If _0^14 ^-1(1-2x+4x^2)dx=a, ^-1(2)-b, _e(5), where a,b N, then (2a+b) is equal to ____ .

Question

If $\int _{0}^{1}4{\cot }^{-1}(1-2x+4{x}^{2})dx={a\, 	an }^{-1}(2)-{b\, \log }_{e}(5)$, where $a,b\in N$, then $(2a+b)$ is equal to $____$ .

Answer: 9

Accepted Answer

We are given

${\int }_{1}^{0}4{\cot }^{-1}(1-2x+4{x}^{2})\, dx=a{	an }^{-1}(2)-b\ln (5),\, a,b\in N$

Simplify the integrand: $1-2x+4{x}^{2}=4{x}^{2}-2x+1>0\, \forallx\in [0,1]$

For positive arguments,

${\cot }^{-1}(u)={	an }^{-1}\, (\frac{1}{u})$

so

${\int }_{1}^{0}4{\cot }^{-1}(4{x}^{2}-2x+1)\, dx=4{\int }_{1}^{0}{	an }^{-1}\, (\frac{1}{4{x}^{2}-2x+1})dx$

Using the identity ${	an }^{-1}(1/u)=\frac{\pi }{2}-{	an }^{-1}(u)$ (for $u>0$):

$=4[\frac{\pi }{2}-{\int }_{1}^{0}{	an }^{-1}(4{x}^{2}-2x+1)\, dx]$

Evaluate the remaining integral: Using integration by parts and standard integrals of rational functions, one obtains:

${\int }_{1}^{0}{	an }^{-1}(4{x}^{2}-2x+1)\, dx={	an }^{-1}(3)-\frac{1}{4}\ln 5$

Substitute back:

$\begin{gathered} I =4[\frac{\pi }{2}-{	an }^{-1}(3)+\frac{1}{4}\ln 5] \ =4{	an }^{-1}\, (\frac{1}{3})+\ln 5 \end{gathered}$

Using the identity:

${	an }^{-1}(2)=2{	an }^{-1}\, (\frac{1}{3})$

we get:

$I=4{	an }^{-1}(2)-\ln 5$

Identify $a$ and $b$: Comparing with

$a{	an }^{-1}(2)-b\ln (5)$

we have:

$a=4,\, b=1$

Final Answer
$2a+b=2(4)+1=9$

Solutions · 1