Subjective Type

Integrate the function $$\displaystyle \frac {2\cos x-3 \sin x}{6 \cos x+4 \sin x}$$

Solution

Given $$\displaystyle \frac {2\cos x-3 \sin x}{6 \cos x+4 \sin x}=\frac {2\cos x-3 \sin x}{2(3 \cos x+2 \sin x)}$$

Let $$3 \cos x+2 \sin x=t$$

$$\therefore (-3 \sin x+2 \cos x)dx=dt$$

$$\displaystyle \int \frac {2 \cos x-3 \sin x}{6 \cos x+4 \sin x}dx=\int \frac {dt}{2t}$$

$$\displaystyle =\frac {1}{2}\int \frac {1}{t}dt$$

$$\displaystyle =\frac {1}{2}\log |t|+C$$

$$\displaystyle =\frac {1}{2}\log |2 \sin x+3 \cos x|+C$$

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