Question

Plzzzz help me...... ​

Answer 1

$\rm \longrightarrow \displaystyle \int \rm \frac{{sec}^{2} x}{1 + tan \: x} \: dx \\ \\ \rm let \: \: \: 1 + tan \: x = t \\ \rm {sec}^{2}\: x \: dx=d t \\ \\ \rm \longrightarrow \displaystyle \int \rm \frac{dx}{t} \\ \\ \rm \longrightarrow \displaystyle \rm ln(t) + c \:.......... (c \: is \: constant) \\ \\ \rm now \: put \: t \: =1 + tan \: x\\ \\ \rm \longrightarrow \displaystyle \rm ln(1 + tan \: x) + c$

Additional-Information :-

$\boxed{\boxed{\begin{minipage}{4cm}\displaystyle\circ\sf\:\int{1\:dx}=x+c\\\\\circ\sf\:\int{a\:dx}=ax+c\\\\\circ\sf\:\int{x^n\:dx}=\dfrac{x^{n+1}}{n+1}+c\\\\\circ\sf\:\int{sin\:x\:dx}=-cos\:x+c\\\\\circ\sf\:\int{cos\:x\:dx}=sin\:x+c\\\\\circ\sf\:\int{sec^2x\:dx}=tan\:x+c\\\\\circ\sf\:\int{e^x\:dx}=e^x+c\end{minipage}}}$

Answer 2

Answer:

⟶∫

1+tanx

sec

2

x

dx

let1+tanx=t

sec

2

xdx=dt

⟶∫

t

dx

⟶ln(t)+c..........(cisconstant)

nowputt=1+tanx

⟶ln(1+tanx)+c

Additional-Information :-

\begin{gathered}\boxed{\boxed{\begin{minipage}{4cm}\displaystyle\circ\sf\:\int{1\:dx}=x+c\\\\\circ\sf\:\int{a\:dx}=ax+c\\\\\circ\sf\:\int{x^n\:dx}=\dfrac{x^{n+1}}{n+1}+c\\\\\circ\sf\:\int{sin\:x\:dx}=-cos\:x+c\\\\\circ\sf\:\int{cos\:x\:dx}=sin