JAMIA 2023 — Mathematics PYQ

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Question

JAMIA 2023 — Mathematics PYQ

JAMIA | Mathematics | 2023

The value of $\int_{0}^{π} x (sin^{4} x cos^{4} x) d x$ is

Choose the correct answer:

Explanation

Given:

$I = \int_{0}^{\pi} x (\sin^4 x \cos^4 x)\, dx$

Step 1: Property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ ka use karke:

I = \int_{0}^{\pi} (\pi - x) \sin^4(\pi - x) \cos^4(\pi - x) \, dx

I = \int_{0}^{\pi} (\pi - x) \sin^4 x \cos^4 x \, dx \quad \dots (1)

I = \int_{0}^{\pi} x \sin^4 x \cos^4 x \, dx \quad \dots (2)

Step 2: Dono equations ko add karne par:

2I = \int_{0}^{\pi} (x + \pi - x) \sin^4 x \cos^4 x \, dx

2I = \pi \int_{0}^{\pi} \sin^4 x \cos^4 x \, dx

I = \frac{\pi}{2} \int_{0}^{\pi} \sin^4 x \cos^4 x \, dx

Step 3: Symmetry property $\int_{0}^{2a} f(x) dx = 2\int_{0}^{a} f(x) dx$ use karke:

I = \frac{\pi}{2} \cdot 2 \int_{0}^{\pi/2} \sin^4 x \cos^4 x \, dx

I = \pi \int_{0}^{\pi/2} \sin^4 x \cos^4 x \, dx

Step 4: Wallis' Formula apply karne par:

Formula: $\int_{0}^{\pi/2} \sin^m x \cos^n x \, dx = \frac{(m-1)!!(n-1)!!}{(m+n)!!} \cdot \frac{\pi}{2}$

Yahan $m=4, n=4$ :

\int_{0}^{\pi/2} \sin^4 x \cos^4 x \, dx = \frac{(3 \cdot 1)(3 \cdot 1)}{8 \cdot 6 \cdot 4 \cdot 2} \cdot \frac{\pi}{2}

\int_{0}^{\pi/2} \sin^4 x \cos^4 x \, dx = \frac{9}{384} \cdot \frac{\pi}{2} = \frac{3}{128} \cdot \frac{\pi}{2} = \frac{3\pi}{256}

Step 5: Final calculation:

I = \pi \left( \frac{3\pi}{256} \right)

I = \frac{3\pi^2}{256}

Choose the correct answer:

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