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Question Number 23796 by tapan das last updated on 06/Nov/17

∫((2sinx+3cosx)/(3sinx+4cosx)) dx

$$\int\frac{\mathrm{2sinx}+\mathrm{3cosx}}{\mathrm{3sinx}+\mathrm{4cosx}}\:\mathrm{dx} \\ $$

Answered by ajfour last updated on 06/Nov/17

2sin x+3cos x=A(3sin x+4cos x)                               +B(3cos x−4sin x)  ⇒ 3A−4B=2  and        4A+3B=3  ⇒  A=((18)/(25))  , B=(1/(25))  ; So  ∫((2sin x+3cos x)/(3sin x+4cos x))dx =((18)/(25))∫dx+                                     +(1/(25))∫((3cos x−4sin x)/(3sin x+4cos x))dx     =((18x)/(25))+(1/(25))ln ∣3sin x+4cos x∣+C .

$$\mathrm{2sin}\:{x}+\mathrm{3cos}\:{x}={A}\left(\mathrm{3sin}\:{x}+\mathrm{4cos}\:{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{B}\left(\mathrm{3cos}\:{x}−\mathrm{4sin}\:{x}\right) \\ $$$$\Rightarrow\:\mathrm{3}{A}−\mathrm{4}{B}=\mathrm{2}\:\:{and} \\ $$$$\:\:\:\:\:\:\mathrm{4}{A}+\mathrm{3}{B}=\mathrm{3} \\ $$$$\Rightarrow\:\:{A}=\frac{\mathrm{18}}{\mathrm{25}}\:\:,\:{B}=\frac{\mathrm{1}}{\mathrm{25}}\:\:;\:{So} \\ $$$$\int\frac{\mathrm{2sin}\:{x}+\mathrm{3cos}\:{x}}{\mathrm{3sin}\:{x}+\mathrm{4cos}\:{x}}{dx}\:=\frac{\mathrm{18}}{\mathrm{25}}\int{dx}+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{25}}\int\frac{\mathrm{3cos}\:{x}−\mathrm{4sin}\:{x}}{\mathrm{3sin}\:{x}+\mathrm{4cos}\:{x}}{dx} \\ $$$$\:\:\:=\frac{\mathrm{18}{x}}{\mathrm{25}}+\frac{\mathrm{1}}{\mathrm{25}}\mathrm{ln}\:\mid\mathrm{3sin}\:{x}+\mathrm{4cos}\:{x}\mid+{C}\:. \\ $$

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