Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIndefinite Integrals5 Marks
Question
Evaluate the following integrals:$\int\frac{\sqrt{1-\sin\text{x}}}{1+\cos\text{x}}\text{e}^{\frac{-\text{x}}{2}}\text{dx}$
✓
Answer
Let $\text{I}=\int\frac{\sqrt{1-\sin\text{x}}}{1+\cos\text{x}}\text{e}^{\frac{-\text{x}}{2}}\text{dx}$
Put $=\frac{\text{x}}{2}=\text{t}$
$\Rightarrow\text{x}=2\text{t}$
$\text{dx}=2\text{dt}$
$\therefore\int\frac{\sqrt{1-\sin\text{x}}}{1+\cos\text{x}}\text{e}^{-\frac{\text{x}}{2}}\text{dx}$
$=2\int\frac{\sqrt{1-\sin2\text{t}}}{1+\cos2\text{t}}\text{e}^{-\text{t}}\text{dt}$ $\big[\because\sin^2\text{t}+\cos^2\text{t}=1\big]$
$=2\int\frac{\sqrt{\sin^2\text{t}+\cos^2\text{t}-2\sin\text{t}\cos\text{t}}}{1+\cos2\text{t}}\text{e}^{-\text{t}}\text{dt}$
$=2\int\frac{\sqrt{(\cos\text{t}-\sin\text{t})^2}}{2\cos^2\text{t}}\text{e}^{-\text{t}}\text{dt}$
$=2\int\frac{(\cos\text{t}-\sin\text{t})}{2\cos^2\text{t}}\text{e}^{-\text{t}}\text{dt}$
$=\int(\sec\text{t}-\tan\text{t}\sec\text{t})\text{e}^{-\text{t}}\text{dt}$
$=\int\sec\text{e}^{-\text{t}}\text{dt}-\int\tan\text{t}\sec\text{e}^{-\text{t}}\text{dt}$
Integrating by parts
$=\text{e}^{-\text{t}}\sec\text{t}+\int\text{e}^{-\text{t}\frac{\text{d}}{\text{dt}}}(\sec\text{t})\text{dt}-\int\tan\text{t}\sec\text{t}\text{ e}^{-\text{t}}\text{dt}$
$=-\text{e}^{-\text{t}}\sec\text{t}+\int\text{e}^{-\text{t}}\sec\text{t}\tan\text{t dt}-\int\sec\text{t}\tan\text{t}\text{ e}^{-\text{}t}\text{dt}$
$=-\text{e}^{-\text{t}}\sec\text{t+C}$
Putting the value of t
$=\text{-e}^{-\frac{\text{x}}{2}}\sec\frac{\text{x}}{2}+\text{C}$
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