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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS1 Mark
Question
Integrate the functions in Exercises:
$\frac{1}{1+\cot\text{x}}$
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Answer

$\text{Let I}=\int\frac{1}{1+\cot\text{x}}\text{ dx} =\int\frac{1}{1+\frac{\cos\text{x}}{\sin\text{x}}}\text{ dx} $
$=\int\frac{1}{\bigg(\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}}\bigg)}\text{ dx} =\int\frac{\sin\text{x}}{\sin\text{x}+\cot\text{x}}\text{ dx} $
$=\frac{1}{2}\int\frac{2\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}=\frac{1}{2}\int\frac{\sin\text{x}+\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx} $
Adding and subtracting $\cos\text{x}$ in the numerator,
$=\frac{1}{2}\int\frac{\sin\text{x}+\cos\text{x}-\cos\text{x}+\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx} $
$=\frac{1}{2}\int\frac{(\sin\text{x}+\cos\text{x})-(\cos\text{x}-\sin\text{x})}{\sin\text{x}+\cos\text{x}}\text{ dx} $
$=\frac{1}{2}\int\bigg(\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}+\cos\text{x}}-\frac{\cos\text{x}-\sin\text{x}}{\sin \text{x}+\cos\text{x}}\bigg)\text{ dx} $
$=\frac{1}{2}\int\bigg(1-\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}+\cos\text{x}}\bigg)\text{ dx} $
$=\frac{1}{2}\bigg[\int1\text{ dx}-\int\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx}\bigg]=\frac{1} {2}[\text{x}-\text{I}_{1}]$$\ \ \ \ \ \text{where }\text{I}_{1}=\int\frac{\cos\text{x}-\sin\text{x}}{\sin\text{x}+\cos\text{x}}\text{ dx} \ \ \ ...\text{(i)}$
Putting $\sin\text{ x + cos x = t}\ \ \ \Rightarrow\ \ \ \cos\text{x}-\sin\text{x}=\frac{\text{dt}}{\text{dx}}\ \ \ \Rightarrow\ \ \ \ (\cos\text{x}-\sin\text{x})\text{ dx = dt} $
$ \therefore\ \ \ \ \ $$\text{I}_1=\int\frac{\text{dt}}{\text{t}}=\log\begin{vmatrix}\text{t}\end{vmatrix}=\log\begin{vmatrix}\sin\text{x}+\cos\text{x}\end{vmatrix} $
Putting this value in eq. (i), we get required integral,
$=\frac{1}{2}[\text{x}-\log\begin{vmatrix}\sin\text{x + cosx}\end{vmatrix}]+\text{c} $

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