Question

Integrate \[\int {\dfrac{1}{{1 + \tan x}}dx} \].

Answer 1

Hint: Here we have to find the value of the given integral. We use many integration formulas to find the solution of the given integral. On doing simplification we will split the equation to solve it. Finally we get the required answer.

Formula used: We will use the following trigonometry formulas: \[{\sec ^2}x = {\tan ^2}x + 1\]
Following integration formulas:
 \[\int {\dfrac{1}{v}dv = \log v + C} \]
\[\int {\dfrac{1}{{{u^2} + 1}}dx} = {\tan ^{ - 1}}u + C\], Where \[C\] is a constant.

Complete step-by-step solution:
It is given that the integral \[\int {\dfrac{1}{{\tan x + 1}}dx} ....\left( 1 \right)\]
We will substitute by the following equation in the given integral: \[{\sec ^2}x = {\tan ^2}x + 1\]
The integration will be: \[\int {\dfrac{1}{{\tan x + 1}}dx} \]
On multiply and divide \[{\sec ^2}x\]we get,
\[ \Rightarrow \int {{{\sec }^2}x\dfrac{1}{{{{\sec }^2}x(\tan x + 1)}}dx} \]
On putting the formula and we get,
\[ \Rightarrow \int {{{\sec }^2}x\dfrac{1}{{(\tan x + 1)({{\tan }^2}x + 1)}}dx} ....\left( 2 \right)\]
Now, we will substitute \[u = \tan x\],
Then we get by using simple derivation with respect to\[x\]:
\[ \Rightarrow \dfrac{{du}}{{dx}} = {\sec ^2}x\]
Taking cross multiply we get,
\[ \Rightarrow du = {\sec ^2}x.dx\]
We can re-write \[\left( 2 \right)\]then the integration as following:
\[ \Rightarrow \int {\dfrac{1}{{(u + 1)({u^2} + 1)}}du} \]
Now, we can perform partial decomposition by the split up the numerator in following way:
\[ \Rightarrow \int {\dfrac{1}{{(u + 1)({u^2} + 1)}}du} \]
Then we can write it as
\[ \Rightarrow \int {\dfrac{1}{2} \times \dfrac{{{u^2} + 1 - {u^2} + 1}}{{(u + 1)({u^2} + 1)}}} du\]
Now if we apply linearity of the equation:
\[ \Rightarrow \dfrac{1}{2}\int {\left( {\dfrac{{{u^2} + 1}}{{(u + 1)({u^2} + 1)}} - \dfrac{{{u^2} - 1}}{{(u + 1)({u^2} + 1)}}} \right)} du\]
Cancel the terms in numerator and denominator in the first term and in the second we can split up the square terms:
\[ \Rightarrow \dfrac{1}{2}\int {\left( {\dfrac{1}{{(u + 1)}} - \dfrac{{(u - 1)(u + 1)}}{{(u + 1)({u^2} + 1)}}} \right)} du\]
Again, cancel the terms:
\[ \Rightarrow \dfrac{1}{2}\int {\left( {\dfrac{1}{{(u + 1)}} - \dfrac{{(u - 1)}}{{({u^2} + 1)}}} \right)} du\]
On splitting the integration we can write it as,
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{1}{{(u + 1)}}du - \dfrac{1}{2}\int {\dfrac{{(u - 1)}}{{({u^2} + 1)}}du} } \]
On splitting the term and we get,
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{1}{{(u + 1)}}du - \dfrac{1}{2}\int {\dfrac{u}{{({u^2} + 1)}}du - \dfrac{1}{2}\int {\dfrac{1}{{{u^2} + 1}}du} } } ....\left( 3 \right)\]
Now we have to find one by one and we get,
So, we take \[\dfrac{1}{2}\int {\dfrac{1}{{(u + 1)}}du} ....\left( 4 \right)\]
Here we will substitute \[v = u + 1\] and we get
Now take the derivation on the both side of the equation with respect to\[u\], we get,
\[ \Rightarrow \dfrac{{dv}}{{du}} = 1\]
Taking cross multiply and we get,
\[ \Rightarrow dv = du\]
Now integrate \[\left( 4 \right)\], we get:
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{{du}}{{u + 1}}} \]
Putting \[v = u + 1\] and we get,
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{{dv}}{v}} \]
Using the formula and we get,
\[ \Rightarrow \dfrac{1}{2}\ln (v) + k\], Here \[k\] is a constant.
Again, putting the value of \[v = u + 1\], we get:
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{{du}}{{u + 1}}} = \dfrac{1}{2}\ln (u + 1) + k\] ,where k is an arbitrary constant.
