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CBSE Sample Paper 2025-26 · Section D · 5 marks

Q33: Evaluate ∫01log⁡(1+x)1+x2 dx\int_0^1\frac{\log(1 + x)}{1 + x^2}\,dx⁠, or find ∫(3sin⁡θ−2)cos⁡θ5−cos⁡2θ−4sin⁡θ dθ\int\frac{(3\sin\theta - 2)\cos\theta}{5 - \cos^2\theta - 4\sin\theta}\,d\theta

Sample 2025-26Ch 7 · Integrals
The question

(a) Evaluate: ∫01log⁡(1+x)1+x2 dx\displaystyle\int_0^1\frac{\log(1 + x)}{1 + x^2}\,dx

OR

(b) Find ∫(3sin⁡θ−2)cos⁡θ5−cos⁡2θ−4sin⁡θ dθ\displaystyle\int\frac{(3\sin\theta - 2)\cos\theta}{5 - \cos^2\theta - 4\sin\theta}\,d\theta

Exam answer

(a)

Let I=∫01log⁡(1+x)1+x2 dxI = \displaystyle\int_0^1\frac{\log(1 + x)}{1 + x^2}\,dx⁠.

Let x=tan⁡θ⇒dx=sec⁡2θ dθx = \tan\theta \Rightarrow dx = \sec^2\theta\,d\theta⁠.

Why?

The factor 11+x2\frac{1}{1 + x^2} suggests x=tan⁡θx = \tan\theta⁠, because then

1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta

and this cancels with the sec⁡2θ\sec^2\theta from dxdx⁠. Also ddθtan⁡θ=sec⁡2θ\frac{d}{d\theta}\tan\theta = \sec^2\theta⁠.

When x=0x = 0⁠, θ=0\theta = 0⁠; when x=1x = 1⁠, θ=π4\theta = \frac{\pi}{4}⁠.

Why?

A definite integral must change its limits with the variable: tan⁡θ=0\tan\theta = 0 gives θ=0\theta = 0⁠, and tan⁡θ=1\tan\theta = 1 gives θ=π4\theta = \frac{\pi}{4}⁠.

After this we stay with θ\theta and never go back to xx⁠.

I=∫0π/4log⁡(1+tan⁡θ)1+tan⁡2θ⋅sec⁡2θ dθ=∫0π/4log⁡(1+tan⁡θ) dθ…(i)\begin{aligned} I &= \int_0^{\pi/4}\frac{\log(1 + \tan\theta)}{1 + \tan^2\theta}\cdot\sec^2\theta\,d\theta \\ &= \int_0^{\pi/4}\log(1 + \tan\theta)\,d\theta\quad\ldots(\text{i}) \end{aligned}
Why?

The denominator 1+tan⁡2θ1 + \tan^2\theta equals sec⁡2θ\sec^2\theta⁠, so it cancels the sec⁡2θ\sec^2\theta from dxdx⁠.

We label this form (i), because the same integral will appear again.

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

I=∫0π/4log⁡[1+tan⁡(π4−θ)]dθ=∫0π/4log⁡[1+1−tan⁡θ1+tan⁡θ]dθ=∫0π/4log⁡[21+tan⁡θ]dθ\begin{aligned} I &= \int_0^{\pi/4}\log\left[1 + \tan\left(\frac{\pi}{4} - \theta\right)\right]d\theta \\ &= \int_0^{\pi/4}\log\left[1 + \frac{1 - \tan\theta}{1 + \tan\theta}\right]d\theta \\ &= \int_0^{\pi/4}\log\left[\frac{2}{1 + \tan\theta}\right]d\theta \end{aligned}
Why?

There is no standard antiderivative of log⁡(1+tan⁡θ)\log(1 + \tan\theta)⁠. The property with a=π4a = \frac{\pi}{4} replaces θ\theta by π4−θ\frac{\pi}{4} - \theta and leaves the value unchanged.

Tangent of a difference, with tan⁡π4=1\tan\frac{\pi}{4} = 1⁠:

tan⁡(π4−θ)=1−tan⁡θ1+tan⁡θ\tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan\theta}{1 + \tan\theta} tan⁡(π4−θ)=tan⁡π4−tan⁡θ1+tan⁡π4tan⁡θ=1−tan⁡θ1+(1)tan⁡θ\begin{aligned} \tan\left(\frac{\pi}{4} - \theta\right) &= \frac{\tan\frac{\pi}{4} - \tan\theta}{1 + \tan\frac{\pi}{4}\tan\theta} \\ &= \frac{1 - \tan\theta}{1 + (1)\tan\theta} \end{aligned}

Then add with a common denominator: 1+tan⁡θ+1−tan⁡θ=21 + \tan\theta + 1 - \tan\theta = 2 on top.

