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Indefinite Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIndefinite Integrals5 Marks
Question
Evaluate the following intregals:
$\int\frac{1}{\sin\text{x}-\cos\text{x}}\ \text{dx}$
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Answer

Let $\text{I}=\int\frac{1}{\sin\text{x}-\cos\text{x}}\ \text{dx}$
Putting $\sin\text{x}=\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}},\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$\text{I}=\int\frac{1}{\frac{2\tan\frac{\text{x}}{2}}{1+\tan\frac{\text{x}}{2}}}+\frac{1-\tan^\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$=\int\frac{1+\tan^2\frac{\text{x}}{2}}{2\tan\frac{\text{x}}{2}+1-\tan^2\frac{\text{x}}{2}}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{2\tan\frac{\text{x}}{2}+1-\tan^2\frac{\text{x}}{2}}\ \text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$\text{I}=\int\frac{2\text{dt}}{2\text{t}+1-\text{t}^2}$
$==-2\int\frac{\text{dt}}{\text{t}^2-2\text{t}+1-1-1}$
$\text{I}=-2\int\frac{\text{dt}}{(\text{t}-1)^2-(\sqrt{2})^2}$
$=2\int\frac{2\text{dt}}{(2\sqrt{2})^2-(\text{t}-1)}$
$=\frac{2}{2\sqrt{2}}\log\Big|\frac{\sqrt{2}+\text{t}-1}{\sqrt{2}-\text{t}+1}\Big|+\text{C}$
$\text{I}=\frac{1}{\sqrt{2}}\log\Big|\frac{\sqrt{2}+\tan\frac{\text{x}}{2}-1}{\sqrt{2}-\tan\frac{\text{x}}{2}+1}\Big|+\text{C}$

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