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प्रश्न
Using second fundamental theorem, evaluate the following:
`int_0^(pi/2) sqrt(1 + cos x) "d"x`
योग
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उत्तर
We know cos 2x = `2cos^2x - 1`
⇒ cos x = `2cos^2 x/2 - 1`
⇒ 1 + cos x = `2cos^2 x/2`
`int_0^(pi/2) sqrt(2 cos^2 x/2) "d"x = int_0^(pi/2) sqrt(2) cos x/2 "d"x`
= `[(sqrt(2) sin x/2)/(1/2)]_0^(pi/2)`
= `2sqrt(2) sin pi/4 - 2sqrt(2) sin 0`
= `2sqrt(2) (1/sqrt(2))`
= 2
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