Definite IntegrationhardFree

Determinant of aₙ=∫(1−cos2nx)/(1−cos2x)dx | JEE

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

Question

Let

a_{n}=\displaystyle\int_{0}^{\pi/2}\dfrac{1-\cos 2nx}{1-\cos 2x}\,dx

. Then the value of

\begin{vmatrix}a_{1} & a_{2} & a_{3}\\[2pt] a_{4} & a_{5} & a_{6}\\[2pt] a_{7} & a_{8} & a_{9}\end{vmatrix}

equals

1

2

3

0

correct

Solution

Step 1: Show the sequence is arithmetic. Consider the combination

a_{n}+a_{n+2}-2a_{n+1}=\int_{0}^{\pi/2}\dfrac{\big(1-\cos 2nx\big)+\big(1-\cos(2n+4)x\big)-2\big(1-\cos(2n+2)x\big)}{1-\cos 2x}\,dx.

The numerator collapses to

2\cos(2n+2)x-\cos 2nx-\cos(2n+4)x

. Step 2: Apply the product-to-sum identity

\cos 2nx+\cos(2n+4)x=2\cos(2n+2)x\cos 2x

2\cos(2n+2)x-2\cos(2n+2)x\cos 2x=2\cos(2n+2)x\big(1-\cos 2x\big).

Therefore the integrand reduces, and the denominator cancels:

a_{n}+a_{n+2}-2a_{n+1}=\int_{0}^{\pi/2}2\cos(2n+2)x\,dx.

Step 3: Evaluate this integral:

\int_{0}^{\pi/2}2\cos(2n+2)x\,dx=2\left[\dfrac{\sin(2n+2)x}{2n+2}\right]_{0}^{\pi/2}=\dfrac{2\sin\big((n+1)\pi\big)}{2n+2}=0.

Hence

a_{n}+a_{n+2}-2a_{n+1}=0

, so

\{a_{n}\}

is an arithmetic progression (with

a_{1}=\displaystyle\int_{0}^{\pi/2}1\,dx=\dfrac{\pi}{2}

). Step 4: Use the A.P. property in the determinant. Apply the column operation

C_{1}\to C_{1}+C_{3}-2C_{2}

. Each entry of the new first column has the form

a_{k}+a_{k+2}-2a_{k+1}=0

, producing a zero column. A determinant with a zero column is

0

. Correct answer: (4)

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.