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Evaluate :$\int \large\frac{\sin 2x}{a^2\sin^2x+b^2\cos^2x}$$dx$

1 Answer

  • Method of substitution:
  • Given $\int f(x)dx$ 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 $\frac{dx}{dx}=g'(t).$
  • dx=g'(t)dt.
  • Thus $ I=\int f(g(t).g'(t))dt.$
  • $\int \sqrt{a^2+x^2}dx=\large\frac{x}{2}$$\sqrt{x^2+a^2}+\large\frac{a^2}{2}$$\log \mid x+\sqrt{x^2+a^2}\mid+c$
Step 1:
Let $I=\int \large\frac{\sin 2x}{a^2\sin^2 x+b^2\cos^2x}$$dx$
Put $a^2\sin^2x+b^2\cos^2x=t$
Differentiating with respect to $x$
$(2a^2\sin x\cos x-2b^2\sin x\cos x)dx=dt$
$\Rightarrow dx=\large\frac{dt}{2\sin x\cos x(a^2-b^2)}$
Substituting this we get,
$I=\int \large\frac{\sin 2x}{t}.\frac{dt}{2\sin x\cos x(a^2-b^2)}$
But $2\sin x\cos x=\sin 2x$
Step 2:
$I=\int \large\frac{dt}{t(a^2-b^2)}$
On integrating we get,
$\;\;=\large\frac{1}{a^2-b^2}$$\log \mid t\mid+c$
Substituting for t we get,
$I=\large\frac{1}{a^2-b^2}$$\log \mid a^2\sin^2 x+b^2\cos ^2 x\mid+c$
answered Sep 20, 2013 by sreemathi.v
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