How do you use partial fractions to find the integral #int (sinx)/(cosx+cos^2x)dx#?

Question

Ezekiel Alexander · Accepted Answer

#ln|cosx + 1| - ln|cosx| + C#

We start with a substitution. Let #u = cosx#. Then #du = -sinxdx -> dx = (du)/-sinx#.

#=>intsinx/(u + u^2) xx (du)/-sinx#

#=> int -1/(u + u^2)du#

We now factor the denominator to #u(1 + u)#. Let's now write #-1/(u + u^2)# in partial fractions.

#A/u + B/(u + 1) = -1/(u(u +1))#

#A(u + 1) + B(u) = -1#

#Au + A + Bu = -1#

#(A + B)u + A = -1#

Now write a system of equations:

#{(A + B = 0), (A = -1):}#

Solving, we get: #A = -1# and #B = 1#.

Thus, the partial fraction decomposition is #-1/u + 1/(u + 1)#. The integral becomes #int(1/(u + 1) - 1/u)du#.

We can now integrate using the rule #int(1/u)du = ln|u| + C#.

#=> ln|u + 1| - ln|u| + C#

Finally, reinsert the value of #u# to make the function defined by #x# instead of #u#:

#=>ln|cosx + 1| - ln|cosx| +C#

Hopefully this helps!

Samantha Adkison · Answer

undefined To integrate $ \int \frac{\sin x}{\cos x + \cos^2 x} \, dx $ using partial fractions, follow these steps:

1. Rewrite the integrand as a sum of partial fractions.
2. Integrate each partial fraction separately.
3. Combine the results to find the final integral.

First, rewrite $ \frac{\sin x}{\cos x + \cos^2 x} $ as partial fractions:

$$ \frac{\sin x}{\cos x + \cos^2 x} = \frac{A}{\cos x} + \frac{B}{\cos^2 x} $$

To find $ A $ and $ B $, multiply both sides by the denominator and then equate coefficients of like terms.

$$ \sin x = A\cos x + B $$

Now, solve for $ A $ and $ B $ by substituting convenient values of $ x $. For instance, setting $ x = 0 $ gives:

$$ 0 = A\cos 0 + B $$

$$ 0 = A + B $$

Setting $ x = \frac{\pi}{2} $ gives:

$$ 1 = A\cos \frac{\pi}{2} + B $$

$$ 1 = 0 + B $$

$$ B = 1 $$

$$ A = -1 $$

Now, rewrite the integral with the partial fractions:

$$ \int \frac{\sin x}{\cos x + \cos^2 x} \, dx = \int \frac{-1}{\cos x} \, dx + \int \frac{1}{\cos^2 x} \, dx $$

Now, integrate each term separately:

$$ \int \frac{-1}{\cos x} \, dx = -\ln|\sec x + \tan x| + C_1 $$

$$ \int \frac{1}{\cos^2 x} \, dx = \tan x + C_2 $$

Finally, combine the results to get the final integral:

$$ \int \frac{\sin x}{\cos x + \cos^2 x} \, dx = -\ln|\sec x + \tan x| + \tan x + C $$

Where $ C $ is the constant of integration.