Physics, asked by terain, 3 months ago

Plzzzz help me...... ​

Attachments:

Answers

Answered by BrainlyEmpire
2

 \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}}}

Answered by akanksha2614
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

Similar questions