How do you use partial fractions to find the integral #int (sec^2x)/(tanx(tanx+1))dx#?

Question

Isabella Ackerman · Accepted Answer

#int sec^2x/(tanx(tanx+1))dx = ln abs (sinx/(sinx+cosx)) +C#

Substitute:

#t = tanx#

#dt = sec^2x dx#

so we have:

#int sec^2x/(tanx(tanx+1))dx = int (dt)/(t(t+1))#

Now we can solve the integral using partial fractions:

#1/(t(t+1)) = A/t+B/(t+1)#

#1/(t(t+1)) = (A(t+1)+ Bt)/(t(t+1))#

#1= (A+B)t +A#

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

#int (dt)/(t(t+1)) = int (1/t-1/(t+1))dt#

#int (dt)/(t(t+1)) = int (dt)/t-int (dt)/(t+1)#

#int (dt)/(t(t+1)) = ln abs(t) -ln abs(t+1) +C =ln abs (t/(t+1)) +C#

and substituting back #x#:

#int sec^2x/(tanx(tanx+1))dx = ln abs (tanx/(tanx+1)) +C#

#int sec^2x/(tanx(tanx+1))dx = ln abs (sinx/(sinx+cosx)) +C#

Isabella Ackerman · Answer

undefined To integrate the given expression, we first rewrite it using partial fractions. We express $ \frac{\sec^2 x}{\tan x(\tan x + 1)} $ as the sum of two fractions:

$$ \frac{\sec^2 x}{\tan x(\tan x + 1)} = \frac{A}{\tan x} + \frac{B}{\tan x + 1} $$

Next, we find the values of $ A $ and $ B $ by equating numerators:

$$ \sec^2 x = A(\tan x + 1) + B(\tan x) $$

$$ \sec^2 x = A\tan x + A + B\tan x $$

By rearranging terms, we have:

$$ \sec^2 x = (A + B)\tan x + A $$

Comparing coefficients of $ \tan x $ and constants on both sides:

$$ A + B = 0 $$
$$ A = 1 $$

From the first equation, we find $ B = -1 $.

So, we rewrite the integral as:

$$ \int \frac{\sec^2 x}{\tan x(\tan x + 1)} dx = \int \frac{1}{\tan x} dx - \int \frac{1}{\tan x + 1} dx $$

Each of these integrals can be solved separately:

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

$$ \int \frac{1}{\tan x + 1} dx = \int \frac{\cos x}{\sin x + \cos x} dx $$

To solve this, let $ u = \sin x + \cos x $, then $ du = (\cos x - \sin x) dx $, and rearrange for $ dx = \frac{du}{\cos x - \sin x} $:

$$ \int \frac{1}{u} \frac{du}{\cos x - \sin x} $$

$$ = \int \frac{1}{u} du $$

$$ = \ln|u| + C_2 $$

$$ = \ln|\sin x + \cos x| + C_2 $$

Therefore, the integral of $ \frac{\sec^2 x}{\tan x(\tan x + 1)} dx $ is:

$$ \ln|\sin x| - \ln|\sin x + \cos x| + C $$

$$ = \ln\left|\frac{\sin x}{\sin x + \cos x}\right| + C $$