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

To solve the integral \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \tan^{18}x}\), we can use the symmetry property of definite integrals. Let's examine the function within the integrand: \(f(x) = \frac{1}{1 + \tan^{18}x}\). We look for properties that might simplify this. 

Consider the transformation \(x = \frac{\pi}{3} - t\). Consequently, \(dx = -dt\) and changing the limits transforms them as follows: when \(x = \frac{\pi}{6}, t = \frac{\pi}{6}\) and when \(x = \frac{\pi}{3}, t = 0\). Therefore:

\[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \tan^{18}x} = -\int_{\frac{\pi}{6}}^{0} \frac{dt}{1 + \tan^{18}(\frac{\pi}{3} - t)}\]

Reversing the limits gives:

\[\int_{0}^{\frac{\pi}{6}} \frac{dt}{1 + \tan^{18}(\frac{\pi}{3} - t)} = \int_{0}^{\frac{\pi}{6}} \frac{dt}{1 + \left(\frac{1}{\tan t}\right)^{18}}\]

This simplifies to:

\[\int_{0}^{\frac{\pi}{6}} \frac{\tan^{18}t}{1 + \tan^{18}t} dt\]

Now, consider the original integral:

\[\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{1 + \tan^{18}x} = \int_{0}^{\frac{\pi}{6}} \frac{dt}{1 + \tan^{18}t} + \int_{0}^{\frac{\pi}{6}} \frac{\tan^{18}t}{1 + \tan^{18}t} dt\]

When these are added, they form:

\[\int_{0}^{\frac{\pi}{6}} \left( \frac{1}{1 + \tan^{18}t} + \frac{\tan^{18}t}{1 + \tan^{18}t} \right) dt = \int_{0}^{\frac{\pi}{6}} dt = \frac{\pi}{6}\]

Thus:

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

The correct answer is \(\frac{\pi}{12}\).

Answer 2

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:

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