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Tuesday, September 2, 2014

2014/074) integrate \dfrac{\sin\ x}{a \cos \ x + b\sin\ x}

Back ground information
∫ \dfrac{\sin\ x}{\sin \ x + \ cos\ x} dx
= ∫\dfrac{1}{2} \dfrac{2\sin\ x}{\sin \ x + \ cos\ x} dx
= ∫\dfrac{1}{2} \dfrac{(\sin\ x+\cos \ x) +(\sin\ x-\cos \ x) }{\sin \ x + \ cos\ x} dx
= ∫\dfrac{1}{2} (1+ \dfrac{\sin\ x-\cos \ x }{\sin \ x + \ cos\ x} dx
=  ∫\dfrac{1}{2}dx + \dfrac{1}{2} \dfrac{\sin\ x-\cos \ x }{\sin \ x + \ cos\ x} dx
= \dfrac{1}{2}x - \dfrac{1}{2} ln \mid (\sin \ x + \ cos\ x)\mid 

as \dfrac{\cos\ x}{\sin \ x + \ cos\ x} 
1-  \dfrac{\sin\ x}{\sin \ x + \ cos\ x} 

so ∫ \dfrac{\cos \ x}{\sin \ x + \ cos\ x} dx
\dfrac{1}{2}x + \dfrac{1}{2} ln \mid (\sin \ x + \ cos\ x)\mid 
Now integral

now we need to inetgrate
\dfrac{\sin\ x}{a \cos \ x + b\ sin\  x}

as derivate of  a \cos \ x + b\sin \  x is - a \sin  \ x + b\cos \   x

we can put  \sin\ x= m(a \cos \ x + b\ sin\  x) + n (- a \sin  \ x + b\cos \   x)

comparing coefficients we get 1 = ma + nb and 0=mb-na

solving these 2 we get

m=\dfrac{b}{b^2+a^2} and n=  \dfrac{- a}{b^2+a^2}

so  \sin\ x=\dfrac{b}{b^2+a^2}(a \cos \ x + b\ sin\  x) - \dfrac{a}{b^2+a^2} (- a \sin  \ x + b\cos \   x)

so \dfrac{\sin\ x}{a \ cos \ x + b\sin \ x }= \dfrac{b}{b^2+a^2} - \dfrac{a}{b^2+a^2} \dfrac{- a \sin  \ x + b\cos \   x} {a \cos \ x + b\sin \ x }

so integral of above = \dfrac{b}{b^2+a^2} x - \dfrac{a}{b^2+a^2} ln \mid a \cos \ x + b\sin \ x \mid

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