1+1−tan⁡θ1+tan⁡θ=1+tan⁡θ1+tan⁡θ+1−tan⁡θ1+tan⁡θ=1+tan⁡θ+1−tan⁡θ1+tan⁡θ=21+tan⁡θ\begin{aligned} &1 + \frac{1 - \tan\theta}{1 + \tan\theta} \\ &= \frac{1 + \tan\theta}{1 + \tan\theta} + \frac{1 - \tan\theta}{1 + \tan\theta} \\ &= \frac{1 + \tan\theta + 1 - \tan\theta}{1 + \tan\theta} \\ &= \frac{2}{1 + \tan\theta} \end{aligned}
I=∫0π/4log⁡2 dθ−∫0π/4log⁡(1+tan⁡θ) dθ=log⁡2[θ]0π/4−I(using (i))\begin{aligned} I &= \int_0^{\pi/4}\log 2\,d\theta \\ &\quad - \int_0^{\pi/4}\log(1 + \tan\theta)\,d\theta \\ &= \log 2\Big[\theta\Big]_0^{\pi/4} - I\quad(\text{using (i)}) \end{aligned}
Why?

The log of a quotient is a difference:

log⁡mn=log⁡m−log⁡n\log\frac{m}{n} = \log m - \log n

The second integral is exactly the one labelled (i), that is II itself. This is the point of the method: II now appears on both sides.

log⁡2\log 2 is a constant, so its integral is log⁡2\log 2 times θ\theta⁠.

⇒2I=π4log⁡2\Rightarrow 2I = \frac{\pi}{4}\log 2
Why?

[θ]0π/4=π4−0=π4\Big[\theta\Big]_0^{\pi/4} = \frac{\pi}{4} - 0 = \frac{\pi}{4}⁠. Then add II to both sides, and divide by 2 at the end.

I=log⁡2⋅π4−II+I=π4log⁡22I=π4log⁡2I=π4×2log⁡2=π8log⁡2\begin{aligned} I &= \log 2 \cdot \frac{\pi}{4} - I \\ I + I &= \frac{\pi}{4}\log 2 \\ 2I &= \frac{\pi}{4}\log 2 \\ I &= \frac{\pi}{4 \times 2}\log 2 = \frac{\pi}{8}\log 2 \end{aligned}

∴ I=π8log⁡2\therefore\ I = \frac{\pi}{8}\log 2

(b)

I=∫(3sin⁡θ−2)cos⁡θ5−cos⁡2θ−4sin⁡θ dθ=∫(3sin⁡θ−2)cos⁡θ5−(1−sin⁡2θ)−4sin⁡θ dθ\begin{aligned} I &= \int\frac{(3\sin\theta - 2)\cos\theta}{5 - \cos^2\theta - 4\sin\theta}\,d\theta \\ &= \int\frac{(3\sin\theta - 2)\cos\theta}{5 - (1 - \sin^2\theta) - 4\sin\theta}\,d\theta \end{aligned}
Why?

Everything except cos⁡θ\cos\theta can be written in terms of sin⁡θ\sin\theta⁠, using

cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1 - \sin^2\theta

The lone cos⁡θ\cos\theta is kept, because cos⁡θ dθ\cos\theta\,d\theta is the differential of sin⁡θ\sin\theta⁠.

Let sin⁡θ=t⇒cos⁡θ dθ=dt\sin\theta = t \Rightarrow \cos\theta\,d\theta = dt⁠.

I=∫3t−25−(1−t2)−4t dt=∫3t−2t2−4t+4 dt=∫3t−2(t−2)2 dt\begin{aligned} I &= \int\frac{3t - 2}{5 - (1 - t^2) - 4t}\,dt \\ &= \int\frac{3t - 2}{t^2 - 4t + 4}\,dt \\ &= \int\frac{3t - 2}{(t - 2)^2}\,dt \end{aligned}
Why?

Since ddθsin⁡θ=cos⁡θ\frac{d}{d\theta}\sin\theta = \cos\theta⁠, the factor cos⁡θ dθ\cos\theta\,d\theta becomes dtdt⁠.

In the denominator, 5−1+t2−4t=t2−4t+45 - 1 + t^2 - 4t = t^2 - 4t + 4⁠. This is a perfect square:

t2−4t+4=(t−2)2t^2 - 4t + 4 = (t - 2)^2

Check by expanding:

(t−2)2=t2−2(t)(2)+22=t2−4t+4\begin{aligned} (t - 2)^2 &= t^2 - 2(t)(2) + 2^2 \\ &= t^2 - 4t + 4 \end{aligned}

Let 3t−2(t−2)2=At−2+B(t−2)2\dfrac{3t - 2}{(t - 2)^2} = \dfrac{A}{t - 2} + \dfrac{B}{(t - 2)^2}⁠.

Why?

The denominator is a repeated linear factor, so partial fractions need one term for each power of t−2t - 2⁠: At−2\frac{A}{t - 2} and B(t−2)2\frac{B}{(t - 2)^2}⁠.

Watch out: writing only B(t−2)2\frac{B}{(t - 2)^2} cannot work, because 3t−23t - 2 is not a constant.

