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Evaluate the following Integral:
$\int {\left( {{\text{cosec }}x} \right)} \log \left( {{\text{cosec }}x - \cot x} \right)dx$

Last updated date: 16th Jul 2024
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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

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
$\dfrac{1}{{{\text{cosec }}x - \cot x}}\left( {\dfrac{d}{{dx}}\left( {{\text{cosec }}x - \cot x}
\right)} \right) = \dfrac{{dt}}{{dx}}$
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.