Question:
\(\int \frac{\sin 7 x}{\cos 9 x \cos 2 x} \mathrm{~d} x\) is equal to
\(\int \frac{\sin 7 x}{\cos 9 x \cos 2 x} \mathrm{~d} x\) is equal to
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Write sin 7x as sin(9x - 2x) and split the fraction into two tangent terms.
Updated On: Nov 13, 2026
\(\mathrm{logsec}(9x)-\mathrm{logsec}(2x)+\mathrm{c}\), where c is the constant of integration
\(\mathrm{logsec}(9x)+\mathrm{logsec}(2x)+\mathrm{c}\), where c is the constant of integration
\(\frac{1}{9}\mathrm{logsec}(9x)-\frac{1}{2}\mathrm{logsec}(2x)+\mathrm{c}\), where c is the constant of integration
\(\frac{1}{9}\mathrm{logsec}(9x)+\frac{1}{2}\mathrm{logsec}(2x)+\mathrm{c}\), where c is the constant of integration
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Since \(7x=9x-2x\), \(\sin7x=\sin9x\cos2x-\cos9x\sin2x\).
Step 2: Split the integrand
\(\dfrac{\sin7x}{\cos9x\cos2x}=\dfrac{\sin9x}{\cos9x}-\dfrac{\sin2x}{\cos2x}=\tan9x-\tan2x\)
Step 3: Calculation
\(\int\tan9x\,dx=\dfrac{1}{9}\log\sec9x\) and \(\int\tan2x\,dx=\dfrac{1}{2}\log\sec2x\).
The result is \(\dfrac{1}{9}\log\sec9x-\dfrac{1}{2}\log\sec2x+c\).
Final Answer:
The integral is \(\frac{1}{9}\log\sec9x-\frac{1}{2}\log\sec2x+c\). \[ \boxed{\text{(C) }\dfrac{1}{9}\log\sec(9x)-\dfrac{1}{2}\log\sec(2x)+c} \]
Since \(7x=9x-2x\), \(\sin7x=\sin9x\cos2x-\cos9x\sin2x\).
Step 2: Split the integrand
\(\dfrac{\sin7x}{\cos9x\cos2x}=\dfrac{\sin9x}{\cos9x}-\dfrac{\sin2x}{\cos2x}=\tan9x-\tan2x\)
Step 3: Calculation
\(\int\tan9x\,dx=\dfrac{1}{9}\log\sec9x\) and \(\int\tan2x\,dx=\dfrac{1}{2}\log\sec2x\).
The result is \(\dfrac{1}{9}\log\sec9x-\dfrac{1}{2}\log\sec2x+c\).
Final Answer:
The integral is \(\frac{1}{9}\log\sec9x-\frac{1}{2}\log\sec2x+c\). \[ \boxed{\text{(C) }\dfrac{1}{9}\log\sec(9x)-\dfrac{1}{2}\log\sec(2x)+c} \]
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