Question

integration of
tanx/sinx dx​

Answer 1

Solution:

Given Integral:

$\displaystyle \rm \longrightarrow I = \int \dfrac{ \tan(x) }{ \sin(x) } \: dx$

Can be written as:

$\displaystyle \rm \longrightarrow I = \int \dfrac{ \sin(x) \div \cos(x) }{ \sin(x) } \: dx$

$\displaystyle \rm \longrightarrow I = \int \dfrac{ \sin(x) }{ \cos(x) } \times \dfrac{1}{ \sin(x) } \: dx$

$\displaystyle \rm \longrightarrow I = \int \dfrac{1}{ \cos(x) } \: dx$

$\displaystyle \rm \longrightarrow I = \int \sec(x) \: dx$

This is a standard integral whose value is ln(|tan(x) + sec(x)|)

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

Therefore:

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

Answer:

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

Learn More:

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

Answer 2

Answer:

integration of tan x /Sinx is log |secx + tanx| + C

Step-by-step explanation:

steps are given in image