Question

indefinite integration sec²x/3+thanx dx​

Answer 1

Answer:

Evaluate :∫(sec2x)/(3+tanx) dxRead more on Sarthaks.com - https://www.sarthaks.com/49205/evaluate-sec-2x-3-tanx-dx?show=49207#a49207

Answer 2

Solution:

Given Integral:

$\displaystyle \rm \longrightarrow I = \int \dfrac{ { \sec}^{2}(x) }{3 + \tan(x) } \: dx$

Let us assume that:

$\displaystyle \rm \longrightarrow u = 3 + \tan(x)$

$\displaystyle \rm \longrightarrow du = \sec^{2} (x) \: dx$

Therefore, the integral becomes:

$\displaystyle \rm \longrightarrow I = \int \dfrac{du}{u}$

$\displaystyle \rm \longrightarrow I = \ln(u) + C$

$\displaystyle \rm \longrightarrow I = \ln( |3 + \tan(x) | ) + C$

Therefore:

$\displaystyle \rm \longrightarrow \int \dfrac{ \sec^{2} (x) }{3 + \tan(x) } \: dx = \ln( |3 + \tan(x) | ) + C$

★ Which is our required answer.

Answer:

$\displaystyle \rm \hookrightarrow \int \dfrac{ \sec^{2} (x) }{3 + \tan(x) } \: dx = \ln( |3 + \tan(x) | ) + C$

Learn More:

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \rm k&\rm kx+C\\ \\ \rm sin(x)&\rm-cos(x)+C\\ \\ \rm cos(x)&\rm sin(x)+C\\ \\ \rm{sec}^{2}(x)&\rm tan(x)+C\\ \\ \rm{cosec}^{2}(x)&\rm-cot(x)+C\\ \\ \rm sec(x)\ tan(x)&\rm sec(x)+C\\ \\ \rm cosec(x)\ cot(x)&\rm-cosec(x)+C\\ \\ \rm tan(x)&\rm log(sec(x))+C\\ \\ \rm\dfrac{1}{x}&\rm log(x)+C\\ \\ \rm{e}^{x}&\rm{e}^{x}+C\\ \\ \rm x^{n},n\neq-1&\rm\dfrac{x^{n+1}}{n+1}+C\end{array}}$