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Find the value of \[\int {\dfrac{1}{{3 + 2\sin x + \cos x}}dx} \] .

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Last updated date: 17th Jun 2024
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Answer
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Hint: Here we will write all the terms in the form of \[\tan x\] by using the below-mentioned formulas:
\[\sin x = \dfrac{{2\tan \dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}\] , \[\cos x = \dfrac{{1 - {{\tan }^2}\dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}\] and \[\tan x = \dfrac{{2\tan \dfrac{x}{2}}}{{1 - {{\tan }^2}\dfrac{x}{2}}}\] .

Complete step-by-step solution:
Step 1: We will replace the terms \[\sin x\] and \[\cos x\] by substituting the values of them in the form of \[\tan x\] as shown below:
\[I = \int {\dfrac{1}{{3 + 2\left( {\dfrac{{2\tan \dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}} \right) + \left( {\dfrac{{1 - {{\tan }^2}\dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}} \right)}}dx} \]
By opening the brackets and multiplying \[2\]inside the term \[\left( {\dfrac{{2\tan \dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}} \right)\] , we get:
\[ \Rightarrow I = \int {\dfrac{1}{{\dfrac{{4\tan \dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}} + \dfrac{{1 - {{\tan }^2}\dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}} + 3}}dx} \] ……………………….. (1)
Step 2: By taking \[1 + {\tan ^2}\dfrac{x}{2}\] common from the denominator and adding the numerator terms in the above expression (1), we get:
\[ \Rightarrow I = \int {\dfrac{1}{{\dfrac{{4\tan \dfrac{x}{2} + 3\left( {1 + {{\tan }^2}\dfrac{x}{2}} \right) + \left( {1 - {{\tan }^2}\dfrac{x}{2}} \right)}}{{1 + {{\tan }^2}\dfrac{x}{2}}}}}dx} \]
By bringing \[1 + {\tan ^2}\dfrac{x}{2}\] into the numerator position, we get:
\[ \Rightarrow I = \int {\dfrac{{1 + {{\tan }^2}\dfrac{x}{2}}}{{4\tan \dfrac{x}{2} + 3\left( {1 + {{\tan }^2}\dfrac{x}{2}} \right) + \left( {1 - {{\tan }^2}\dfrac{x}{2}} \right)}}dx} \] ………….. (2)
As we know that \[1 + {\tan ^2}\dfrac{x}{2} = {\sec ^2}\dfrac{x}{2}\] , so by replacing it in the above expression (2), we get:
\[ \Rightarrow I = \int {\dfrac{{{{\sec }^2}\dfrac{x}{2}}}{{4\tan \dfrac{x}{2} + 3\left( {1 + {{\tan }^2}\dfrac{x}{2}} \right) + \left( {1 - {{\tan }^2}\dfrac{x}{2}} \right)}}dx} \]
By opening the brackets in the denominator of the above expression we get:
\[ \Rightarrow I = \int {\dfrac{{{{\sec }^2}\dfrac{x}{2}}}{{4\tan \dfrac{x}{2} + 3 + 3{{\tan }^2}\dfrac{x}{2} + 1 - {{\tan }^2}\dfrac{x}{2}}}dx} \]
By doing simple addition and subtraction in the denominator part of the above expression we get:
\[ \Rightarrow I = \int {\dfrac{{{{\sec }^2}\dfrac{x}{2}}}{{4\tan \dfrac{x}{2} + 2{{\tan }^2}\dfrac{x}{2} + 4}}dx} \] …………………… (3)
Step 3: Now, we will assume that \[\tan \dfrac{x}{2} = t\] and differentiating it w.r.t \[t\] , we get:
\[ \Rightarrow {\sec ^2}\left( {\dfrac{x}{2}} \right) \times \dfrac{1}{2}dx = dt\]
We can write the above expression as below by bringing \[2\] into the RHS side:
\[ \Rightarrow {\sec ^2}\left( {\dfrac{x}{2}} \right)dx = 2dt\]
By substituting these values in the expression (3), we get:
\[ \Rightarrow I = \int {\dfrac{{2dt}}{{4t + 2{\operatorname{t} ^2} + 4}}} \] \[\left( {\because \tan \dfrac{x}{2} = t,{{\sec }^2}\dfrac{x}{2}dx = 2dt} \right)\]
By dividing the RHS side with \[2\], we get:
\[ \Rightarrow I = \int {\dfrac{{dt}}{{2t + {\operatorname{t} ^2} + 2}}} \]
By writing the term \[2t + {\operatorname{t} ^2} + 2 = {\left( {t + 1} \right)^2} + {\left( 1 \right)^2}\]in the RHS side of the expression \[I = \int {\dfrac{{dt}}{{2t + {\operatorname{t} ^2} + 2}}} \] , we get:
\[ \Rightarrow I = \int {\dfrac{{dt}}{{{{\left( {t + 1} \right)}^2} + {{\left( 1 \right)}^2}}}} \] …………………. (4)
Step 4: As we know that \[\dfrac{1}{{{x^2} + {a^2}}}dx = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + {\text{c}}\] , where \[{\text{c}}\] is an arbitrary constant. Comparing the above expression (4) with this formula, we get:
\[ \Rightarrow I = \dfrac{1}{1}{\tan ^{ - 1}}\dfrac{{\left( {t + 1} \right)}}{1} + {\text{c}}\]
By substituting the value of \[t = \tan \dfrac{x}{2}\] , in the above expression we get:
\[ \Rightarrow I = {\tan ^{ - 1}}\left( {\tan \dfrac{x}{2} + 1} \right) + {\text{c}}\]