⇒3t−2=A(t−2)+B\Rightarrow 3t - 2 = A(t - 2) + B

Comparing the coefficients of tt and the constant terms on both sides,

A=3,−2A+B=−2⇒B=4A = 3,\quad -2A + B = -2 \Rightarrow B = 4
Why?

Multiplying both sides by (t−2)2(t - 2)^2 clears the fractions. The right side is At+(−2A+B)At + (-2A + B)⁠.

Coefficient of tt⁠: 3=A3 = A⁠. Constant terms: −2=−2A+B=−6+B-2 = -2A + B = -6 + B⁠, so B=4B = 4⁠.

A(t−2)+B=At−2A+B−2=−2(3)+B−2=−6+BB=−2+6=4\begin{aligned} A(t - 2) + B &= At - 2A + B \\[4pt] -2 &= -2(3) + B \\ -2 &= -6 + B \\ B &= -2 + 6 = 4 \end{aligned}

Quick check: put t=2t = 2⁠. Then 3(2)−2=B3(2) - 2 = B⁠, so B=4B = 4 again.

I=∫3t−2 dt+∫4(t−2)2 dt=3log⁡∣t−2∣−4t−2+C\begin{aligned} I &= \int\frac{3}{t - 2}\,dt + \int\frac{4}{(t - 2)^2}\,dt \\ &= 3\log|t - 2| - \frac{4}{t - 2} + C \end{aligned}
Why?

Standard integrals:

∫dtt−2=log⁡∣t−2∣∫(t−2)−2 dt=−(t−2)−1\begin{aligned} \int\frac{dt}{t - 2} &= \log|t - 2| \\ \int(t - 2)^{-2}\,dt &= -(t - 2)^{-1} \end{aligned}

The second one is the power rule with n=−2n = -2⁠:

∫(t−2)−2 dt=(t−2)−2+1−2+1=(t−2)−1−1=−1t−2\begin{aligned} \int(t - 2)^{-2}\,dt &= \frac{(t - 2)^{-2 + 1}}{-2 + 1} \\ &= \frac{(t - 2)^{-1}}{-1} = -\frac{1}{t - 2} \end{aligned}

Watch the minus sign in the second one: 4×(−1t−2)=−4t−24 \times \left(-\frac{1}{t - 2}\right) = -\frac{4}{t - 2}⁠.

Finally put t=sin⁡θt = \sin\theta back, so the answer is in θ\theta⁠.

∴ I=3log⁡∣sin⁡θ−2∣\therefore\ I = 3\log|\sin\theta - 2| − 4sin⁡θ−2+C-\ \dfrac{4}{\sin\theta - 2} + C

Where this comes from

Chapter 7, Integrals: Definite integral by property; partial fractions.

Question type: (a) Substitution + Limits from 0 to aa (combined); (b) Substitution + Partial fractions (combined)

How to spot it: click to see

(a) Type 2. Substitution: when the derivative is hiding in the question

Spot it: a function and its derivative both appear, like log⁡x\log x and 1x\frac{1}{x}⁠, or x2+1x^2 + 1 and xx⁠.

Here: First, 11+x2\frac{1}{1 + x^2} is the derivative of tan⁡−1x\tan^{-1}x⁠, so put x=tan⁡θx = \tan\theta⁠. Then dx1+x2\frac{dx}{1 + x^2} becomes dθd\theta⁠, and the limits 0 and 1 become 0 and π4\frac{\pi}{4}⁠.

(a) 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 integral left, ∫0π/4log⁡(1+tan⁡θ) dθ\int_0^{\pi/4}\log(1 + \tan\theta)\,d\theta⁠, has no card. Replace θ\theta by π4−θ\frac{\pi}{4} - \theta⁠: 1+tan⁡θ1 + \tan\theta becomes 21+tan⁡θ\frac{2}{1 + \tan\theta}⁠, II comes back with a minus sign, and 2I=π4log⁡22I = \frac{\pi}{4}\log 2⁠.

(b) Type 2. Substitution: when the derivative is hiding in the question

Spot it: a function and its derivative both appear, like log⁡x\log x and 1x\frac{1}{x}⁠, or x2+1x^2 + 1 and xx⁠.

Here: First, the lone cos⁡θ\cos\theta is the derivative of sin⁡θ\sin\theta⁠, and cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1 - \sin^2\theta turns the rest into sin⁡θ\sin\theta⁠. So put sin⁡θ=t\sin\theta = t⁠, and the bottom becomes (t−2)2(t - 2)^2⁠.

(b) Type 7. Partial fractions: break a fraction into simple pieces

Spot it: a fraction of two polynomials whose bottom splits into factors.

Here: The fraction left, 3t−2(t−2)2\frac{3t - 2}{(t - 2)^2}⁠, has a repeated factor at the bottom. Split it into 3t−2+4(t−2)2\frac{3}{t - 2} + \frac{4}{(t - 2)^2}⁠, which gives a log and −4t−2-\frac{4}{t - 2}⁠.

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