Please give me the answer of RS Aggarwal :Qno 47 of Exersise 5'B' - Maths - Integrals

Question

Please give me the answer of RS Aggarwal :Qno 47 of Exersise
5'B'

Aakash Sharma · Accepted Answer

$Let I = \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{\cos θ}{(\cos \frac{θ}{2} + \sin \frac{θ}{2})^{3}} d θ \Rightarrow I = \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{\cos^{2} \frac{θ}{2} - \sin^{2} \frac{θ}{2}}{(\cos \frac{θ}{2} + \sin \frac{θ}{2})^{3}} d θ = \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{(\cos \frac{θ}{2} - \sin \frac{θ}{2}) (\cos \frac{θ}{2} + \sin \frac{θ}{2}) d θ}{(\cos \frac{θ}{2} + \sin \frac{θ}{2})^{3}} \Rightarrow I = \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{(\cos \frac{θ}{2} - \sin \frac{θ}{2})}{(\cos \frac{θ}{2} + \sin \frac{θ}{2})^{2}} d θ Let cos \frac{θ}{2} + \sin \frac{θ}{2} = t$

$Then \frac{1}{2} (\cos \frac{θ}{2} - \sin \frac{θ}{2}) = d t \Rightarrow (\cos \frac{θ}{2} - \sin \frac{θ}{2}) d θ = 2 d t \Rightarrow \int \frac{\cos \frac{θ}{2} - \sin \frac{θ}{2}}{(\cos \frac{θ}{2} + \sin \frac{θ}{2})^{2}} d θ = \int \frac{2 d t}{t^{2}} = \frac{- 2}{t} + C = \frac{- 2}{\cos \frac{θ}{2} + \sin \frac{θ}{2}} + C \Rightarrow I = \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{\cos \frac{θ}{2} - \sin \frac{θ}{2}}{(\cos \frac{θ}{2} + \sin \frac{θ}{2})^{2}} = [\frac{- 2}{\cos \frac{θ}{2} + \sin \frac{θ}{2}}]_{\frac{π}{4}}^{\frac{π}{2}} \Rightarrow I = [\frac{- 2}{\cos \frac{\frac{π}{2}}{2} + \frac{\sin \frac{π}{2}}{2}} - \frac{(- 2)}{\cos \frac{\frac{π}{4}}{2} + \sin \frac{\frac{π}{4}}{4}}]$

$= \frac{- 2}{(\cos \frac{π}{4} + \sin \frac{π}{4})} + \frac{2}{(\cos \frac{π}{8} + \sin \frac{π}{8})} = \frac{2}{\cos \frac{π}{8} + \sin \frac{π}{8}} - \frac{2}{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} = \frac{2}{\cos \frac{π}{8} + \sin \frac{π}{8}} - \frac{2}{\frac{2}{\sqrt{2}}} = \frac{2}{\cos \frac{π}{8} + \sin \frac{π}{8}} - \sqrt{2}$

Please give me the answer of RS Aggarwal :Qno. 47 of Exersise 5'B'.