Question

Integration of :-
$\frac{1 + \tan(x) }{ \cos {}^{2} (x) } dx$
Step by step explanation and will be marked as brainliest.
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Answer 1

Step-by-step explanation:

$Given: \int \frac{1+tanx}{cos^2x}dx$

$\Longrightarrow \int \frac{1}{cos^2x} + \frac{tanx}{cos^2x} dx$

$\Longrightarrow \int \frac{1}{cos^2x} + \int \frac{tanx}{cos^2x} dx$

$\Longrightarrow tanx + \int tanx \ sec^2x \ dx$

$\Longrightarrow tan x + \int u \ du$

$\Longrightarrow tan x + \frac{u^2}{2}$

$\Longrightarrow \boxed{tan x + \frac{1}{2} tan^2 \ x + C}$

Hope it helps!

Answer 2

Integration of ((cos²x) /(1+tanx))

multiply and divide by sec^4 x

integration of ((cos²x)sec^4 x /(1+tanx)sec^4x )

integration of (sec^2 x /(sec^4x + tanx sec^4x ))

let u = tanx

du = - ln|cox| dx

substituting nad solving

Integration (cos²x/(1+tanx)) dx = 1/8(4x+sin(2x)+cos(2x)+2log(sin x+cos x))+C