JAMIA 2024 — Mathematics PYQ

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Question

JAMIA 2024 — Mathematics PYQ

JAMIA | Mathematics | 2024

$\int_{0}^{π} sin^{2} x d x =$

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Explanation

Solution:

Hume is integral ki value nikalni hai:

\int_{0}^{\pi} \sin^2 x \, dx

Step 1: Trigonometric identity ka use karein

Hum jaante hain ki:

\sin^2 x = \frac{1 - \cos 2x}{2}

Step 2: Integral mein value rakhein

\int_{0}^{\pi} \frac{1 - \cos 2x}{2} \, dx

\frac{1}{2} \int_{0}^{\pi} (1 - \cos 2x) \, dx

Step 3: Integrate karein

\frac{1}{2} \left[ x - \frac{\sin 2x}{2} \right]_{0}^{\pi}

Step 4: Limits apply karein

Upper limit $(\pi)$ aur lower limit $(0)$ rakhne par:

\frac{1}{2} \left[ \left( \pi - \frac{\sin 2\pi}{2} \right) - \left( 0 - \frac{\sin 0}{2} \right) \right]

Kyonki $\sin 2\pi = 0$ aur $\sin 0 = 0$ :

\frac{1}{2} [ (\pi - 0) - (0 - 0) ]

\frac{1}{2} \times \pi = \frac{\pi}{2}

Final Answer:

\mathbf{\frac{\pi}{2}}

Choose the correct answer:

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