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Simple Integration Q2
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$\int_{0}^{\frac{\pi}{4}}( \sec 2x \tan 2x )\ dx$
calculus
asked
Jan 5
in
Calculus
by
PEKKA
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edited
Jan 5
by
srestha

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answer
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log2.

please explain with steps.
Ans given as (1/2) ln 2
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1
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Best answer
$\int$sec2x = 1/2 * lnsec2x + tan2x
$\int$tan2x = 1/2 * lncos2x
$\int_{0}^{\prod/4}$(sec2x  tan2x) = $\int_{0}^{\prod/4}$(1/2lnsec2x + tan2x+1/2lncos2x) = $\int_{0}^{\prod/4}$(1/2ln1+sin2x) = 1/2ln2
answered
Jan 5
by
Samujjal Das
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edited
Jan 5
by
Samujjal Das
comment
please check it if there is any mistake
answer given as (1/2) ln 2
∫sec2x = 1/2 * logsec2x + tan2x
∫tan2x =  1/2 * logcos2x
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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