int {frac{{1 + {{cos }^2}x}}{{{{sin }^2}x}}} | vedclass

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Question

(C) We are given the integral $I = \int {\frac{{1 + {{\cos }^2}x}}{{{{\sin }^2}x}}} \,dx$. Splitting the fraction,we get $I = \int {(\frac{1}{{{{\sin

Vedclass · Accepted Answer

$ - 2\cot x - x + c$

Vedclass · Answer

(C) We are given the integral $I = \int {\frac{{1 + {{\cos }^2}x}}{{{{\sin }^2}x}}} \,dx$. Splitting the fraction,we get $I = \int {(\frac{1}{{{{\sin }^2}x}} + \frac{{{{\cos }^2}x}}{{{{\sin }^2}x}})} \,dx$. Using trigonometric identities,$\frac{1}{{{{\sin }^2}x}} = \csc^2 x$ and $\frac{{{{\cos }^2}x}}{{{{\sin }^2}x}} = \cot^2 x$. So,$I = \int {(\csc^2 x + \cot^2 x)} \,dx$. Since $\cot^2 x = \csc^2 x - 1$,we substitute this into the integral: $I = \int {(\csc^2 x + \csc^2 x - 1)} \,dx = \int {(2\csc^2 x - 1)} \,dx$. Integrating term by term,we get $I = 2 \int \csc^2 x \,dx - \int 1 \,dx$. Since $\int \csc^2 x \,dx = -\cot x$,the result is $I = -2\cot x - x + c$.

$\int {\frac{{1 + {{\cos }^2}x}}{{{{\sin }^2}x}}} \,dx = $

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