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$$
\int \frac{\sin 2 x}{4 \sin ^2 x+9 \cos ^2 x} \mathrm{~d} x=
$$
(Where $C$ is a constant of integration).
Options:
\int \frac{\sin 2 x}{4 \sin ^2 x+9 \cos ^2 x} \mathrm{~d} x=
$$
(Where $C$ is a constant of integration).
Solution:
1740 Upvotes
Verified Answer
The correct answer is:
$-\frac{1}{5} \log \left(4 \sin ^2 x+9 \cos ^2 x\right)+C$
Let $4 \sin ^2 x+9 \cos ^2 x=t$
$\begin{aligned} & \Rightarrow(4 \times 2 \sin x \cdot \cos x-9 \times 2 \cos x \sin x) \mathrm{d} x=\mathrm{d} t \\ & \Rightarrow-5 \sin 2 x \mathrm{~d} x=\mathrm{d} t \\ & \Rightarrow-\frac{1}{5} \int \frac{\mathrm{d} t}{t}=-\log |t|+c \\ & =-\frac{1}{5} \log \left|4 \sin ^2 x+9 \cos ^2 x\right|+c\end{aligned}$
$\begin{aligned} & \Rightarrow(4 \times 2 \sin x \cdot \cos x-9 \times 2 \cos x \sin x) \mathrm{d} x=\mathrm{d} t \\ & \Rightarrow-5 \sin 2 x \mathrm{~d} x=\mathrm{d} t \\ & \Rightarrow-\frac{1}{5} \int \frac{\mathrm{d} t}{t}=-\log |t|+c \\ & =-\frac{1}{5} \log \left|4 \sin ^2 x+9 \cos ^2 x\right|+c\end{aligned}$
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