Question

The value of \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \tan^{18}x}\) is:

• A. \(\frac{\pi}{4}\)
• B. \(\frac{\pi}{12}\)
• C. \(\frac{\pi}{9}\)
• D. \(\frac{\pi}{6}\)

Answer 1

The Correct Option is B

The given integral is:

\[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{1+\tan^{18}x}.\]

Let us simplify the denominator \(1+\tan^{18}x\). The integral's limits are symmetric about \(\frac{\pi}{4}\), and we use the property of symmetry for trigonometric functions:

\[f(x)=\frac{1}{1+\tan^{18}x}\]

Using the property \(\tan(\frac{\pi}{2}-x)=\cot(x)\):

\[f(\frac{\pi}{2}-x)=\frac{1}{1+\cot^{18}x}.\]

Adding both expressions:

\[f(x)+f(\frac{\pi}{2}-x)=\frac{1}{1+\tan^{18}x}+\frac{1}{1+\cot^{18}x}.\]

Simplify \(1+\cot^{18}x=\frac{\tan^{18}x+1}{\tan^{18}x}\):

\[f(x)+f(\frac{\pi}{2}-x)=\frac{\tan^{18}x}{\tan^{18}x+1}+\frac{1}{\tan^{18}x+1}=1.\]

Thus, the integral becomes:

\[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{1+\tan^{18}x}=\frac{1}{2}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}1~dx.\]

Evaluate the integral:

\[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}1~dx=[x]_{\frac{\pi}{6}}^{\frac{\pi}{3}}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}.\]

Thus:

\[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{dx}{1+\tan^{18}x}=\frac{1}{2}\times \frac{\pi}{6}=\frac{\pi}{12}.\]

Final Answer:

\[\frac{\pi}{12}\]