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Integrate : $\int \sec^2 x \log (\tan x) .dx$

$(a)\;\tan x [\tan x -1]+c \\(b)\; \cos x[\cos x-1]+c\\(c)\;\sin x [\sin x-1]+c \\ (d)\;None$

1 Answer

$\tan x=t$
By differentiate with respect to x
$\sec^2 x dx =dt$
$\int 1. \log t .dt$
=> $\int u.v dx= u. \int v.dx-\int \bigg [ \large\frac{dy}{dx} \int v.dx\bigg]$$dx$
=> $t. \log t -t+c$
=> $\tan x [\log (\tan x )-1]+c$
Hence d is the correct answer.
answered Jan 3, 2014 by meena.p