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$\int_{0}^{\frac{\pi}{4}}( \sec 2x -\tan 2x )\ dx$
asked in Calculus by Loyal (2.5k points) 5 27 62
edited by | 120 views
log2.
-----
please explain with steps.
Ans given as (1/2) ln 2

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Best answer
$\int$sec2x = 1/2 * ln|sec2x + tan2x|

$\int$tan2x = -1/2 * ln|cos2x|

$\int_{0}^{\prod/4}$(sec2x - tan2x) = $\int_{0}^{\prod/4}$(1/2ln|sec2x + tan2x|+1/2ln|cos2x|) = $\int_{0}^{\prod/4}$(1/2ln|1+sin2x|) = 1/2ln2
answered by Veteran (10.2k points) 6 43 108
edited by
please check it if there is any mistake
answer given as (1/2) ln 2
∫sec2x = 1/2 * log|sec2x + tan2x|

∫tan2x = - 1/2 * log|cos2x|

these are correct
@pavan is correct . @gate u did make a mistake :)  it willbe 2 ln 2
thanks :)
yeah corrected!! thanks

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