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

1 Answer

  • (i)Method of substitution :
  • Given f(x)dx can be transformed into another form by changing independent variable x to t by substituting x=g(t).
  • Consider $I=\int f(x)dx.$
  • Put x=g(t) so that $\frac{dx}{dt}=g'(t).$
  • $\Rightarrow $dx=g'(t)dt.
  • Thus $I=\int f(g(t).g'(t))dt.$
  • (ii)$\int \sin xdx=-\cos x.$
  • (iii)\int\cos xdx=\sin x+c.$
Given $I=\int \frac{2\cos x-3\sin x}{6\cos x+4\sin x}dx.$
$\;\;\;\;\qquad=\int \frac{2\cos x-3\sin x}{2(3\cos x+2\sin x)}dx.$
Put $3\cos x+2\sin x=t.$
$-3\sin x+2\cos x=dt.$
Now substituting for x and dx we get,
$I=\frac{1}{2}\int \frac{dt}{t} .$
On integrating we get,
$\frac{1}{2}[log |t|]+c.$
Substituting back for t we get,
$\int \frac{2\cos x-3\sin x}{6\cos x+4\sin x}dx=\frac{1}{2}log (3\cos x+2\sin x)+c.$


answered Jan 29, 2013 by sreemathi.v