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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals3 Marks
Question
Evaluate the following integrals:

$\int\sqrt{\text{cosec}\text{x}-1}\text{ dx}$

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Answer

$\int\sqrt{\text{cosec}\text{x}-1}\text{ dx}$
$=\int\sqrt{\frac{1}{\sin\text{x}}-1}\text{ dx}$
$=\int\frac{\sqrt{1-\sin\text{x}}}{\sqrt{\sin\text{x}}}\text{ dx}$
$=\int\frac{\sqrt{(1-\sin\text{x})(1+\sin\text{x})}}{\sqrt{\sin\text{x}(1+\sin\text{x})}}\text{ dx}$
$=\int\frac{\sqrt{1-\sin^2\text{x}}}{\sqrt{\sin^2\text{x}+\sin\text{x}}}\text{ dx}$
$=\int\frac{\cos\text{x}\text{ dx}}{\sqrt{\sin^2\text{x}+\sin\text{x}}}$
Let $\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}\text{ dx}=\text{dt}$
Now, $=\int\frac{\cos\text{x}\text{ dx}}{\sqrt{\sin^2\text{x}+\sin\text{x}}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}}}$
$=\int\frac{\text{dt}}{\sqrt{\text{t}^2+\text{t}+\big(\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}}$
$=\int\frac{\text{dt}}{\sqrt{\big(\text{t}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}}$
$=\log\Big|\Big(\text{t}+\frac{1}{2}\Big)+\sqrt{\big(\text{t}+\frac{1}{2}\big)^2-\big(\frac{1}{2}\big)^2}\Big|+\text{C}$
$=\log\Big|\text{t}+\frac{1}{2}+\sqrt{\text{t}^2+\text{t}}\Big|+\text{C}$
$=\log\Big|\sin\text{x}+\frac{1}{2}\sqrt{\sin^2\text{x}+\sin\text{x}}\Big|+\text{C}$

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