Answer to Question #114630 in Calculus for ANJU JAYACHANDRAN

Question

Evaluvate integral 0 to 1 tan inverse xdx.

Answer 1

Here, the objective is evaluate the definite integral using integration by parts.

Let $u=\arctan(x),dv=dx$ , then

$du=\frac{1}{1+x^2}dx$

$v=x$

So, use the integration by parts as,

$\int_{0}^1(\arctan(x))dx=uv-\int(v)du$

$=[x\arctan(x)]_{0}^1-\int_{0}^1\frac{x}{1+x^2}dx$

$=\frac{\pi}{4}-\frac{1}{2}\int_{0}^1\frac{2x}{1+x^2}dx$

Here, $g'(x)=2x,$ if $g(x)=1+x^2$ , so using fundamental theorem of calculus, the integral is,

$=\frac{\pi}{4}-\frac{1}{2}[ln|1+x^2|]_{0}^1$

$=\frac{\pi}{4}-\frac{1}{2}[ln|1+1|-ln|1|]$

$=\frac{\pi}{4}-\frac{1}{2}[ln2-0|]$

$=\frac{\pi}{4}-\frac{1}{2}ln2$

Therefore, the value of the definite integral is $\int_{0}^1(\arctan(x))dx=\frac{\pi}{4}-\frac{1}{2}ln2$