Answer
Verified
481.5k+ views
Hint – In this question Substitute $\log \left( {{\text{cosec }}x - \cot x} \right)$ to
any other variable then differentiate this function w.r.t. x so, use this method to reach the
answer.
Given integration is
$I = \int {\left( {{\text{cosec }}x} \right)} \log \left( {{\text{cosec }}x - \cot x}
\right)dx$…………….. (1)
Now, substitute $\log \left( {{\text{cosec }}x - \cot x} \right) = t$
Differentiate above equation w.r.t. x
$\dfrac{d}{{dx}}\log \left( {{\text{cosec }}x - \cot x} \right) = \dfrac{d}{{dx}}t$
As we know that the differentiation of $\log \left( {a + bx} \right) = \dfrac{1}{{a + bx}}\left(
{\dfrac{d}{{dx}}\left( {a + bx} \right)} \right)$ use this property in above differentiation we
have
$\dfrac{1}{{{\text{cosec }}x - \cot x}}\left( {\dfrac{d}{{dx}}\left( {{\text{cosec }}x - \cot x}
\right)} \right) = \dfrac{{dt}}{{dx}}$
Now we know differentiation of $\dfrac{d}{{dx}}{\text{cosec }}x = - {\text{cosec }}x\cot
x,{\text{ }}\dfrac{d}{{dx}}\cot x = - {\text{cose}}{{\text{c}}^2}x$ so, substitute these values in
above equation we have,
$\dfrac{1}{{{\text{cosec }}x - \cot x}}\left( { - {\text{cosec }}x\cot x - \left( { -
{\text{cose}}{{\text{c}}^2}x} \right)} \right)\dfrac{{dx}}{{dx}} = \dfrac{{dt}}{{dx}}$
Now take $\left( {{\text{cosec }}x} \right)$ common from numerator we have
$
\dfrac{{{\text{cosec }}x}}{{{\text{cosec }}x - \cot x}}\left( { - \cot x + {\text{cosec}}x} \right)dx
= dt \\
\Rightarrow {\text{cosec }}x{\text{ }}dx = dt \\
$
Now substitute this value in equation (1) we have.
$
I = \int {\left( {{\text{cosec }}x} \right)} \log \left( {{\text{cosec }}x - \cot x} \right)dx \\
I = \int {t{\text{ }}dt} \\
$
Now integrate the above equation we have
$I = \dfrac{{{t^2}}}{2} + c$ , where c is some arbitrary integration constant.
Now re-substitute the value of $t$ we have
$I = \dfrac{{{{\left[ {\log \left( {{\text{cosec }}x - \cot x} \right)} \right]}^2}}}{2} + c$
So, this is the required integration of the given integral.
Note – In such types of questions the key concept we have to remember is to always
substitute some part of integration to any other variable as above then differentiate this
equation w.r.t. given variable and re-substitute this value in the given integral then integrate
using basic property of integration, we will get the required answer.
any other variable then differentiate this function w.r.t. x so, use this method to reach the
answer.
Given integration is
$I = \int {\left( {{\text{cosec }}x} \right)} \log \left( {{\text{cosec }}x - \cot x}
\right)dx$…………….. (1)
Now, substitute $\log \left( {{\text{cosec }}x - \cot x} \right) = t$
Differentiate above equation w.r.t. x
$\dfrac{d}{{dx}}\log \left( {{\text{cosec }}x - \cot x} \right) = \dfrac{d}{{dx}}t$
As we know that the differentiation of $\log \left( {a + bx} \right) = \dfrac{1}{{a + bx}}\left(
{\dfrac{d}{{dx}}\left( {a + bx} \right)} \right)$ use this property in above differentiation we
have
$\dfrac{1}{{{\text{cosec }}x - \cot x}}\left( {\dfrac{d}{{dx}}\left( {{\text{cosec }}x - \cot x}
\right)} \right) = \dfrac{{dt}}{{dx}}$
Now we know differentiation of $\dfrac{d}{{dx}}{\text{cosec }}x = - {\text{cosec }}x\cot
x,{\text{ }}\dfrac{d}{{dx}}\cot x = - {\text{cose}}{{\text{c}}^2}x$ so, substitute these values in
above equation we have,
$\dfrac{1}{{{\text{cosec }}x - \cot x}}\left( { - {\text{cosec }}x\cot x - \left( { -
{\text{cose}}{{\text{c}}^2}x} \right)} \right)\dfrac{{dx}}{{dx}} = \dfrac{{dt}}{{dx}}$
Now take $\left( {{\text{cosec }}x} \right)$ common from numerator we have
$
\dfrac{{{\text{cosec }}x}}{{{\text{cosec }}x - \cot x}}\left( { - \cot x + {\text{cosec}}x} \right)dx
= dt \\
\Rightarrow {\text{cosec }}x{\text{ }}dx = dt \\
$
Now substitute this value in equation (1) we have.
$
I = \int {\left( {{\text{cosec }}x} \right)} \log \left( {{\text{cosec }}x - \cot x} \right)dx \\
I = \int {t{\text{ }}dt} \\
$
Now integrate the above equation we have
$I = \dfrac{{{t^2}}}{2} + c$ , where c is some arbitrary integration constant.
Now re-substitute the value of $t$ we have
$I = \dfrac{{{{\left[ {\log \left( {{\text{cosec }}x - \cot x} \right)} \right]}^2}}}{2} + c$
So, this is the required integration of the given integral.
Note – In such types of questions the key concept we have to remember is to always
substitute some part of integration to any other variable as above then differentiate this
equation w.r.t. given variable and re-substitute this value in the given integral then integrate
using basic property of integration, we will get the required answer.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Change the following sentences into negative and interrogative class 10 english CBSE
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE