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

Last updated date: 17th Jun 2024
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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:
$\Rightarrow \int {\dfrac{{dx}}{{{x^2} + {a^2}}} = \dfrac{1}{a}{{\tan }^{ - 1}}\dfrac{x}{a} + {\text{c}}}$