Definite IntegrationmediumFree

Definite Integration: Value

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

Question
The value of
[π⋅21⋅23⋅43⋅45⋅65⋅67⋅87⋅89⋯∞]\left[\pi\cdot\frac21\cdot\frac23\cdot\frac43\cdot\frac45\cdot\frac65\cdot\frac67\cdot\frac87\cdot\frac89\cdots\infty\right]
 is Where [ ⋅ ]\left[\ \cdot\ \right] is G.I.F.
Solution
Answer: 4
Step 1:
21⋅23⋅43⋅45⋅65⋅67⋯=∏k=1∞2k2k−1⋅2k2k+1\frac21\cdot\frac23\cdot\frac43\cdot\frac45\cdot\frac65\cdot\frac67\cdots = \prod_{k=1}^{\infty}\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}
 is Wallis's product, whose value is π2\dfrac\pi2. Steps 2 to 4 prove this, so nothing is quoted blind. Step 2: Let
In=∫0π/2sin⁡nx dxI_n = \int_0^{\pi/2}\sin^nx\,dx
 Integrating by parts gives the reduction In=n−1nIn−2I_n = \dfrac{n-1}{n}I_{n-2}. With I0=π2I_0 = \dfrac\pi2 and I1=1I_1 = 1,
I2n=(2n−1)(2n−3)⋯12n(2n−2)⋯2⋅π2,I2n+1=2n(2n−2)⋯2(2n+1)(2n−1)⋯3I_{2n} = \frac{\left(2n-1\right)\left(2n-3\right)\cdots1}{2n\left(2n-2\right)\cdots2}\cdot\frac\pi2 , \qquad I_{2n+1} = \frac{2n\left(2n-2\right)\cdots2}{\left(2n+1\right)\left(2n-1\right)\cdots3}
 Step 3: The ratio is the partial product, inverted:
I2nI2n+1=π2/[21⋅23⋅43⋅45⋯2n2n−1⋅2n2n+1]\frac{I_{2n}}{I_{2n+1}} = \frac\pi2\Big/\left[\frac21\cdot\frac23\cdot\frac43\cdot\frac45\cdots \frac{2n}{2n-1}\cdot\frac{2n}{2n+1}\right]
 Step 4: 0≤sin⁡x≤10 \le \sin x \le 1 on [0,π2]\left[0,\tfrac\pi2\right], so InI_n is decreasing in nn, and
I2n+1≤I2n≤I2n−1=2n+12nI2n+1⟹1≤I2nI2n+1≤2n+12nI_{2n+1} \le I_{2n} \le I_{2n-1} = \frac{2n+1}{2n}I_{2n+1} \quad\Longrightarrow\quad 1 \le \frac{I_{2n}}{I_{2n+1}} \le \frac{2n+1}{2n}
 Letting n→∞n\to\infty, I2nI2n+1→1\dfrac{I_{2n}}{I_{2n+1}}\to1, hence the product equals π2\dfrac\pi2. Step 5:
π⋅π2=π22\pi\cdot\frac\pi2 = \frac{\pi^2}{2}
π22=9.8696…2=4.9348…⟹[π22]=4\frac{\pi^2}{2} = \frac{9.8696\ldots}{2} = 4.9348\ldots \quad\Longrightarrow\quad \left[\frac{\pi^2}{2}\right] = 4
 Answer: 44
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Definite Integration · easy
∫0π/2sin⁡x1+cos⁡x+sin⁡x dx\displaystyle\int_{0}^{\pi/2}\dfrac{\sin x}{1+\cos x+\sin x}\,dx equals
Definite Integration · easy
If f(x)=∫1/xxcos⁡t2 dtf(x)=\displaystyle\int_{1/x}^{\sqrt{x}}\cos t^{2}\,dt for x>0x>0, then df(x)dx\dfrac{df(x)}{dx} is
Definite Integration · easy
If ∫0π/2sin⁡10x cos⁡10x dx=2−n∫0π/2(sin⁡10x+cos⁡10x) dx\displaystyle\int_{0}^{\pi/2}\sin^{10}x\,\cos^{10}x\,dx=2^{-n}\displaystyle\int_{0}^{\pi/2}\big(\sin^{10}x+\cos^{10}x\big)\,dx, n∈Nn\in\mathbb{N}, then n=n=
Definite Integration · easy
If ∫sin⁡x1t2f(t) dt=1−sin⁡x\displaystyle\int_{\sin x}^{1}t^{2}f(t)\,dt=1-\sin x for all x∈(0,π2]x\in\left(0,\dfrac{\pi}{2}\right], then the value of f ⁣(13)f\!\left(\dfrac{1}{\sqrt3}\right) is

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.

Sign up freeExplore doMath