Evaluate: `int_0^pi cos^2 x "d"x`

`int_0^pi cos^2 x "d"x = int_0^pi((1 + cos 2x)/2) "d"x` = `1/2[int_0^pi "d"x + int_0^pi cos 2x "d"x]` = `1/2[[x]_0^pi + [(sin 2x)/2]_0^pi]` = `1/2[(pi - 0) + 1/2(sin 2pi - sin 0)]` = `1/2[pi + 1/2(0 - 0)]` = `pi/2`

Evaluate: ∫0πcos2x dx - Mathematics and Statistics

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Evaluate: $\int_{0}^{π} \cos^{2} x d x$

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उत्तर

$\int_{0}^{π} \cos^{2} x d x = \int_{0}^{π} (\frac{1 + \cos 2 x}{2}) d x$

= $\frac{1}{2} [\int_{0}^{π} d x + \int_{0}^{π} \cos 2 x d x]$

= $\frac{1}{2} [{[x]}_{0}^{π} + {[\frac{\sin 2 x}{2}]}_{0}^{π}]$

= $\frac{1}{2} [(π - 0) + \frac{1}{2} (\sin 2 π - \sin 0)]$

= $\frac{1}{2} [π + \frac{1}{2} (0 - 0)]$

= $\frac{π}{2}$

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Methods of Evaluation and Properties of Definite Integral

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Q 4 Q 3 Q 5

पाठ 2.4: Definite Integration - Short Answers I

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एससीईआरटी महाराष्ट्र Mathematics and Statistics (Arts and Science) [English] 12 Standard HSC

पाठ 2.4 Definite Integration
Short Answers I | Q 4

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Evaluate: ∫0πcos2x dx - Mathematics and Statistics

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Evaluate: ∫0πcos2x dx - Mathematics and Statistics

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