by #2# to integrate with respect to #u#: #int\ cos(2x)\ dx=1/2int\ cos(u)\ du=1/2sin(u)+C#

Undoing the substitution, we get: #1/2sin(u)+C=1/2sin(2x)+C#

Completing the original integral Now that we know Integral 1 and Integral 2, we can complete teh original integral: #int\ 1/sin^2(x)+cos(2x)\ dx=1/2sin(2x)-cot(x)+C#

Christopher Agresta · Answer

undefined To find the integral of $ \frac{1}{\sin^2(x)} + \cos(2x) $, you can split the integral into two parts and then integrate each part separately.

First, let's consider $ \frac{1}{\sin^2(x)} $. Rewrite it as $ \csc^2(x) $, where $ \csc(x) $ is the cosecant function.

The integral of $ \csc^2(x) $ is $ -\cot(x) $.

Next, let's consider $ \cos(2x) $. The integral of $ \cos(2x) $ is $ \frac{1}{2}\sin(2x) $.

So, the integral of $ \frac{1}{\sin^2(x)} + \cos(2x) $ is:

$$ \int \left(\frac{1}{\sin^2(x)} + \cos(2x)\right) \, dx = \int \csc^2(x) \, dx + \int \cos(2x) \, dx $$

$$ = -\cot(x) + \frac{1}{2}\sin(2x) + C $$

Where $ C $ is the constant of integration. Therefore, the integral of $ \frac{1}{\sin^2(x)} + \cos(2x) $ is $ -\cot(x) + \frac{1}{2}\sin(2x) + C $.