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# How do you integrate this $\int {\left( {\dfrac{1}{{1 + \sin x + \cos x}}} \right)} dx$ ?

Last updated date: 27th Mar 2023
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Hint: To solve and take out the integral of this question , we need to solve it step by step . Here we are going to transform the fraction apart and perform some calculations and formulae of integration to simplify the given question with the help of the concept of Weierstrass substitution . It is used for converting rational expressions of trigonometric functions into algebraic rational functions, which may be easier to integrate . Any rational expression of trigonometric functions can be always reduced to integrating a rational function by making the Weierstrass substitution . And then again adding back together when the individual fraction would be solved to make our question solved easily . Some basic trigonometry formulae will come into existence while solving the integration part and Don’t forget to place Constant of integration $C$ at the end of the integral.

We are given an expression $\int {\left( {\dfrac{1}{{1 + \sin x + \cos x}}} \right)} dx$ and we have to calculate its integral .
The Weierstrass substitution is very useful for integrals involving a simple rational expression in $\sin x$ and/or $\cos x$ in the denominator .
In order to calculate this integral we may use the following to substitute –
$t = \tan \left( {\dfrac{x}{2}} \right)$
Hence,
$\sin x = \dfrac{{2t}}{{1 + {t^2}}}$ , $\cos x = \dfrac{{1 - {t^2}}}{{1 + {t^2}}}$ , $dx = \dfrac{2}{{1 + {t^2}}}dt$
Now we are going to substitute these in our given question .
$\int {\left( {\dfrac{1}{{1 + \sin x + \cos x}}} \right)} dx$
$= \int {\left( {\dfrac{1}{{1 + \sin x + \cos x}}} \right)} dx \\ = \int {\left( {\dfrac{1}{{1 + \dfrac{{2t}}{{1 + {t^2}}} + \dfrac{{1 - {t^2}}}{{1 + {t^2}}}}}} \right)} \dfrac{2}{{1 + {t^2}}}dt \;$
To solve this further we will now try to cancel out the common terms by taking LCM as –
$= \int {\left( {\dfrac{1}{{\dfrac{{1 + {t^2}}}{{1 + {t^2}}} + \dfrac{{2t}}{{1 + {t^2}}} + \dfrac{{1 - {t^2}}}{{1 + {t^2}}}}}} \right)} \dfrac{2}{{1 + {t^2}}}dt \\ = \int {\left( {\dfrac{1}{{\dfrac{{1 + {t^2} + 2t + 1 - {t^2}}}{{1 + {t^2}}}}}} \right)} \dfrac{2}{{1 + {t^2}}}dt \\ = \int {\left( {\dfrac{2}{{1 + {t^2} + 2t + 1 - {t^2}}}} \right)} dt \;$
Now perform simple calculations like addition and subtraction to make it simplified .
$= \int {\left( {\dfrac{2}{{1 + {t^2} + 2t + 1 - {t^2}}}} \right)} dt \\ = \int {\left( {\dfrac{2}{{2 + 2t}}} \right)} dt \\ = \int {\left( {\dfrac{2}{{2(1 + t)}}} \right)} dt \\ = \int {\left( {\dfrac{1}{{1 + t}}} \right)} dt \;$
Now it is easier for us to integrate this , we get –
$= \int {\left( {\dfrac{1}{{1 + t}}} \right)} dt \\ = \ln |1 + t| + C \;$
Now , we will simply substitute the value of t , we get-
$= \ln |1 + t| + C \\ = \ln \left| {1 + \tan \left( {\dfrac{x}{2}} \right)} \right| + C \;$
$\ln \left| {1 + \tan \left( {\dfrac{x}{2}} \right)} \right| + C$
This is the required answer $\ln \left| {1 + \tan \left( {\dfrac{x}{2}} \right)} \right| + C$ .
So, the correct answer is “ $\ln \left| {1 + \tan \left( {\dfrac{x}{2}} \right)} \right| + C$ ”.

Note:
I.Always make use of the Weierstrass substitution for integrals involving a simple rational expression in $\sin x$ and/or $\cos x$ in the denominator.
II.Use standard formula carefully while evaluating the integrals.
III.Indefinite integral=Let $f(x)$ be a function .Then the family of all its primitives (or antiderivatives) is called the indefinite integral of $f(x)$ and is denoted by $\int {f(x)} dx$
IV.The symbol $\int {f(x)dx}$ is read as the indefinite integral of $f(x)$ with respect to x.
V.C is known as the constant of integration and Don’t forget to place Constant of integration $C$ at the end of the integral.
VI.Cross check the answer and always keep the final answer simplified .
VII.Remember the algebraic identities and apply appropriately .