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1.	Evaluate: \(\int\frac{3sinx}{3 + 4cosx}dx\)
Answer» \(\int\frac{3sinx}{3 + 4cosx}dx\) = \(\int\frac{3\frac{dt}{-4}}{t} = \frac{-3}{4}\int\frac{1}{t}dt\) = \(\frac{-3}{4}\)log t + c = \(\frac{-3}{4}\)log(3 + 4 cos x) + c put 3 + 4 cos x = t – 4 sinx dx = dt sin x dx = \(\frac{dt}{-4}\)

Evaluate: \(\int\frac{3sinx}{3 + 4cosx}dx\)

Answer»

\(\int\frac{3sinx}{3 + 4cosx}dx\)

= \(\int\frac{3\frac{dt}{-4}}{t} = \frac{-3}{4}\int\frac{1}{t}dt\)

= \(\frac{-3}{4}\)log t + c

= \(\frac{-3}{4}\)log(3 + 4 cos x) + c

put 3 + 4 cos x = t

– 4 sinx dx = dt

sin x dx = \(\frac{dt}{-4}\)

Discussion