int frac{sin 7 x}{cos 9 x cos 2 x} ,d x is eq | vedclass

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(C) We have the integral $I = \int \frac{\sin 7 x}{\cos 9 x \cos 2 x} \,d x$. Since $7x = 9x - 2x$, we can write $\sin 7x = \sin(9x - 2x)$. Using the

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$\frac{1}{9} \log |\sec (9 x)| - \frac{1}{2} \log |\sec (2 x)| + c$, where $c$ is the constant of integration

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(C) We have the integral $I = \int \frac{\sin 7 x}{\cos 9 x \cos 2 x} \,d x$. Since $7x = 9x - 2x$, we can write $\sin 7x = \sin(9x - 2x)$. Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$, we get: $I = \int \frac{\sin 9x \cos 2x - \cos 9x \sin 2x}{\cos 9x \cos 2x} \,d x$. Splitting the fraction: $I = \int \left( \frac{\sin 9x \cos 2x}{\cos 9x \cos 2x} - \frac{\cos 9x \sin 2x}{\cos 9x \cos 2x} ight) \,d x$. $I = \int (	an 9x - 	an 2x) \,d x$. Integrating term by term: $I = \int 	an 9x \,d x - \int 	an 2x \,d x$. Using the formula $\int 	an(ax) \,d x = \frac{1}{a} \ln |\sec(ax)| + c$: $I = \frac{1}{9} \ln |\sec 9x| - \frac{1}{2} \ln |\sec 2x| + c$. Thus, the correct option is $C$.

$\int \frac{\sin 7 x}{\cos 9 x \cos 2 x} \,d x$ is equal to

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