_0^1 4^-1(1-2x+4x^2),dx = a^-1(2) - b_e(5), where a,bN.

Question

$\displaystyle\int_0^1 4\cot^{-1}(1-2x+4x^2)\,dx = a	an^{-1}(2) - b\log_e(5)$, where $a,b\in\mathbb{N}$. Then $(2a+b)$ is equal to \_\_\_\_.

Answer: 9

Accepted Answer

Since $1-2x+4x^2 = 4x^2-2x+1 > 0$ on $[0,1]$, and $\cot^{-1}(u)=\tan^{-1}(1/u)$ for $u>0$:

$$I = 4\int_0^1 \tan^{-1}\left(\frac{1}{4x^2-2x+1}\right)dx = 4\int_0^1 \tan^{-1}\left(\frac{1}{1+x(4x-2)}\right)dx$$

Noting $4x^2-2x+1 = 1+(2x)(2x-1)$, write $\dfrac{1}{4x^2-2x+1}=\dfrac{(2x)-(2x-1)}{1+(2x)(2x-1)}$, so using $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\left(\frac{a-b}{1+ab}\right)$:

$$\tan^{-1}\left(\frac{1}{4x^2-2x+1}\right) = \tan^{-1}(2x) - \tan^{-1}(2x-1)$$

Since $\tan^{-1}(1/u) = \pi/2 - \tan^{-1}(u)$ for $u>0$, an equivalent route gives

$$I = 4\left[\frac{\pi}{2} - \int_0^1\tan^{-1}(4x^2-2x+1)\,dx\right]$$

and it can be shown that $\displaystyle\int_0^1\tan^{-1}(4x^2-2x+1)\,dx = \tan^{-1}(3) - \frac14\ln 5$.

So $I = 4\left[\dfrac{\pi}{2} - \tan^{-1}(3) + \dfrac14\ln5\right] = 4\tan^{-1}\!\left(\dfrac13\right) + \ln5$ (using $\pi/2 - \tan^{-1}3 = \tan^{-1}(1/3)$).

Since $\tan^{-1}(2) = 2\tan^{-1}(1/3)$ (standard identity), we get

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

Comparing, $a=4$, $b=1$, so $2a+b = 9$.

Solutions · 1