Question

Evaluate the integral of \[\int {{\text{cosec x log}}\left( {{\text{cosec x - cot x}}} \right){\text{dx}}} \]

Answer 1

Hint: In this type of problem you need to find the value of given expression by using integration and differentiation methods.
The key point in these questions is to find the integral of a given question by using integration and differentiation methods.
By using integration and differentiation methods, the given equation has been separated as a ${\text{t}}$. From that we find ${\text{dt}}$.
Next we substitute the values of ${\text{t}}$and ${\text{dt}}$ in the given equation.
By integrating the values, we get the values then substitute the value of ${\text{t}}$.
In this question we have to evaluate the question to find the value of integral of \[\int {{\text{cosec x log}}\left( {{\text{cosec x - cot x}}} \right){\text{dx}}} \].

Formula used: ${\text{log x = }}\dfrac{{\text{1}}}{{\text{x}}}$ where ${\text{x}}$ is a constant.
$\dfrac{{\text{d}}}{{{\text{dx}}}}{\text{uv = u }}\dfrac{{{\text{dv}}}}{{{\text{dx}}}}{\text{ + v }}\dfrac{{{\text{du}}}}{{{\text{dx}}}}$

Complete step-by-step answer:
Here it is a given that integral of \[\int {{\text{cosec x log}}\left( {{\text{cosec x - cot x}}} \right){\text{dx}}} \]. We have to find the integral of the given expression.
By using integration and differentiation method we are going to find the value of given expression.
Now, Consider given as
${\text{I = }}\int {{\text{cosec x }}{\text{. log}}\left( {{\text{cosec x - cot x}}} \right)} {\text{ dx}}$
Let us taking the value as,
Let ${\text{log}}\left( {{\text{cosec x - cot x}}} \right) = {\text{t}}$
First we differentiate $\log $ then using formula ${\text{uv}}$we get this step,
$ \Rightarrow \dfrac{{\left( {{\text{ - cosec x cot x + cose}}{{\text{c}}^{\text{2}}}{\text{x}}} \right)}}{{\left( {{\text{cosec}}{\text{.cosec x - cotx}}} \right)}}{\text{ = }}\dfrac{{{\text{dt}}}}{{{\text{dx}}}}$
Dividing the terms, we get
$ \Rightarrow \left( {\dfrac{{{\text{cosec x - cot x}}}}{{{\text{cosec x - cot x}}}}} \right){{ \times cosec x dx = dt}}$
Dividing the terms, we get
$ \Rightarrow {\text{cosec x dx = dt}}$
Now, ${\text{I = }}\int {{\text{cosec x }}{\text{. log}}\left( {{\text{cosec x - cot x}}} \right)} {\text{ dx}}$
Here we are substitute the values ${\text{t}}$ and ${\text{dt}}$
$ \Rightarrow \int {{\text{t}}{\text{.dt}}} $
Now, the integration values become,
$ \Rightarrow \left[ {\dfrac{{{{\text{t}}^{\text{2}}}}}{{\text{2}}}} \right] + {\text{c}}$
Here we are substituting the values of ${\text{t}}$ and squaring the values,
We have, ${\text{I = }}\dfrac{{{{\left[ {{\text{log}}\left( {{\text{cosec x - cot x}}} \right)} \right]}^{\text{2}}}}}{{\text{2}}}{\text{ + c}}$

Hence the integral of \[\int {{\text{cosec x log}}\left( {{\text{cosec x - cot x}}} \right){\text{dx}}} \] ${\text{ = }}\dfrac{{{{\left[ {{\text{log}}\left( {{\text{cosec x - cot x}}} \right)} \right]}^{\text{2}}}}}{{\text{2}}}{\text{ + c}}$

Note: We use definite integrals when the upper and lower limits of that function are given. We use indefinite integrals when no limits are given to a particular function. Integration is the inverse of differentiation.