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Integrate the function\[\frac{\sin x}{1+\cos x}\]

$\begin{array}{1 1} -\log(1+\cos x)+c. \\ -\log(1+\sin x)+c \\ \log(1+\cos x)+c \\ \log(1+\sin x)+c. \end{array}$

1 Answer

  • 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 $I=\int f(g(t).g'(t))dt.$
Given $I=\int \frac{\sin x}{1+\cos x}dx.$
Put $1+\cos x=t.$
$-\sin xdx=dt\Rightarrow \sin xdx=-dt.$
Now substituting for x and dx we get,
$I=\int\frac{-dt} {t}=-\int\frac{dt}{t}.$
On integrating we get,
$-log t+c.$
Substituting back for t we get,
$-log(1+\cos x)+c.$
Hence $\int \frac{\sin x}{1+\cos x}dx=-log(1+\cos x)+c.$


answered Jan 29, 2013 by sreemathi.v