Board 2023 Class 12 Maths Q26 Solution: ∫₀^(π/2) [log(sin x) − log(2 cos x)] dx

The question

Evaluate :

\int_0^{\frac{\pi}{2}}[\log(\sin x) - \log(2\cos x)]\,dx.

Exam answer

Let I be the given integral. Then

\begin{aligned} I &= \int_0^{\pi/2}[\log\sin x \\ &\qquad - \log(2\cos x)]\,dx \\ &= \int_0^{\pi/2}\log\left(\frac{\tan x}{2}\right)dx \quad\ldots(\text{i}) \end{aligned}

Why?

Combine the two logs with

\log a - \log b = \log\frac{a}{b}

Then

\frac{\sin x}{2\cos x} = \frac12\cdot\frac{\sin x}{\cos x} = \frac{\tan x}{2}

One log is easier to handle than two. Label this form (i), because it will be used again.

Using $\displaystyle\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx$ ⁠,

\begin{aligned} I &= \int_0^{\pi/2}\log\left(\frac{\tan\left(\frac{\pi}{2} - x\right)}{2}\right)dx \\ &= \int_0^{\pi/2}\log\left(\frac{\cot x}{2}\right)dx \quad\ldots(\text{ii}) \end{aligned}

Why?

Here $a = \frac{\pi}{2}$ ⁠, so every $x$ is replaced by $\frac{\pi}{2} - x$ ⁠, and

\tan\left(\frac{\pi}{2} - x\right) = \cot x

This property is the right tool here. It turns $\tan x$ into $\cot x$ ⁠, and $\tan x\cot x = 1$ ⁠, so adding (i) and (ii) will remove $x$ from the log.

Adding (i) and (ii),

\begin{aligned} 2I &= \int_0^{\pi/2}\log\left(\frac{\tan x}{2}\times\frac{\cot x}{2}\right)dx \\ &= \int_0^{\pi/2}\log\left(\frac14\right)dx \\ &= \log\left(\frac14\right)\big[x\big]_0^{\pi/2} = \frac{\pi}{2}\log\frac14 \end{aligned}

Why?

Add the logs with $\log a + \log b = \log(ab)$ ⁠. The product inside is a constant:

\begin{aligned} \frac{\tan x}{2}\times\frac{\cot x}{2} &= \frac{\tan x\cot x}{4} \\ &= \frac{1}{4} \end{aligned}

because $\tan x\cot x = 1$ ⁠. A constant comes outside the integral, and $\big[x\big]_0^{\pi/2}$ is $\frac{\pi}{2} - 0$ ⁠.

\begin{aligned} &\log\left(\frac14\right)\left(\frac{\pi}{2} - 0\right) \\ &= \log\left(\frac14\right) \times \frac{\pi}{2} \\ &= \frac{\pi}{2}\log\frac14 \end{aligned}

Dividing by 2,

Why?

\begin{aligned} I &= \frac12\cdot\frac{\pi}{2}\log\frac14 \\ &= \frac{1 \times \pi}{2 \times 2}\log\frac14 \\ &= \frac{\pi}{4}\log\frac14 \end{aligned}

The second form uses

\begin{aligned} \frac14 &= \frac{1}{2^2} = 2^{-2} \\ \log\frac14 &= \log 2^{-2} = -2\log 2 \end{aligned}

(by the rule $\log a^n = n\log a$ ⁠), so that $\frac{\pi}{4}\times(-2\log 2) = -\frac{\pi}{2}\log 2$ ⁠. Either form is accepted.

\begin{aligned} \frac{\pi}{4}\times(-2\log 2) &= -\frac{2\pi}{4}\log 2 \\ &= -\frac{\pi}{2}\log 2 \end{aligned}

$\therefore\ I = \dfrac{\pi}{4}\log\dfrac14$ or $-\dfrac{\pi}{2}\log 2$ ⁠.

Where this comes from

Chapter 7, Integrals: Definite integrals: the f(a − x) property.

Question type: Limits from 0 to $a$

How to spot it: click to see

Type 12. Limits from 0 to $a$ ⁠: the $f(a - x)$ trick

Spot it: replacing $x$ by $a - x$ swaps the parts of the question, like sin and cos when $a$ is $\frac{\pi}{2}$ ⁠.

Here: After joining the logs into $\log\frac{\tan x}{2}$ ⁠, the limits are $0$ and $\frac{\pi}{2}$ ⁠, and replacing $x$ by $\frac{\pi}{2} - x$ turns $\tan x$ into $\cot x$ ⁠. Adding the two forms gives $\log\frac14$ ⁠, a constant.

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Q26: ∫0π/2[log⁡(sin⁡x)−log⁡(2cos⁡x)] dx\int_0^{\pi/2}[\log(\sin x) - \log(2\cos x)]\,dx∫0π/2​[log(sinx)−log(2cosx)]dx

Q26: $\int_0^{\pi/2}[\log(\sin x) - \log(2\cos x)]\,dx$