Question

نشان دهید که: $$I= \int _0^ \frac{ \pi }{2} \frac{sinx}{ \sqrt{1+4sinxcosx} } dx= \frac{1}{ \sqrt{2} } Arctan ( \sqrt{2} )$$

Comments

$$x= \frac{ \pi }{2} -t \Longrightarrow I=- \int _ \frac{ \pi }{2} ^0 \frac{cost}{ \sqrt{1+sintcost} } dt= \int _0^ \frac{ \pi }{2}  \frac{cosx}{ \sqrt{1+4sinxcosx} } dx \Longrightarrow 2I= \int _0^ \frac{ \pi }{2}  \frac{sinx+cosx}{ \sqrt{1+4sinxcosx} } dx \Longrightarrow sin x-cosx=t \Longrightarrow [cosx +sinx]dx=dt \wedge sinx-cos x=t \Longrightarrow  [sin x-cosx]^{2} = t^{2}  \Longrightarrow 1-2sin xcosx= t^{2}  \Longrightarrow  2sin xcosx=1- t^{2}  \Longrightarrow 2I=  \int _ {-1} ^ {1} \frac{dt}{ \sqrt{3-2 t^{2} } }  \Longrightarrow 2I= \frac{1}{ \sqrt{3} }  \int _ {-1} ^ {1} \frac{dt}{ \sqrt{1- ( \sqrt{ \frac{2}{3} t}  )^{2} } } \Longrightarrow  2I=\frac{1}{ \sqrt{3} }   \int _ { -\sqrt{ \frac{2}{3} } } ^ { \sqrt{ \frac{2}{3} } }   \frac{1}{ \sqrt{1- y^{2} } }  \frac{ \sqrt{3} }{ \sqrt{2} } dy \Longrightarrow 2I=\frac{1}{ \sqrt{2} }  \int _ { -\sqrt{ \frac{2}{3} } } ^ { \sqrt{ \frac{2}{3} } } \frac{1}{ \sqrt{1- y^{2} } } dy \Longrightarrow 2I=\frac{1}{ \sqrt{3} } [Arcsin(y)]_ { -\sqrt{ \frac{2}{3} } } ^ { \sqrt{ \frac{2}{3} } }  = \frac{2}{ \sqrt{2} } Arcsin ( \sqrt{ \frac{2}{3} } ) \Longrightarrow I \Longrightarrow \frac{1}{ \sqrt{2} } Arcsin ( \sqrt{ \frac{2}{3} } )= \frac{1}{ \sqrt{2} } Arctan ( \sqrt{2} )$$

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$$x= \frac{ \pi }{2} -t \Longrightarrow I=- \int _ \frac{ \pi }{2} ^0 \frac{cost}{ \sqrt{1+sintcost} } dt= \int _0^ \frac{ \pi }{2}  \frac{cosx}{ \sqrt{1+4sinxcosx} } dx \Longrightarrow 2I= \int _0^ \frac{ \pi }{2}  \frac{sinx+cosx}{ \sqrt{1+4sinxcosx} } dx \Longrightarrow sin x-cosx=t \Longrightarrow [cosx +sinx]dx=dt \wedge sinx-cos x=t \Longrightarrow  [sin x-cosx]^{2} = t^{2}  \Longrightarrow 1-2sin xcosx= t^{2}  \Longrightarrow  2sin xcosx=1- t^{2}  \Longrightarrow 2I=  \int _ {-1} ^ {1} \frac{dt}{ \sqrt{3-2 t^{2} } }  \Longrightarrow 2I= \frac{1}{ \sqrt{3} }  \int _ {-1} ^ {1} \frac{dt}{ \sqrt{1- ( \sqrt{ \frac{2}{3} t}  )^{2} } } \Longrightarrow  2I=\frac{1}{ \sqrt{3} }   \int _ { -\sqrt{ \frac{2}{3} } } ^ { \sqrt{ \frac{2}{3} } }   \frac{1}{ \sqrt{1- y^{2} } }  \frac{ \sqrt{3} }{ \sqrt{2} } dy \Longrightarrow 2I=\frac{1}{ \sqrt{2} }  \int _ { -\sqrt{ \frac{2}{3} } } ^ { \sqrt{ \frac{2}{3} } } \frac{1}{ \sqrt{1- y^{2} } } dy \Longrightarrow 2I=\frac{1}{ \sqrt{3} } [Arcsin(y)]_ { -\sqrt{ \frac{2}{3} } } ^ { \sqrt{ \frac{2}{3} } }  = \frac{2}{ \sqrt{2} } Arcsin ( \sqrt{ \frac{2}{3} } ) \Longrightarrow I \Longrightarrow \frac{1}{ \sqrt{2} } Arcsin ( \sqrt{ \frac{2}{3} } )= \frac{1}{ \sqrt{2} } Arctan ( \sqrt{2} ) $$

Answer 1

$$x= \frac{ \pi }{2} -t \Longrightarrow I=- \int _ \frac{ \pi }{2} ^0 \frac{cost}{ \sqrt{1+sintcost} } dt= \int _0^ \frac{ \pi }{2} \frac{cosx}{ \sqrt{1+4sinxcosx} } dx \Longrightarrow 2I= \int _0^ \frac{ \pi }{2} \frac{sinx+cosx}{ \sqrt{1+4sinxcosx} } dx \Longrightarrow sin x-cosx=t \Longrightarrow [cosx +sinx]dx=dt \wedge sinx-cos x=t \Longrightarrow [sin x-cosx]^{2} = t^{2} \Longrightarrow 1-2sin xcosx= t^{2} \Longrightarrow 2sin xcosx=1- t^{2} \Longrightarrow 2I= \int _ {-1} ^ {1} \frac{dt}{ \sqrt{3-2 t^{2} } } \Longrightarrow 2I= \frac{1}{ \sqrt{3} } \int _ {-1} ^ {1} \frac{dt}{ \sqrt{1- ( \sqrt{ \frac{2}{3} t} )^{2} } } \Longrightarrow 2I=\frac{1}{ \sqrt{3} } \int _ { -\sqrt{ \frac{2}{3} } } ^ { \sqrt{ \frac{2}{3} } } \frac{1}{ \sqrt{1- y^{2} } } \frac{ \sqrt{3} }{ \sqrt{2} } dy \Longrightarrow 2I=\frac{1}{ \sqrt{2} } \int _ { -\sqrt{ \frac{2}{3} } } ^ { \sqrt{ \frac{2}{3} } } \frac{1}{ \sqrt{1- y^{2} } } dy \Longrightarrow 2I=\frac{1}{ \sqrt{3} } [Arcsin(y)]_ { -\sqrt{ \frac{2}{3} } } ^ { \sqrt{ \frac{2}{3} } } = \frac{2}{ \sqrt{2} } Arcsin ( \sqrt{ \frac{2}{3} } ) \Longrightarrow I \Longrightarrow \frac{1}{ \sqrt{2} } Arcsin ( \sqrt{ \frac{2}{3} } )= \frac{1}{ \sqrt{2} } Arctan ( \sqrt{2} ) $$