Now substitute the \[u = \tan x\] in the above equation:
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{{du}}{{u + 1}}} = \dfrac{1}{2}\ln (\tan x + 1) + k....\left( 5 \right)\], where k is a constant.
Now we take \[\dfrac{1}{2}\int {\dfrac{u}{{({u^2} + 1)}}du} ...\left( 6 \right)\]
Now substitute \[p = {u^2} + 1\] in \[\left( 6 \right)\], we get:
Taking the derivation on \[p\] with respect to \[u\]:
\[ \Rightarrow \dfrac{{dp}}{{du}} = 2u\].
Taking cross multiply we get,
\[ \Rightarrow dp = 2udu\]
On dividing \[2\] on both sides and we get,
\[ \Rightarrow \dfrac{1}{2}dp = udu\].
So, the equation \[\left( 6 \right)\]would become
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{1}{2}\dfrac{{dp}}{p}} \]
On multiply the terms and we get,
\[ \Rightarrow \dfrac{1}{4}\int {\dfrac{{dp}}{p}} \]
By using the formula we get:
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{u}{{{u^2} + 1}}du = \dfrac{1}{4}\ln (p) + z} \], where \[z\]is a constant.
Now putting the value of \[p = {u^2} + 1\]:
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{u}{{{u^2} + 1}}du = \dfrac{1}{4}\ln ({u^2} + 1) + z} \], where \[z\] is a constant.
Again, putting the value of \[u = \tan x\]:
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{u}{{{u^2} + 1}}du = \dfrac{1}{4}\ln ({{\tan }^2}x + 1) + z} ....\left( 7 \right)\].
Now we take\[\dfrac{1}{2}\int {\dfrac{1}{{{u^2} + 1}}du} \]
Now the integration of the following iteration, by using the formula:
\[ \Rightarrow \dfrac{1}{2}{\tan ^{ - 1}}u + g\], where \[g\] is a constant.
Now putting the value of \[u = \tan x\]:
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{1}{{{u^2} + 1}}du = \dfrac{1}{2}{{\tan }^{ - 1}}(\tan x)} + g....\left( 8 \right)\], where \[g\] is a constant.
On putting\[\left( 5 \right)\], \[\left( 7 \right)\] and \[\left( 8 \right)\]in \[\left( 3 \right)\] we get,
\[ \Rightarrow \dfrac{1}{2}\int {\dfrac{1}{{(u + 1)}}du - \dfrac{1}{2}\int {\dfrac{u}{{({u^2} + 1)}}du - \dfrac{1}{2}\int {\dfrac{1}{{{u^2} + 1}}du} } } \]
\[ \Rightarrow \dfrac{{\ln (\left| {\tan x\left. { + 1} \right|} \right.)}}{2} - \dfrac{{\ln (\left| {{{\tan }^2}x\left. { + 1} \right|} \right.)}}{4} + \dfrac{{{{\tan }^{ - 1}}(\tan (x))}}{2} + k - z + g\]
On rewriting the term and we get,
\[ \Rightarrow \dfrac{{\ln (\left| {\tan x\left. { + 1} \right|} \right.)}}{2} - \dfrac{{\ln (\left| {{{\tan }^2}x\left. { + 1} \right|} \right.)}}{4} + \dfrac{{{{\tan }^{ - 1}}(\tan (x))}}{2} + {C_1}\], where \[{C_1}\] is constant and \[{C_1} = k - z + g\].

The value of the given integral \[ \int {\dfrac{1}{{\tan x + 1}}dx = } \dfrac{{\ln (\left| {\tan x\left. { + 1} \right|} \right.)}}{2} - \dfrac{{\ln (\left| {{{\tan }^2}x\left. { + 1} \right|} \right.)}}{4} + \dfrac{{{{\tan }^{ - 1}}(\tan (x))}}{2} + {C_1}\]

Note: Integration is a technique of combining two or more parts into a single function by taking two different ranges. By using this technique we try to find a function \[g(x)\], whose derivative is \[Dg(x)\], is equal to the given function \[f(x)\].
Integration is indicated by integral signs.
If integration is indicated by, it states an indefinite integration.
If integration is indicated by a limit of, it states a definite integration.
The symbol represents an infinitesimal displacement along \[x\].
Thus \[\int {f(x)dx} \] is the summation or concatenation of \[f(x)\] and \[dx\] at a given close circuit.
Always divide the whole part into the small sub parts to do the integration in ease.