`int_(0)^(pi//2)(1)/(1+tan^(3)x)dx=`

Options

A. 0
B. 1
C. `(pi)/(2)`
D. `(pi)/(4)`

Answer

To solve the definite integral

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{1}{1 + \tan^3 x} \, dx, \]

we can use a property of definite integrals. This property states that

\[ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a - x) \, dx. \]

In our case, we set \( a = \frac{\pi}{2} \). Thus, we can write:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{1}{1 + \tan^3\left(\frac{\pi}{2} - x\right)} \, dx. \]

Using the identity \( \tan\left(\frac{\pi}{2} - x\right) = \cot x \), we have:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{1}{1 + \cot^3 x} \, dx. \]

Now, we can express \( \cot^3 x \) in terms of \( \tan x \):

\[ \cot^3 x = \frac{1}{\tan^3 x}. \]

Thus, we can rewrite the integral as:

\[ I = \int_{0}^{\frac{\pi}{2}} \frac{1}{1 + \frac{1}{\tan^3 x}} \, dx = \int_{0}^{\frac{\pi}{2}} \frac{\tan^3 x}{\tan^3 x + 1} \, dx. \]

Now we have two expressions for \( I \):

1. \( I = \int_{0}^{\frac{\pi}{2}} \frac{1}{1 + \tan^3 x} \, dx \) (Equation 1)
2. \( I = \int_{0}^{\frac{\pi}{2}} \frac{\tan^3 x}{\tan^3 x + 1} \, dx \) (Equation 2)

Next, we can add these two equations:

\[ 2I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1}{1 + \tan^3 x} + \frac{\tan^3 x}{\tan^3 x + 1} \right) \, dx. \]

Simplifying the integrand:

\[ \frac{1}{1 + \tan^3 x} + \frac{\tan^3 x}{\tan^3 x + 1} = \frac{1 + \tan^3 x}{1 + \tan^3 x} = 1. \]

Thus, we have:

\[ 2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx = \left[ x \right]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} - 0 = \frac{\pi}{2}. \]

Now, dividing both sides by 2 gives:

\[ I = \frac{\pi}{4}. \]

Therefore, the value of the integral is:

\[ \int_{0}^{\frac{\pi}{2}} \frac{1}{1 + \tan^3 x} \, dx = \frac{\pi}{4}. \]