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Evaluate : $ \int\large \frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}$$dx$

1 Answer

Toolbox:
  • 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 $I=\int f(g(t).g'(t))dt.$
  • $\int e^{ax}dx=\frac{1}{a}e^{ax}+c.$
Step 1:
Given $I=\int \large\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}$$dx $.
Let $e^{2x}+e^{-2x}=t.$
$(2e^2x-2e^{-2x})dx=dt.$
$\Rightarrow 2(e^{2x}-e^{-2x})dx=dt.\Rightarrow (e^{2x}-e^{-2x})dx=\large\frac{dt}{2}.$
Step 2:
Now substituting for x and dx we get,
$I=\int\frac{dt}{2t}=\frac{1}{2}\int\frac{dt}{t}.$
On integrating we get,
$\large\frac{1}{2}$$\log|t|+c.$
Substituting back for t we get,
$\int \large\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}$$dx=\large\frac{1}{2}$$\log|e^{2x}+e^{-2x}|+c.$
answered Nov 11, 2013 by sreemathi.v
 
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