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Q24  Integrate the functions  \frac{2 \cos x - 3\sin x }{6 \cos x + 4 \sin x }

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Given function  \frac{2 \cos x - 3\sin x }{6 \cos x + 4 \sin x },

or simplified as \frac{2 \cos x - 3\sin x }{2(3 \cos x + 2 \sin x) }

Assume the 3\cos x +2\sin x =t

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

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

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

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

Now, back substituted the value of t.

= \frac{1}{2}\log|3\cos x +2\sin x| +C, where C is any constant value.

Posted by

Divya Prakash Singh

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