Question

Integrate: root of [1+2tanx(secx+tanx)]

Answer 1

$\int\sqrt{1+2tanx(secx+tanx)}dx$
we know,
sec²x - tan²x = 1
(secx + tanx) = 1/(secx - tanx) put it in integration .
$\int\sqrt{1+\frac{2tanx}{secx-tanx}}dx\\\\=\int\sqrt{\frac{secx-tanx+2tanx}{secx-tanx}}dx\\\\=\int\sqrt{ \frac{secx+tanx}{secx-tanx}}dx \\ \\ = \int \sqrt{ {( secx +tanx)}^{2} } dx$
$=\int |secx + tanx |dx$

= ∫secx.dx + ∫tanx.dx
= ln|secx-tanx| + ln|secx|+C