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3.2(A) · Q57

Q.Integrate: ∫12+3tan⁡x dx\int \dfrac{1}{2+3\tan x}\,dx

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Rewrite 12+3tan⁡x=cos⁡x2cos⁡x+3sin⁡x\dfrac{1}{2+3\tan x}=\dfrac{\cos x}{2\cos x+3\sin x}.

Write cos⁡x=A(2cos⁡x+3sin⁡x)+B(−2sin⁡x+3cos⁡x)\cos x=A(2\cos x+3\sin x)+B(-2\sin x+3\cos x), the second bracket being the derivative of the denominator.

Matching: 2A+3B=12A+3B=1 (cos), 3A−2B=03A-2B=0 (sin). Solving gives A=213A=\dfrac{2}{13}, B=313B=\dfrac{3}{13}. …

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