Question

Evaluate the following integral

$\int \: \frac{dx}{1 + cotx}$
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Answer 1

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

Given integral is

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

can be rewritten as

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

$\rm \: = \: \displaystyle \int \rm \: \frac{dx}{\dfrac{sinx + cosx}{sinx} }$

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

$\rm \: = \: \frac{1}{2} \displaystyle \int \rm \: \frac{2sinx}{sinx + cosx} \: dx$

$\rm \: = \: \frac{1}{2} \displaystyle \int \rm \: \frac{sinx + sinx}{sinx + cosx} \: dx$

$\rm \: = \: \frac{1}{2} \displaystyle \int \rm \: \frac{sinx + sinx + cosx - cosx}{sinx + cosx} \: dx$

$\rm \: = \: \frac{1}{2} \displaystyle \int \rm \: \frac{(sinx + cosx) + (sinx - cosx)}{sinx + cosx} \: dx$

$\rm \: = \: \frac{1}{2} \displaystyle \int \rm \: \frac{sinx + cosx}{sinx + cosx} \: dx + \frac{1}{2} \displaystyle \int \rm \: \frac{sinx - cosx}{sinx + cosx} \: dx$

$\rm \: = \: \frac{1}{2} \displaystyle \int \rm \: 1 \: dx - \frac{1}{2} \displaystyle \int \rm \: \frac{cosx - sinx}{sinx + cosx} \: dx$

In second integral, we use method of Substitution.

So, Substitute

$\rm \: sinx + cosx = y$

$\rm \: (cosx - sinx)dx = dy$

So, on substituting the values, we get

$\rm \: = \: \frac{1}{2} \displaystyle \int \rm \:dx \: - \: \frac{1}{2}\displaystyle \int \rm \: \frac{dy}{y}$

$\rm \: = \: \dfrac{1}{2}x - \dfrac{1}{2} log |y| + c$

$\rm \: = \: \dfrac{x}{2} - \dfrac{1}{2} log |sinx + cosx| + c$

Hence,

$\boxed{ \rm{ \:\rm \:\displaystyle \int \rm \: \frac{dx}{1 + cotx} = \: \dfrac{x}{2} - \dfrac{1}{2} log |sinx + cosx| + c \: \: }}$

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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

$\small\colorbox{lightyellow} {\text{ \bf♕ Brainliest answer }}$

Question :-

Evaluate the following integral $\displaystyle \rm\int \: \frac{dx}{1 + \cot x}$

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$\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}$

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$\[ \begin{array}{l} \\ \displaystyle\rm I=\int \frac{d x}{1+\cot x}=\int \frac{\sin x}{\sin x+\cos x} d x \\ \\ \displaystyle\begin{aligned} \displaystyle\rm\sin x \displaystyle\rm=M \frac{d}{d x}(\sin x+\cos x)+N(\sin x+\cos x) \\ &\\ \displaystyle\rm=M(-\sin x+\cos x)+N(\sin x+\cos x) \end{aligned} \end{array} \]$

Comparing the coefficients of sin x and cos x , we have

$\rm\[ -M+N=1, M+N=0 \]$

Solving these equations, we have

$\\ \rm M=-\frac{1}{2}, N=\frac{1}{2}$

$\[ \begin{array}{l} \displaystyle\rm\sin x=-\frac{1}{2}(-\sin x+\cos x)+\frac{1}{2}(\sin x+\cos x) \\ \\ \displaystyle\rm I=\int \frac{\sin x}{\sin x+\cos x} d x \\ \\ \displaystyle\rm \quad=-\frac{1}{2} \int \frac{(-\sin x+\cos x)}{(\sin x+\cos x)} d x+\frac{1}{2} \int \frac{(\sin x+\cos x)}{(\sin x+\cos x)} d x \\ \\ \displaystyle\rm \quad=-\frac{1}{2} \ln |(\sin x+\cos x)|+\frac{1}{2} x+c \end{array} \]$