CBSE Board Examination 2026 · Section C · 3 marks

Q26: Evaluate $\int_0^1 x\tan^{-1}x\,dx$

Board 2026Ch 7 · Integrals

The question

Evaluate : $\displaystyle\int_0^1 x\tan^{-1}x\,dx$ ⁠.

Exam answer

Let $I = \displaystyle\int_0^1 x\tan^{-1}x\,dx$ ⁠.

Integrating by parts, with $\tan^{-1}x$ as the first function and $x$ as the second function,

Why?

The integrand is a product of two different kinds of function, so integration by parts is the method:

\begin{aligned} \int u\,v\,dx &= u\int v\,dx \\ &\quad - \int\left(\frac{du}{dx}\int v\,dx\right)dx \end{aligned}

Choose the first function $u$ by ILATE (Inverse trig, Log, Algebraic, Trig, Exponential). $\tan^{-1}x$ is inverse trig, so it comes first. That is also the only sensible choice: we can differentiate $\tan^{-1}x$ ⁠, but we cannot easily integrate it.

\begin{aligned} I &= \left[\tan^{-1}x\cdot\frac{x^2}{2}\right]_0^1 \\ &\quad - \int_0^1\frac{1}{1 + x^2}\cdot\frac{x^2}{2}\,dx \end{aligned}

Why?

Here the two pieces are:

\begin{aligned} u = \tan^{-1}x &\Rightarrow \frac{du}{dx} = \frac{1}{1 + x^2} \\ v = x &\Rightarrow \int v\,dx = \frac{x^2}{2} \end{aligned}

For a definite integral the first part is evaluated between the limits, in square brackets.

\begin{aligned} I &= \frac{\pi}{8} - \frac12\int_0^1\frac{x^2}{1 + x^2}\,dx \end{aligned}

Why?

At $x = 1$ ⁠: $\tan^{-1}1 = \frac{\pi}{4}$ ⁠, so the bracket is $\frac{\pi}{4}\cdot\frac12 = \frac{\pi}{8}$ ⁠.

\begin{aligned} \tan^{-1}1\cdot\frac{1^2}{2} &= \frac{\pi}{4}\cdot\frac{1}{2} \\ &= \frac{\pi}{4 \times 2} = \frac{\pi}{8} \end{aligned}

At $x = 0$ ⁠: $\tan^{-1}0 = 0$ ⁠, so the bracket is $0$ ⁠.

The constant $\frac12$ moves outside the integral.

\begin{aligned} I &= \frac{\pi}{8} - \frac12\int_0^1\frac{1 + x^2 - 1}{1 + x^2}\,dx \\ &= \frac{\pi}{8} - \frac12\int_0^1\left(1 - \frac{1}{1 + x^2}\right)dx \end{aligned}

Why?

$\frac{x^2}{1 + x^2}$ has no standard integral as it stands. Adding and subtracting $1$ in the numerator splits it into two pieces we know:

\begin{aligned} \frac{x^2}{1 + x^2} &= \frac{1 + x^2 - 1}{1 + x^2} \\ &= \frac{1 + x^2}{1 + x^2} - \frac{1}{1 + x^2} \\ &= 1 - \frac{1}{1 + x^2} \end{aligned}

\begin{aligned} I &= \frac{\pi}{8} - \frac12\Big[x - \tan^{-1}x\Big]_0^1 \end{aligned}

Why?

Standard integrals: $\int 1\,dx = x$ and $\int\frac{1}{1 + x^2}\,dx = \tan^{-1}x$ ⁠.

\begin{aligned} I &= \frac{\pi}{8} - \frac12\left(1 - \frac{\pi}{4}\right) \\ &= \frac{\pi}{8} - \frac12 + \frac{\pi}{8} \end{aligned}

Why?

Upper limit minus lower limit: $\left(1 - \frac{\pi}{4}\right) - (0 - 0)$ ⁠.

\begin{aligned} &\Big[x - \tan^{-1}x\Big]_0^1 \\ &= (1 - \tan^{-1}1) - (0 - \tan^{-1}0) \\ &= \left(1 - \frac{\pi}{4}\right) - (0 - 0) \\ &= 1 - \frac{\pi}{4} \end{aligned}

Watch the sign: $-\frac12\cdot\left(-\frac{\pi}{4}\right) = +\frac{\pi}{8}$ ⁠.

\begin{aligned} -\frac12\cdot 1 &= -\frac12 \\ -\frac12\cdot\left(-\frac{\pi}{4}\right) &= +\frac{\pi}{8} \end{aligned}

Last, add the two $\pi$ terms:

\frac{\pi}{8} + \frac{\pi}{8} = \frac{2\pi}{8} = \frac{\pi}{4}

$\therefore\ I = \dfrac{\pi}{4} - \dfrac12$ ⁠.

Where this comes from

Chapter 7, Integrals: Definite integral by parts.

Question type: Integration by parts

How to spot it: click to see

Type 8. Integration by parts: a product of two different kinds

Spot it: a product such as $x\sin x$ ⁠, $xe^x$ or $x\log x$ ⁠. A lone $\log x$ or $\tan^{-1}x$ works too: take the other function as 1.

Here: In $x\tan^{-1}x$ the inverse trig part comes first in ILATE, so take $u = \tan^{-1}x$ and $v = x$ ⁠. The new integral $\int\frac{x^2}{1 + x^2}\,dx$ splits as $\int\left(1 - \frac{1}{1 + x^2}\right)dx$ ⁠.

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