Question

Intergrate with respect to t 1/ (cos^2t)√1+tant

Answer 1

Answer:

Step-by-step explanation:

We have,

$\displaystyle\tt{\int\dfrac{dt}{{cos}^{2}(t)\sqrt{1+tan(t)}}}$

$\displaystyle=\tt{\int\dfrac{{sec}^{2}(t)}{\sqrt{1+tan(t)}}\,dt}$

$\bf{Put\,\,\,1+tan(t)=x}$

$\bf{\implies\,sec^2(t)\,dt=dx}$

$\displaystyle=\tt{\int\dfrac{dx}{\sqrt{x}}}$

$\displaystyle=\tt{2\sqrt{x}+C}$

$\displaystyle=\tt{2\sqrt{1+tan(t)}+C}$