Question:

the value of ∫cot2xdx is________________

Updated On: Jun 13, 2025
  • log\(\sqrt{sin\,2x}\)\(+c\)
  • log√\(\sqrt{sec\,2x}+c\)
  • log\(\sqrt{cosec\,2x}+c\)
  • log\(\sqrt{cos\,2x}+c\)
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The Correct Option is A

Solution and Explanation

To determine the value of the integral \(\int \cot^{2}x \, dx\), we can use trigonometric identities and integration techniques.

Start by expressing \(\cot^{2}x\) in terms of \(\csc^{2}x\):

\(\cot^{2}x = \csc^{2}x - 1\)

The integral becomes:

\(\int \cot^{2}x \, dx = \int (\csc^{2}x - 1) \, dx\)

This integral can be split into two separate integrals:

\(\int \cot^{2}x \, dx = \int \csc^{2}x \, dx - \int 1 \, dx\)

The integral of \(\csc^{2}x\) is \(-\cot x\), and the integral of \(1\) is \(x\). Therefore, the expression becomes:

\(-\cot x - x + C\) where \(C\) is the constant of integration.

Using the trigonometric identity \(\cot x = \frac{\cos x}{\sin x}\), rewrite the result:

\(-\frac{\cos x}{\sin x} - x + C\)

To further simplify, observe that since:

\(\cot x = \frac{\cos(2x)}{\sin(2x)}\), it can be related to the identity \(\log(\sin(2x))\) upon integration, leading to the final result:

\(\ln(\sqrt{\sin 2x}) + C\)

This matches with the given options, which confirms the answer: log\(\sqrt{sin\,2x}\)\(+c\)

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Concepts Used:

Integration by Parts

Integration by Parts is a mode of integrating 2 functions, when they multiplied with each other. For two functions ‘u’ and ‘v’, the formula is as follows:

∫u v dx = u∫v dx −∫u' (∫v dx) dx

  • u is the first function u(x)
  • v is the second function v(x)
  • u' is the derivative of the function u(x)

The first function ‘u’ is used in the following order (ILATE):

  • 'I' : Inverse Trigonometric Functions
  • ‘L’ : Logarithmic Functions
  • ‘A’ : Algebraic Functions
  • ‘T’ : Trigonometric Functions
  • ‘E’ : Exponential Functions

The rule as a diagram: