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CBSE Board Examination 2022 (Term 2) · Section C · 4 marks

Q13: ∫0πx1+sin⁡x dx\int_0^{\pi} \frac{x}{1 + \sin x}\,dx

Board 2022 T2Ch 7 · Integrals
The question

Evaluate : ∫0πx1+sin⁡x dx\displaystyle\int_0^{\pi} \frac{x}{1 + \sin x}\,dx⁠.

Exam answer
Let I=∫0πx1+sin⁡x dx…(i)\text{Let } I = \int_0^{\pi} \frac{x}{1 + \sin x}\,dx \quad\ldots(\text{i})

Using ∫0af(x) dx=∫0af(a−x) dx\int_0^a f(x)\,dx = \int_0^a f(a - x)\,dx⁠,

I=∫0ππ−x1+sin⁡(π−x) dx=∫0ππ−x1+sin⁡x dx…(ii)\begin{aligned} I &= \int_0^{\pi} \frac{\pi - x}{1 + \sin(\pi - x)}\,dx \\ &= \int_0^{\pi} \frac{\pi - x}{1 + \sin x}\,dx \quad\ldots(\text{ii}) \end{aligned}
Why?

The property comes from putting x=a−tx = a - t⁠: then dx=−dtdx = -dt⁠, the limits 0,a0, a become a,0a, 0⁠, and the minus sign swaps them back. Here a=πa = \pi⁠, so xx is replaced by π−x\pi - x⁠.

It helps because sin⁡(π−x)=sin⁡x\sin(\pi - x) = \sin x⁠: the denominator does not change, but the numerator xx becomes π−x\pi - x⁠.

Adding (i) and (ii),

2I=π∫0πdx1+sin⁡x2I = \pi\int_0^{\pi} \frac{dx}{1 + \sin x}
Why?

Both integrals have the same denominator and the same limits, so the numerators add: x+(π−x)=πx + (\pi - x) = \pi⁠. The awkward factor xx is gone, and the constant π\pi comes outside.

I+I=∫0πx+(π−x)1+sin⁡x dx2I=∫0ππ1+sin⁡x dx=π∫0πdx1+sin⁡x\begin{aligned} I + I &= \int_0^{\pi} \frac{x + (\pi - x)}{1 + \sin x}\,dx \\ 2I &= \int_0^{\pi} \frac{\pi}{1 + \sin x}\,dx \\ &= \pi\int_0^{\pi} \frac{dx}{1 + \sin x} \end{aligned}

Then divide both sides by 2 to get II⁠.

∴ I=π2∫0πdx1+sin⁡x=π2∫0πdx1+cos⁡(π2−x)\begin{aligned} \therefore\ I &= \frac{\pi}{2}\int_0^{\pi} \frac{dx}{1 + \sin x} \\ &= \frac{\pi}{2}\int_0^{\pi} \frac{dx}{1 + \cos\left(\frac{\pi}{2} - x\right)} \end{aligned}
Why?

Since sin⁡x=cos⁡(π2−x)\sin x = \cos\left(\frac{\pi}{2} - x\right)⁠, the denominator can be written with a cosine. Then the half-angle identity below turns it into a single square.

Using 1+cos⁡θ=2cos⁡2θ21 + \cos\theta = 2\cos^2\frac{\theta}{2}⁠,

I=π2∫0πdx2cos⁡2(π4−x2)=π4∫0πsec⁡2(π4−x2)dx\begin{aligned} I &= \frac{\pi}{2}\int_0^{\pi} \frac{dx}{2\cos^2\left(\frac{\pi}{4} - \frac{x}{2}\right)} \\ &= \frac{\pi}{4}\int_0^{\pi} \sec^2\left(\frac{\pi}{4} - \frac{x}{2}\right)dx \end{aligned}
Why?

With θ=π2−x\theta = \frac{\pi}{2} - x⁠, half of it is π4−x2\frac{\pi}{4} - \frac{x}{2}⁠.

θ2=12(π2−x)=π4−x2\begin{aligned} \frac{\theta}{2} &= \frac12\left(\frac{\pi}{2} - x\right) \\ &= \frac{\pi}{4} - \frac{x}{2} \end{aligned}

So 1+cos⁡(π2−x)=2cos⁡2(π4−x2)1 + \cos\left(\frac{\pi}{2} - x\right) = 2\cos^2\left(\frac{\pi}{4} - \frac{x}{2}\right)⁠.

Then 1cos⁡2u=sec⁡2u\frac{1}{\cos^2u} = \sec^2u⁠, and the 2 in the denominator joins π2\frac{\pi}{2} to give π4\frac{\pi}{4}⁠.

π2⋅12cos⁡2(π4−x2)=π2×2⋅sec⁡2(π4−x2)=π4sec⁡2(π4−x2)\begin{aligned} &\frac{\pi}{2} \cdot \frac{1}{2\cos^2\left(\frac{\pi}{4} - \frac{x}{2}\right)} \\ &= \frac{\pi}{2 \times 2} \cdot \sec^2\left(\frac{\pi}{4} - \frac{x}{2}\right) \\ &= \frac{\pi}{4}\sec^2\left(\frac{\pi}{4} - \frac{x}{2}\right) \end{aligned}
I=π4[−2tan⁡(π4−x2)]0π\begin{aligned} I &= \frac{\pi}{4}\left[-2\tan\left(\frac{\pi}{4} - \frac{x}{2}\right)\right]_0^{\pi} \end{aligned}
Why?

For a linear expression inside,

∫sec⁡2(ax+b) dx=1atan⁡(ax+b)+C\begin{aligned} &\int \sec^2(ax + b)\,dx \\ &= \frac{1}{a}\tan(ax + b) + C \end{aligned}

Here the coefficient of xx is a=−12a = -\frac12⁠, so 1a=−2\frac{1}{a} = -2⁠. Watch out for this minus sign.

I=π4[−2tan⁡(−π4)+2tan⁡π4]=π4[2−(−2)]\begin{aligned} I &= \frac{\pi}{4}\left[-2\tan\left(-\frac{\pi}{4}\right) + 2\tan\frac{\pi}{4}\right] \\ &= \frac{\pi}{4}[2 - (-2)] \end{aligned}
Why?

At x=πx = \pi⁠: π4−π2=−π4\frac{\pi}{4} - \frac{\pi}{2} = -\frac{\pi}{4}⁠, and tan⁡(−π4)=−1\tan\left(-\frac{\pi}{4}\right) = -1⁠, so the bracket is 22⁠.

At x=0x = 0⁠: −2tan⁡π4=−2-2\tan\frac{\pi}{4} = -2⁠. Upper minus lower is 2−(−2)=42 - (-2) = 4⁠, and π4×4=π\frac{\pi}{4} \times 4 = \pi⁠.

x=π: −2tan⁡(π4−π2)=−2tan⁡(−π4)=−2(−1)=2x=0: −2tan⁡(π4−0)=−2(1)=−22−(−2)=2+2=4π4×4=π\begin{aligned} x = \pi&:\ -2\tan\left(\frac{\pi}{4} - \frac{\pi}{2}\right) \\ &= -2\tan\left(-\frac{\pi}{4}\right) \\ &= -2(-1) = 2 \\[4pt] x = 0&:\ -2\tan\left(\frac{\pi}{4} - 0\right) \\ &= -2(1) = -2 \\[4pt] &2 - (-2) = 2 + 2 = 4 \\ &\frac{\pi}{4} \times 4 = \pi \end{aligned}

∴ I=π\therefore\ I = \pi

Where this comes from

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

Question type: Limits from 0 to aa

How to spot it: click to see

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

Spot it: replacing xx by a−xa - x swaps the parts of the question, like sin and cos when aa is π2\frac{\pi}{2}⁠.

Here: The limits are 00 and π\pi⁠, and sin⁡(π−x)=sin⁡x\sin(\pi - x) = \sin x⁠: replacing xx by π−x\pi - x keeps the bottom but turns the top xx into π−x\pi - x⁠. Adding the two forms leaves just π\pi on top.

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