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Question
Evaluate the definite integral:
`int_0^(pi/2) cos^2 xdx`
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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`
RELATED QUESTIONS
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`int_(-1)^1 (x + 1)dx`
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`int_0^1 (2x + 3)/(5x^2 + 1) dx`
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`int_1^(sqrt3)dx/(1+x^2) ` equals:
`int_0^(2/3) dx/(4+9x^2)` equals:
Evaluate : \[\int\frac{x \cos^{- 1} x}{\sqrt{1 - x^2}}dx\] .
`int_1^sqrt(3) (dx)/(1 + x^2)` equals
`int_6^(2/3) (dx)/(4 + 9x^2)` equals
Evaluate:
`int_0^π(sin^4x + cos^4x)dx`
Hence evaluate:
`int_(-2π)^(2π) (sin^4x + cos^4x)/(1 + e^x)dx`
