Definite IntegrationhardFree

aₙIₙ₊₂+bₙIₙ=cₙ for Iₙ=∫xⁿtan⁻¹x dx | JEE

JEE Maths question with a full step-by-step solution.

Question

Let

I_{n}=\displaystyle\int_{0}^{1}x^{n}\tan^{-1}x\,dx

. If

a_{n}I_{n+2}+b_{n}I_{n}=c_{n}

for all

n\in\mathbb{N},\ n\ge 1

, then

a_{1},a_{2},a_{3},\dots

are in A.P.correct

b_{1},b_{2},b_{3},\dots

are in G.P.

c_{1},c_{2},c_{3},\dots

are in H.P.

a_{1},a_{2},a_{3},\dots

are in H.P.

Solution

Step 1: Integrate

I_{n}

by parts with

u=\tan^{-1}x

dv=x^{n}\,dx

(so

v=\dfrac{x^{n+1}}{n+1}

I_{n}=\left[\dfrac{x^{n+1}}{n+1}\tan^{-1}x\right]_{0}^{1}-\int_{0}^{1}\dfrac{x^{n+1}}{(n+1)\big(1+x^{2}\big)}\,dx=\dfrac{\pi}{4(n+1)}-\dfrac{1}{n+1}\int_{0}^{1}\dfrac{x^{n+1}}{1+x^{2}}\,dx.

Multiplying through by

(n+1)

(n+1)I_{n}=\dfrac{\pi}{4}-\int_{0}^{1}\dfrac{x^{n+1}}{1+x^{2}}\,dx.

Step 2: Write the same relation with

n

replaced by

n+2

(n+3)I_{n+2}=\dfrac{\pi}{4}-\int_{0}^{1}\dfrac{x^{n+3}}{1+x^{2}}\,dx.

Step 3: Add the two equations. The integrals combine neatly because

x^{n+1}+x^{n+3}=x^{n+1}\big(1+x^{2}\big)

(n+1)I_{n}+(n+3)I_{n+2}=\dfrac{\pi}{2}-\int_{0}^{1}\dfrac{x^{n+1}\big(1+x^{2}\big)}{1+x^{2}}\,dx=\dfrac{\pi}{2}-\int_{0}^{1}x^{n+1}\,dx=\dfrac{\pi}{2}-\dfrac{1}{n+2}.

Step 4: Compare with

a_{n}I_{n+2}+b_{n}I_{n}=c_{n}

a_{n}=n+3,\qquad b_{n}=n+1,\qquad c_{n}=\dfrac{\pi}{2}-\dfrac{1}{n+2}.

Here

a_{n}=n+3

gives

4,5,6,\dots

, an arithmetic progression. (Note

b_{n}=n+1

is also A.P., not G.P.;

c_{n}

is in no standard progression.) Thus the correct statement is that the

a_{n}

are in A.P. Correct answer: (1)Step 1: Integrate

I_{n}