$ \therefore$ The value of \[\int {\dfrac{1}{{3 + 2\sin x + \cos x}}dx} = {\tan ^{ - 1}}\left( {\tan \dfrac{x}{2} + 1} \right) + {\text{c}}\]

Note: In solving these types of question students should remember some basic formulas as given below:
\[\sin x = \dfrac{{2\tan \dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}\] , \[\cos x = \dfrac{{1 - {{\tan }^2}\dfrac{x}{2}}}{{1 + {{\tan }^2}\dfrac{x}{2}}}\] and \[\tan x = \dfrac{{2\tan \dfrac{x}{2}}}{{1 - {{\tan }^2}\dfrac{x}{2}}}\] , these are known as tangent half-angle formulas.
\[\dfrac{1}{{{x^2} + {a^2}}}dx = \dfrac{1}{a}{\tan ^{ - 1}}\dfrac{x}{a} + {\text{c}}\] , proof of which is showing below:
Suppose we need to evaluate the integral \[\dfrac{1}{{{x^2} + {a^2}}}dx\] for \[a \ne 0\]. So, by multiplying and dividing the expression with \[{a^2}\] , we get:
\[ \Rightarrow \int {\dfrac{1}{{{x^2} + {a^2}}}dx = \int {\dfrac{1}{{1 + \dfrac{{{x^2}}}{{{a^2}}}}}\dfrac{{dx}}{{{a^2}}}} } \]
Now by writing the terms \[\dfrac{x}{a} = u\] and differentiating it, we get:
\[ \Rightarrow \dfrac{{dx}}{a} = du\]
By substituting this value in the above expression \[\int {\dfrac{1}{{{x^2} + {a^2}}}dx = \int {\dfrac{1}{{1 + \dfrac{{{x^2}}}{{{a^2}}}}}\dfrac{{dx}}{{{a^2}}}} } \] , we get:
\[ \Rightarrow \int {\dfrac{{dx}}{{{x^2} + {a^2}}} = \dfrac{1}{a}\int {\dfrac{{du}}{{1 + {u^2}}}} } \]
We can write the above expression as below:
\[ \Rightarrow \int {\dfrac{{dx}}{{{x^2} + {a^2}}} = \dfrac{1}{a}{{\tan }^{ - 1}}u + {\text{c}}} \] , where \[{\text{c}}\] is an arbitrary constant.
By substituting the value of \[\dfrac{x}{a} = u\] in the above expression, we get:
\[ \Rightarrow \int {\dfrac{{dx}}{{{x^2} + {a^2}}} = \dfrac{1}{a}{{\tan }^{ - 1}}\dfrac{x}{a} + {\text{c}}} \]