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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluvate the following intregals:
$\int\frac{1}{3+4\cot\text{x}}\ \text{dx}$
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Answer

Let $\text{I}=\int\frac{1}{3+4\cot\text{x}}\ \text{dx}$
$=\int\frac{1}{3+\frac{4\cos\text{x}}{\sin\text{x}}}\ \text{dx}$
$=\int\frac{\sin\text{x}}{3\sin\text{x}+4\cos\text{x}}\ \text{dx}$
Let $\sin\text{x}=\text{A}(3\sin\text{x}+4\cos\text{x})+\text{B}(3\cos\text{x}-4\sin\text{x})\dots(1)$
$\Rightarrow\sin\text{x}=(3\text{A}-4\text{B})\sin\text{x}+(4\text{A}+3\text{B})\cos\text{x}$
By compairing the coefficient of both sides we get,
$3\text{A}-4\text{B}=1\dots(2)$
$4\text{A}+3\text{B}=0\dots(3)$
Multiplying eq (2) by (3) and equation (3) by 4, then by adding them we get
$9\text{A}-12\text{B}+16\text{A}+12\text{B}=3+0$
$\Rightarrow25\text{A}=3$
$\text{A}=\frac{3}{25}$
Putting value of A in eq (3) we get,
$4\times\frac{3}{25}+3\text{B}=0$
$\Rightarrow3\text{B}=-\frac{12}{25}$
$\Rightarrow\text{B}=-\frac{4}{25}$
Thus, by substituting the value of A and B in eq (1) we get
$\text{I}=\int\bigg[\frac{\frac{3}{25}(3\sin\text{x}+4\cos\text{x})-\frac{4}{25}(3\cos\text{x}-4\sin\text{x})}{3\sin\text{x}+4\cos\text{x}}\bigg]\text{dx}$
$=\int\text{dx}-\frac{4}{25}\int\Big(\frac{3\cos\text{x}-4\sin\text{x}}{3\sin\text{x}+4\cos\text{x}}\Big)\ \text{dx}$
Putting $3\sin\text{x}+4\cos\text{x}=\text{t}$
$\Rightarrow(3\cos\text{x}-4\sin\text{x})\text{ dx}=\text{dt}$
$\therefore\text{I}=\frac{3}{25}\int\text{dx}-\frac{4}{25}\int\frac{\text{dt}}{\text{t}}$
$=\frac{3}{25}\text{x}-\frac{4}{25}\ln|\text{t}|+\text{C}$
$=\frac{3\text{x}}{25}-\frac{4}{25}\ln|3\sin\text{x}+4\cos\text{x}|+\text{C}$

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