Question

intragate 1 upon (1+sin x) dx
​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm \: \displaystyle\int\rm \frac{1}{1 + sinx} \: dx$

On rationalizing the denominator, we get

$\rm \: = \displaystyle\int\rm \frac{1}{1 + sinx} \times \frac{1 - sinx}{1 - sinx} \: dx$

$\rm \: = \displaystyle\int\rm \frac{1 - sinx}{1 - {sin}^{2} x} \: dx$

$\rm \: = \displaystyle\int\rm \frac{1 - sinx}{{cos}^{2} x} \: dx$

$\rm \: = \displaystyle\int\rm \bigg( \frac{1}{ {cos}^{2} x} - \frac{ sinx}{{cos}^{2} x}\bigg) \: dx$

$\rm \: = \displaystyle\int\rm ( {sec}^{2}x - secx \: tanx) \: dx$

$\rm \: = \: tanx \: - \: secx \: + \: c$

Hence,

$\red{\rm\implies \:\boxed{ \rm{ \:\rm \:\displaystyle\int\rm \frac{1}{1 + sinx}dx = \: tanx \: - \: secx \: + \: c \: }}}$

$\rule{190pt}{2pt}$

Additional information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Step-by-step explanation:

\large\underline{\sf{Solution-}}

Solution−

Given integral is

\begin{gathered}\rm \: \displaystyle\int\rm \frac{1}{1 + sinx} \: dx \\ \end{gathered}

∫

1+sinx

1

dx

On rationalizing the denominator, we get

\begin{gathered}\rm \: = \displaystyle\int\rm \frac{1}{1 + sinx} \times \frac{1 - sinx}{1 - sinx} \: dx \\ \end{gathered}

=∫

1+sinx

1

×

1−sinx

1−sinx

dx

\begin{gathered}\rm \: = \displaystyle\int\rm \frac{1 - sinx}{1 - {sin}^{2} x} \: dx \\ \end{gathered}

=∫

1−sin

2

x

1−sinx

dx

\begin{gathered}\rm \: = \displaystyle\int\rm \frac{1 - sinx}{{cos}^{2} x} \: dx \\ \end{gathered}

=∫

cos

2

x

1−sinx

dx

\begin{gathered}\rm \: = \displaystyle\int\rm \bigg( \frac{1}{ {cos}^{2} x} - \frac{ sinx}{{cos}^{2} x}\bigg) \: dx \\ \end{gathered}

=∫(

cos

2

x

1

−

cos

2

x

sinx

)dx

\begin{gathered}\rm \: = \displaystyle\int\rm ( {sec}^{2}x - secx \: tanx) \: dx \\ \end{gathered}

=∫(sec

2

x−secxtanx)dx

\begin{gathered}\rm \: = \: tanx \: - \: secx \: + \: c \\ \end{gathered}

=tanx−secx+c

Hence,

\begin{gathered}\red{\rm\implies \:\boxed{ \rm{ \:\rm \:\displaystyle\int\rm \frac{1}{1 + sinx}dx = \: tanx \: - \: secx \: + \: c \: }}} \\ \end{gathered}

⟹

∫

1+sinx

1

dx=tanx−secx+c

\rule{190pt}{2pt}

Additional information :-

\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}\end{gathered}

f(x)

k

sinx

cosx

sec

2

x

cosec

2

x

secxtanx

cosecxcotx

tanx

x

1

e

x

∫f(x)dx

kx+c

−cosx+c

sinx+c

tanx+c

−cotx+c

secx+c

−cosecx+c

logsecx+c

logx+c

e

x

+c