What is the antiderivative of $y=arc cotx$ ?

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Answer 1 · 2016-08-18T20:59:42

$int"arccot"(x)dx=x"arccot"(x)+1/2ln(x^2+1)+C$ .

Let $I=intydx=int"arccot"(x)dx=int("arccot"(x))(1)dx$

Now, we use the Rule of Integration by Parts, which states,

$intuvdx=uintvdx-int((du)/dx*intvdx)dx$

We take $u="arccot"(x)rArr (du)/dx=-1/(x^2+1)$ , and,

$v=1 rArr intvdx=x$ .

Hence, $I=x"arccot"(x)+int x/(x^2+1)dx$

$=x"arccot"(x)+1/2int(2x)/(x^2+1)dx$

$=x"arccot"(x)+1/2int(d(x^2+1))/(x^2+1)$

$:. I=x"arccot"(x)+1/2ln(x^2+1)+C, .......[as, intdt/t=ln|t|]$ .

What is the antiderivative of y=arc cotxy=arc cotx?