Question

solve the above attachment guys






For several days I was in offline because of my exams.
Hru all​

Answer 1

Question:-

$\sf \large Find \: \int \frac{sec \: x}{sec \: x + tan \: x} dx.$

Given:-

$\sf \large \int \frac{sec \: x}{sec \: x + tan \: x} dx.$

To Find:-

$\sf \large \int \frac{sec \: x}{sec \: x + tan \: x} dx.$

Solution:-

$\sf \large I = \int\frac{sec \: x}{sec \: x + tan \: x} dx$

$\sf \large I = \int \frac{sec \: x}{sec \: x + tan \: x} \times \frac{sec \: x - tan \: x}{sec \: x - tan \: x} dx$

$\sf \large I = \int \frac{sec \: x(sec \: x - tan \: x)}{sec {}^{2}x - tan {}^{2} x } dx \\ \\ \sf \large \color{red}( \because(a - b)(a + b) = a {}^{2} - b {}^{2} )$

$\sf \large I = \int(sec {}^{2} x - sec \: x \: tan \: x)dx \\ \\ \sf \large \color{red}( \because \: sec {}^{2} x - tan {}^{2} x = 1)$

$\sf \large I = \int sec {}^{2} x \: dx - \int sec \: x \: tan \: x \: \: dx$

$\sf \large I = tan \: x - sec \: x + c \\ \\ \sf \large \color{red}( \because \int sec {}^{2} \theta d \theta = tan \theta \\ \sf \large \color{red} \int sec \theta \: tan \theta \: d \theta = sec \theta )$

Answer:-

$\sf \large \blue {\int \frac{sec \: x}{sec \: x + tan \: x}dx = tan \: x - sec \: x + c.}$

Hope you have satisfied. ⚘