the value of ∫cot2xdx is________________
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\(\sqrt{sin\,2x}\)\(+c\)
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
The first function ‘u’ is used in the following order (ILATE):
The rule as a diagram: