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cos-x-4-x-2-cos-x-4-x-2-

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Question Number 8728 by kuldeep singh raj last updated on 24/Oct/16
∫cos x/4−x^2 ∫((cos x)/(4−x^2 ))
∫cosx/4−x2∫cosx4−x2
Commented by prakash jain last updated on 24/Oct/16
Is it? ∫ ((cos x)/(4−x^2 )) or ∫ ((cos x)/4)−x^2
Isit?∫cosx4−x2or∫cosx4−x2
Commented by prakash jain last updated on 25/Oct/16
∫ ((cos x)/(4−x^2 )) =(1/4)[∫ ((cos x)/(2−x))dx+∫ ((cos x)/(2+x))dx] ∫ ((cos x)/(2−x))dx 2−x=u du=−dx =−∫((cos (2−u))/u)du =−∫((cos 2cos u+sin 2sin u)/u)du =−cos 2∫((cos u)/u)du−sin 2∫((sin u)/u)du =−cos 2∙Ci(u)−sin 2∙Si(u) =−cos 2Ci(2−x)−sin 2∙Si(2−x)+C Similarly ∫((cos x)/(2+x))dx; x+2=u, dx=du ∫((cos (u−2))/u)du=∫((cos 2cos u)/u)du+∫((sin 2sin u)/u)du =cos 2∙Ci(2+x)+sin 2∙Si(2+x)+C Hence ∫((cos x)/(4−x^2 ))dx=cos 2{Ci(2+x)−Ci(2−x)}+ sin 2{Si(x+2)−Si(2−x)}+C Note ∫((sin x)/x)dx=Si(x) ∫((cos x)/x)dx=Ci(x)
∫cosx4−x2=14[∫cosx2−xdx+∫cosx2+xdx]∫cosx2−xdx2−x=udu=−dx=−∫cos(2−u)udu=−∫cos2cosu+sin2sinuudu=−cos2∫cosuudu−sin2∫sinuudu=−cos2⋅Ci(u)−sin2⋅Si(u)=−cos2Ci(2−x)−sin2⋅Si(2−x)+CSimilarly∫cosx2+xdx;x+2=u,dx=du∫cos(u−2)udu=∫cos2cosuudu+∫sin2sinuudu=cos2⋅Ci(2+x)+sin2⋅Si(2+x)+CHence∫cosx4−x2dx=cos2{Ci(2+x)−Ci(2−x)}+sin2{Si(x+2)−Si(2−x)}+CNote∫sinxxdx=Si(x)∫cosxxdx=Ci(x)

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