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

Answer: (B)

$\frac{\pi}{12}$

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:

$\frac{\pi}{12}$