Question
Evalute the following integrals:
$\int\frac{\sin2\text{x}}{\text{a}^2+\text{b}^2\sin^2\text{x}}\text{dx}$
$\int\frac{\sin2\text{x}}{\text{a}^2+\text{b}^2\sin^2\text{x}}\text{dx}$
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Answer
Let $\int\frac{\sin2\text{x}}{\text{a}^2+\text{b}^2\sin^2\text{x}}\text{dx}\ .....(\text{i})$
Let $\text{a}^2+\text{b}^2\sin^2\text{x}=\text{t}$ then,
$\text{d}\big(\text{a}^2+\text{b}^2\sin^2\text{x}\big)=\text{dt}$
$=\text{b}^2(2\sin\text{x}\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{\text{b}^2(2\sin\text{x}\cos\text{x})}$
$=\frac{\text{dt}}{\text{b}^2\sin2\text{x}}$
Putting $\text{a}^2+\text{b}^2\sin^2\text{x}=\text{t}$ and $\text{dx}=\frac{\text{dt}}{\text{b}^2\sin2\text{x}}$ in equation (i), we get,
$\text{I}=\int\frac{\sin2\text{x}}{\text{t}}\times\frac{\text{dt}}{\text{b}^2\sin2\text{x}}$
$=\frac{1}{\text{b}^2}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{\text{b}^2}\log|\text{t}|+\text{C}$
$=\frac{1}{\text{b}^2}\log|\text{a}^2+\text{b}^2\sin^2\text{x}|+\text{C}$
$\Rightarrow\text{I}=\frac{1}{\text{b}^2}\log|\text{a}^2+\text{b}^2\sin^2\text{x}|+\text{C}$
Let $\text{a}^2+\text{b}^2\sin^2\text{x}=\text{t}$ then,
$\text{d}\big(\text{a}^2+\text{b}^2\sin^2\text{x}\big)=\text{dt}$
$=\text{b}^2(2\sin\text{x}\cos\text{x})\text{dx}=\text{dt}$
$\Rightarrow\text{dx}=\frac{\text{dt}}{\text{b}^2(2\sin\text{x}\cos\text{x})}$
$=\frac{\text{dt}}{\text{b}^2\sin2\text{x}}$
Putting $\text{a}^2+\text{b}^2\sin^2\text{x}=\text{t}$ and $\text{dx}=\frac{\text{dt}}{\text{b}^2\sin2\text{x}}$ in equation (i), we get,
$\text{I}=\int\frac{\sin2\text{x}}{\text{t}}\times\frac{\text{dt}}{\text{b}^2\sin2\text{x}}$
$=\frac{1}{\text{b}^2}\int\frac{\text{dt}}{\text{t}}$
$=\frac{1}{\text{b}^2}\log|\text{t}|+\text{C}$
$=\frac{1}{\text{b}^2}\log|\text{a}^2+\text{b}^2\sin^2\text{x}|+\text{C}$
$\Rightarrow\text{I}=\frac{1}{\text{b}^2}\log|\text{a}^2+\text{b}^2\sin^2\text{x}|+\text{C}$
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