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Evaluate the definite integral: ∫0π2cos2xdx

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Question

Evaluate the definite integral:

`int_0^(pi/2) cos^2 xdx`

Evaluate
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Solution

`int_0^(pi/2)  cos^2  x  dx`    ...(i)

= `1/2 int_0^(pi/2) (1 + cos 2x) dx`

`because cos 2x = 1 - 2 cos^2 x`

`=> cos^2 x = 1/2 (1 + cos 2x)`

= `1/2 [x + (sin 2x)/2]_0^(pi/2)`

= `1/2 [pi/2 - (sin pi)/2 - 0 - 1/2 sin 0]`

= `1/2 [pi/2 + 0 - 0]`

= `pi/4`

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Chapter 7: Integrals - Exercise 7.9 [Page 338]

APPEARS IN

NCERT Mathematics Part 1 and 2 [English] Class 12
Chapter 7 Integrals
Exercise 7.9 | Q 12 | Page 338

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