# Evaluate the following Integral:

$\int {\left( {{\text{cosec }}x} \right)} \log \left( {{\text{cosec }}x - \cot x} \right)dx$

Answer

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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.

Last updated date: 24th Sep 2023

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