Browse Questions

# Integrate : $\int \large\frac{\sin 2x }{a \cos ^2 x + b \sin ^2 x }$$dx (a)\;\frac{1}{b-a} \log _e |(a \cos ^2 x + b \sin ^2 x)|+c \qquad(b)\;\frac{1}{b-a} \log _e |(a \sin ^2 x + b \cos ^2 x)|+c \qquad(c)\;\frac{1}{b-a} \log _e |(a \sin ^2 x + b \sin ^2 x)|+c \qquad (d)\;None Can you answer this question? ## 1 Answer 0 votes Toolbox: • \int \large\frac{1}{x}$$.dx=log_e |x|$
Given: $I=\int \large\frac{\sin 2x }{a \cos ^2 x + b \sin ^2 x }$$dx Let a\cos ^2 x + b \sin ^2 x=t differentiating both the sides \Rightarrow\:(-2a\:cosx sinx+2b\:sinxcosx)dx=dt \Rightarrow\:2sinxcosx.(b-a)dx=dt But 2sinxcosx=sin2x \therefore\:(b-a).sin2x\:dx=dt \Rightarrow\:sin2x\:dx=\large\frac{1}{b-a}$$.dt$
Substituting the values of $t$ and $dx$ in $I$ we get
$\Rightarrow\:I=\int \large\frac{1}{(b-a)} \times \frac{1}{t} $$.dt \Rightarrow\:I=\large\frac{1}{(b-a)}$$\int \large\frac{1}{t}$$dt=\large\frac{1}{b-a}$$log_e|t|+c$
Since $\int \large\frac{1}{t}$$.dt=log_e |t| \Rightarrow\:I=\large\frac{1}{b-a}$$ \log _e |(a \cos ^2 x + b \sin ^2 x)|+c$
Hence (a) is the correct answer.
edited Mar 13, 2014