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Evaluate the definite integral: ∫0π4tanxdx

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प्रश्न

Evaluate the definite integral:

`int_0^(pi/4) tan x dx`

योग
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उत्तर

`int_0^(pi/4)  tan x  dx`

`= [log sec x]_0^(pi/4)`

`= log sec  pi/4 - log sec 0`

`= log sqrt2 - log 1`

`= log sqrt2 = log 2^(1/2)`

`= 1/2  log 2`

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अध्याय 7: Integrals - Exercise 7.9 [पृष्ठ ३३८]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
अध्याय 7 Integrals
Exercise 7.9 | Q 7 | पृष्ठ ३३८

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