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Indefinite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integration5 Marks
Question
Evaluate : $\int \frac{1}{3-2 \sin x+5 \cos x} \cdot d x$
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Answer

put $\tan \frac{x}{2}=t$
$
\begin{aligned}
& \therefore \quad d x=\frac{2}{1+t^2} \\
& \therefore \quad \sin x=\frac{2}{1+t^2} \cdot d t \text { and } \cos x=\frac{1-t^2}{1+t^2} \\
& I=\int \frac{1\left(\frac{2}{1+t^2}\right)}{3-2\left(\frac{2}{1+t^2}\right)+5\left(\frac{1-t^2}{1+t^2}\right)} \cdot d t \\
& =\int \frac{\frac{2}{1+t^2}}{\frac{3\left(1+t^2\right)-2(2 t)+5\left(1-t^2\right)}{1+t^2}} \cdot d t \\
& =\int \frac{2}{3+3 t^2-4 t+5-5 t^2} \cdot d t \\
& =\int \frac{2}{8-4 t-2 t^2} \cdot d t \\
& =\int \frac{1}{4-2 t-t^2} \cdot d t \\
& =\int \frac{1}{4-\left(t^2+2 t\right)} \cdot d t \\
& =\int \frac{1}{4-\left(t^2+2 t+1-1\right)} \cdot d t \\
& =\int \frac{1}{5-\left(t^2+2 t+1\right)} \cdot d t \\
& =\int \frac{1}{(\sqrt{5})^2-(t+1)^2} \cdot d t \\
& =\frac{1}{2(\sqrt{5})} \cdot \log \left(\frac{\sqrt{5}+(t+1)}{\sqrt{5}-(t+1)}\right)+c \\
& =\frac{1}{2 \sqrt{5}} \cdot \log \left(\frac{\sqrt{5}+1+\tan \frac{x}{2}}{\sqrt{5}-1-\tan \frac{x}{2}}\right)+c \\
&
\end{aligned}
$

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