Question

Integrate \[\int{\left( \dfrac{1}{\csc (x)+1} \right)dx}\]

Answer 1

Hint: To evaluate the required integral, first we will rationalize the denominator that means multiply \[(\csc (x)-1)\]on numerator and denominator then by simplifying then apply the trigonometry formula that is \[{{\cot }^{2}}(x)=1-{{\csc }^{2}}(x)\]. After simplification, we will use the integration formula to integrate the function and get the required answer.

Complete answer:
In this problem, we have to evaluate the integral that is \[\int{\left( \dfrac{1}{\csc (x)+1} \right)dx}\]
For that we need to rationalize the denominator to make the problem easier that means we multiply numerator as well as denominator by \[(1-\csc (x))\] we get;
\[\Rightarrow \int{\left( \dfrac{\csc (x)-1}{\left( \csc (x)+1 \right)\left( \csc (x)-1 \right)} \right)dx}\]
By using the basic property of mathematics that is \[(a-b)(a+b)={{a}^{2}}-{{b}^{2}}\] substitute this property in above step then we get:
\[\Rightarrow \int{\left( \dfrac{\csc (x)-1}{{{\csc }^{2}}(x)-{{1}^{2}}} \right)dx}\]
By simplifying this we get:
\[\Rightarrow \int{\left( \dfrac{\csc (x)-1}{{{\csc }^{2}}(x)-1} \right)dx}\]
If you kindly notice this above equation then we can see in the denominator that we can apply the formula of trigonometry that is \[{{\cot }^{2}}(x)=1-{{\csc }^{2}}(x)\] that means these formula substitute in this above step to make the problem easier.
\[\Rightarrow \int{\left( \dfrac{\csc (x)-1}{{{\cot }^{2}}(x)} \right)dx}\]
Now, by splitting the denominator we get:
\[\Rightarrow \int{\left( \dfrac{\csc (x)}{{{\cot }^{2}}(x)}-\dfrac{1}{{{\cot }^{2}}(x)} \right)dx}\]
Now, we can write the above expression in the term of sin and cosine for that we also need to use the trigonometry formula that is \[\cot (x)=\dfrac{\cos (x)}{\sin (x)}\], \[\csc (x)=\dfrac{1}{\sin (x)}\] and \[\tan (x)=\dfrac{1}{\cot (x)}\] then substituting this in the above step we get:
\[\Rightarrow \int{\left( \dfrac{\dfrac{1}{\sin (x)}}{\dfrac{{{\cos }^{2}}(x)}{{{\sin }^{2}}(x)}}-{{\tan }^{2}}(x) \right)dx}\]
By further simplification this we get:
\[\Rightarrow \int{\left( \dfrac{\sin (x)}{{{\cos }^{2}}(x)}-{{\tan }^{2}}(x) \right)dx}\]
By further solving and further simplifying this we get:
\[\Rightarrow \int{\left( \left( \dfrac{\sin (x)}{\cos (x)}\times \dfrac{1}{\cos (x)} \right)-{{\tan }^{2}}(x) \right)dx}\]
As we know that, \[\sec (x)=\dfrac{1}{\cos (x)}\] then substituting this we get:
\[\Rightarrow \int{\left( \sec (x)\tan (x)-{{\tan }^{2}}(x) \right)dx}\]
As we know that \[{{\tan }^{2}}(x)={{\sec }^{2}}(x)-1\] then substituting this we get:
\[\Rightarrow \int{\left( \sec (x)\tan (x)-{{\sec }^{2}}(x)+1 \right)dx}\]
Now, we have to perform integration by using formula that is \[\int{\sec (x)\tan (x)=\sec (x)}\] and \[\int{{{\sec }^{2}}(x)\,\,=\,\tan (x)}\] we get:
\[\Rightarrow \sec (x)-\tan (x)+x+c\]
Hence, the value of \[\int{\left( \dfrac{1}{\csc (x)+1} \right)dx}\]is \[\sec (x)-\tan (x)+x+c\].

Note:
In this problem, we have evaluated the indefinite integral. \[F(x)\] is an antiderivative of the function \[f(x)\] then antiderivative of \[f(x)\] is called an indefinite integral and it is denoted by\[\int{f(x)}\,dx=F(x)+c\] where, C is an integration constant. Note that f(x) is called the integrand. An integral of the form \[\int{f(x)}\,dx\] without upper and lower limits is called indefinite integral. In this type of problem of integration, we must remember the trigonometric identities and formulas. Also remember the formulas of integration and also remember the trigonometry formulas.