UPSC 2014 Maths Optional Paper 1 Q1d — Step-by-Step Solution

10 marks · Section A

Indefinite integrals · Calculus · asked 7× in 13 yrs · Read the full method →

Question

Evaluate $\displaystyle\int_0^1\dfrac{\log_e(1+x)}{1+x^2}\,dx$ .

Technique

Trig substitution + the standard “symmetrise via $a-x$ ” trick that exploits $\int_0^a f(x)dx=\int_0^a f(a-x)dx$ .

Solution

Strategy. Classical Putnam-style integral. Substitute $x=\tan\theta$ to convert to trig form, then use the symmetry $\theta\leftrightarrow\pi/4-\theta$ to collapse the integral.

Step 1 — Trig substitution

$x=\tan\theta\Rightarrow dx=\sec^{2}\theta\,d\theta$ , $1+x^{2}=\sec^{2}\theta$ . Limits: $x=0\to\theta=0$ , $x=1\to\theta=\pi/4$ .

I=\int_0^{1}\dfrac{\log(1+x)}{1+x^{2}}\,dx=\int_0^{\pi/4}\dfrac{\log(1+\tan\theta)}{\sec^{2}\theta}\cdot\sec^{2}\theta\,d\theta=\int_0^{\pi/4}\log(1+\tan\theta)\,d\theta.

Step 2 — Symmetry $\theta\to\pi/4-\theta$

Substitute $u=\pi/4-\theta$ , $du=-d\theta$ , limits reverse:

I=\int_0^{\pi/4}\log\!\left(1+\tan(\tfrac{\pi}{4}-u)\right)du.

Use the addition formula:

\tan\!\left(\tfrac{\pi}{4}-u\right)=\dfrac{1-\tan u}{1+\tan u},

1+\tan(\tfrac{\pi}{4}-u)=1+\dfrac{1-\tan u}{1+\tan u}=\dfrac{(1+\tan u)+(1-\tan u)}{1+\tan u}=\dfrac{2}{1+\tan u}.

Take log:

\log\!\left(1+\tan(\tfrac{\pi}{4}-u)\right)=\log 2-\log(1+\tan u).

Step 3 — Self-reference and solve

Substitute back:

I=\int_0^{\pi/4}[\log 2-\log(1+\tan u)]\,du=\dfrac{\pi}{4}\log 2-I.

So $2I=\dfrac{\pi}{4}\log 2$ , giving

Answer

\boxed{\;I=\dfrac{\pi}{8}\log 2\;\bigl(\approx 0.272\bigr).\;}

UPSC 2014 Maths Optional Paper 1 Q1d — Step-by-Step Solution

Question

Technique

Solution

Step 1 — Trig substitution

Step 2 — Symmetry θ→π/4−θ\theta\to\pi/4-\thetaθ→π/4−θ

Step 3 — Self-reference and solve

Answer

This solution is part of the Maths Coverage Map — 13 years, mapped. Get the take-away PDF free.

Step 2 — Symmetry $\theta\to\pi/4-\theta$