Mathematics - Definite Integration Question with Solution | TestHub

We start with the given integral:

$$I={\int }_0^1{\tan }^{-1}(1-x+x^2)dx$$

We use the property of definite integrals:

$${\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x)dx$$

Here, $a=1$.

$$I={\int }_0^1{\tan }^{-1}(1-(1-x)+(1-x{)}^2)dx$$

$$I={\int }_0^1{\tan }^{-1}(1-1+x+1-2x+x^2)dx$$

$$I={\int }_0^1{\tan }^{-1}(x^2-x+1)dx$$

This result is the same integral we started with, so this property alone doesn't solve it.

Step 2: Use trigonometric identities

We use the identity

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

for $u>0$. The expression $1-x+x^2$ is always positive for $x\in [0,1]$ (its minimum is $\frac{3}{4}$ at $x=\frac{1}{2}$).

The original integral can be rewritten as:

$$I={\int }_0^1\left(\frac{\pi }{2}-{\tan }^{-1}\left(\frac{1}{1-x+x^2}\right)\right)dx$$

$$I=\frac{\pi }{2}{\int }_0^{1}dx-{\int }_0^1{\tan }^{-1}\left(\frac{1}{1-x+x^2}\right)dx$$

$$I=\frac{\pi }{2}-{\int }_0^1{\tan }^{-1}\left(\frac{1}{1-x+x^2}\right)dx$$

Let $J={\int }_0^1{\tan }^{-1}\left(\frac{1}{1-x+x^2}\right)dx$.

We can use another trigonometric identity for the integrand in $J$:

$${\tan }^{-1}(a)-{\tan }^{-1}(b)={\tan }^{-1}\left(\frac{a-b}{1+ab}\right)$$

Notice that

$$\frac{1}{1-x+x^2}=\frac{x-(x-1)}{1+x(x-1)}$$

No, the identity gives

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

So,

$$J={\int }_0^1({\tan }^{-1}(x)-{\tan }^{-1}(x-1))dx$$

$$J={\left[x{\tan }^{-1}(x)-\frac{1}{2}\ln |1+x^2|-\left(x{\tan }^{-1}(x-1)-\frac{1}{2}\ln |1+(x-1{)}^2|\right)\right]}_0^1$$

This is getting complicated. Let's use the property that

$${\int }_0^1({\tan }^{-1}(x)-{\tan }^{-1}(x-1))dx={\int }_0^1{\tan }^{-1}(x)dx-{\int }_0^1{\tan }^{-1}(x-1)dx$$

Let $u=x-1$ in the second integral.

$du=dx$. When $x=0$, $u=-1$. When $x=1$, $u=0$.

$$J={\int }_0^1{\tan }^{-1}(x)dx-{\int }_{-1}^0{\tan }^{-1}(u)du$$

Using ${\tan }^{-1}(-u)=-{\tan }^{-1}(u)$, the second integral is

$${\int }_{-1}^0{\tan }^{-1}(u)du={\left[u{\tan }^{-1}(u)-\frac{1}{2}\ln |1+u^2|\right]}_{-1}^0$$

$$J=2{\int }_0^1{\tan }^{-1}(x)dx$$

We know

$$\int {\tan }^{-1}(x)dx=x{\tan }^{-1}(x)-\frac{1}{2}\ln |1+x^2|+C$$

$${\int }_0^1{\tan }^{-1}(x)dx={\left[x{\tan }^{-1}(x)-\frac{1}{2}\ln |1+x^2|\right]}_0^1$$

$$=(1\cdot {\tan }^{-1}(1)-\frac{1}{2}\ln (1+1))-(0\cdot {\tan }^{-1}(0)-\frac{1}{2}\ln (1+0))$$

$$=\left(\frac{\pi }{4}-\frac{1}{2}\ln (2)\right)-(0-0)=\frac{\pi }{4}-\frac{1}{2}\ln (2)$$

So,

$$J=2\left(\frac{\pi }{4}-\frac{1}{2}\ln (2)\right)=\frac{\pi }{2}-\ln (2)$$

$I=\frac{\pi }{2}-J$

$$I=\frac{\pi }{2}-\left(\frac{\pi }{2}-\ln (2)\right)$$

$$I=\ln (2)$$