Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

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

Can you answer this question?

1 Answer

0 votes
  • 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
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App