Question #0b243

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Answer 1 · 2017-11-15T17:27:49

$1/4ln2.$

Let, $I=int_0^(pi/12) tan4xdx.$

By the Fundamental Theorem of Calculus, we know that,

$intf(x)dx=F(x)+C rArr int_a^bf(x)dx=[F(x)]_a^b=F(b)-F(a)$ .

Now, $inttan4xdx=1/4ln|sec4x|+C.$

$:. int_0^(pi/12)tan4xdx=[1/4ln|sec4x|]_0^(pi/12),$

$=1/4[ln|sec(4*pi/12)|-ln|sec0|],$

$=1/4[ln|sec(pi/3)|-ln|1|],$

$=1/4[ln|2|-0],$

$rArr int_0^(pi/12)tan4xdx=1/4ln2.$