Question

Question 1. Choose the correct answer: Question 1. The value of \int { \frac { sec^{ 2 }x }{ 1+tanx } } dx is: (a) loge (l + tan x) + c (b) tan x + c (c) – cot x + c (c) – cot x + c (d) loge x + c
Thas not true​

Answer 1

Solution:

Given Integral:

$\displaystyle \rm \longrightarrow I = \int \dfrac{ \sec^{2} (x) }{1 + \tan(x) } \: dx$

Let us assume that:

$\rm \longrightarrow u =1 + \tan(x)$

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

So, the integral changes to:

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

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

Substituting back u = 1 + tan(x), we get:

$\displaystyle \rm \longrightarrow I = \ln( |1 + \tan(x) | ) + C$

Therefore:

$\displaystyle \rm \longrightarrow \int \dfrac{ \sec^{2} (x) }{1 + \tan(x) } \: dx = \ln( |1 + \tan(x) | ) + C$

★ Which is our required answer.

Learn More:

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