Question

intigral of cot invars x​

Answer 1

(integral) cot x = ln|sin x| + C.

Answer 2

$\boxed{\underline{\textsf{Question :}}}$

$\textsf{Find integration of}\: \int cot^{-1}x\:dx$

$\boxed{\underline{\textsf{Method 1 :}}}$

$\textsf{Let,}\:cot^{-1}x=z$

$\implies \textsf{x = cotz}$

$\textsf{Then,}\:dx=-cosec^{2}z\:dz$

$\textsf{Also,}\:sinz=\frac{1}{\sqrt{x^{2}+1}}$

$\textsf{Now,}\: \int cot^{-1}x\:dx$

$=\int (-z\:cosec^{2}z)dz$

$=\int z\int (-cosec^{2}z)dz$

$-\int \{\frac{d}{dz}(z)*\int (-cosec^{2}z)dz\}dz$

$=z\:cotz-\int cotz\:dz$

$=zcotz-log(sinz)+C$

$\textsf{where C is integral constant.}$

$=x\:cot^{-1}x-log\bigg(\frac{1}{\sqrt{x^{2}+1}}\bigg)+C$

$=x\:cot^{-1}x-log(x^{2}+1)^{-\frac{1}{2}}+C$

$=x\:cot^{-1}x+\frac{1}{2}log(x^{2}+1)+C$

$\textsf{which is the required integral.}$

$\boxed{\underline{\textsf{Method 2 :}}}$

$\textsf{Now,}\: \int cot^{-1}x\:dx$

$=cot^{-1}x \int dx - \int \{\frac{d}{d}(cot^{-1}x)\int dx\}dx$

$=x\:cot^{-1}x - \int \frac{x\:dx}{-(1+x^{2})}+C$

$\textsf{where C is integral constant.}$

$=x\:cot^{-1}x+ \frac{1}{2} \int \frac{2x\:dx}{1+x^{2}}$

$=x\:cot^{-1}x+\frac{1}{2} log(1+x^{2})+C$

$\textsf{which is the required integral.}$

$\boxed{\underline{\textsf{Finding}}\: \int \frac{2x\:dx}{1+x^{2}}}$

$\textsf{Let,}\:1+x^{2}=z$

$\textsf{Then,}\:2x\:dx=dz$

$\textsf{So,}\:\int \frac{2x\:dx}{1+x^{2}}$

$=\int \frac{dz}{z}$

$=logz+C$

$\textsf{where C is integral constant.}$

$=log(1+x^{2})+C$

$\textsf{which is the required integral.}$