The value of `int_(pi//4)^(3pi//4)(phi)/(1+sinphi)d phi` is

Options

A. `pi "tan"(pi)/8`
B. `log "tan"(pi)/8`
C. `"tan"(pi)/8`
D. None of these

Answer

To solve the integral

\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\phi}{1 + \sin \phi} \, d\phi, \]

we can use the property of definite integrals. The property states that

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx. \]

In our case, \( a = \frac{\pi}{4} \) and \( b = \frac{3\pi}{4} \), so \( a + b = \pi \). Thus, we can write:

\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\phi}{1 + \sin \phi} \, d\phi = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi - \phi}{1 + \sin(\pi - \phi)} \, d\phi. \]

Since \( \sin(\pi - \phi) = \sin \phi \), we have:

\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi - \phi}{1 + \sin \phi} \, d\phi. \]

Now, we can express this integral as:

\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi}{1 + \sin \phi} \, d\phi - \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\phi}{1 + \sin \phi} \, d\phi. \]

Let’s denote the second integral as \( I \):

\[ I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi}{1 + \sin \phi} \, d\phi - I. \]

Adding \( I \) to both sides gives:

\[ 2I = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi}{1 + \sin \phi} \, d\phi. \]

Thus, we can solve for \( I \):

\[ I = \frac{1}{2} \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{\pi}{1 + \sin \phi} \, d\phi. \]

Next, we need to evaluate the integral \( \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1}{1 + \sin \phi} \, d\phi \). To simplify this integral, we can multiply the numerator and denominator by \( 1 - \sin \phi \):

\[ \int \frac{1 - \sin \phi}{(1 + \sin \phi)(1 - \sin \phi)} \, d\phi = \int \frac{1 - \sin \phi}{\cos^2 \phi} \, d\phi. \]

This gives us:

\[ \int \sec^2 \phi \, d\phi - \int \frac{\sin \phi}{\cos^2 \phi} \, d\phi = \tan \phi + \frac{1}{\cos \phi} + C. \]

Evaluating the definite integral from \( \frac{\pi}{4} \) to \( \frac{3\pi}{4} \):

\[ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1}{1 + \sin \phi} \, d\phi = \left[ \tan \phi - \ln |\sec \phi + \tan \phi| \right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}. \]

After evaluating the limits, we find:

\[ \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{1}{1 + \sin \phi} \, d\phi = \frac{\pi}{2}. \]

Finally, we substitute back into our equation for \( I \):

\[ I = \frac{1}{2} \cdot \pi \cdot \frac{\pi}{2} = \frac{\pi^2}{4}. \]

Thus, the value of the integral is:

\[ \boxed{\frac{\pi^2}{4}}. \]