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# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x} d x$

Solution:

Taking 2 common in denominator we get

$\Rightarrow \int \frac{2 \cos x-3 \sin x}{2(3 \cos x+2 \sin x)} d x$

Now assume

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

$(-3 \sin x+2 \cos x) d x=d t$

Put $t$ and dt in given equation we get

$\Rightarrow \frac{1}{2} \int \frac{\mathrm{dt}}{\mathrm{t}}$

$=\frac{1}{2} \ln |\mathrm{t}|+\mathrm{c}$

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

$=\frac{1}{2} \ln |3 \cos x+2 \sin x|+c$