If `int (dx )/(cos ^(3) x-sin ^(3))=A tan ^(-1) (f (x)) +bln |(sqrt2+f (x))/(sqrt2-f (x))|+C` where `f (x) = sin x + cos x` find the value of `(12A+9sqrt2V)-3.`

Answer

To solve the integral

\[ I = \int \frac{dx}{\cos^3 x - \sin^3 x} \]

we can use the identity for the difference of cubes:

\[ a^3 - b^3 = (a - b)(a^2 + b^2 + ab) \]

In our case, let \( a = \cos x \) and \( b = \sin x \). Thus, we can rewrite the integral as:

\[ I = \int \frac{dx}{(\cos x - \sin x)(\cos^2 x + \sin^2 x + \cos x \sin x)} \]

Using the Pythagorean identity \( \cos^2 x + \sin^2 x = 1 \), we simplify the expression to:

\[ I = \int \frac{dx}{(\cos x - \sin x)(1 + \cos x \sin x)} \]

Next, we multiply the numerator and denominator by \( \cos x \sin x \):

\[ I = \int \frac{\cos x \sin x \, dx}{(\cos x - \sin x)(\cos x \sin x + \cos^2 x \sin^2 x)} \]

Now, we can rewrite the denominator using the identity:

\[ (\cos x - \sin x)^2 = \cos^2 x + \sin^2 x - 2\cos x \sin x = 1 - 2\cos x \sin x \]

Thus, we have:

\[ I = \int \frac{2(\cos x - \sin x) \, dx}{(1 - 2\cos x \sin x)(1 + \cos x \sin x)} \]

Now, we will substitute \( f(x) = \sin x + \cos x \). Hence, we differentiate \( f(x) \):

\[ f'(x) = \cos x - \sin x \]

This gives us:

\[ dx = \frac{df}{\cos x - \sin x} \]

Substituting this back into the integral, we have:

\[ I = \int \frac{2 \, df}{(1 - 2\cos x \sin x)(1 + \cos x \sin x)} \]

Next, we express \( 1 - 2\cos x \sin x \) and \( 1 + \cos x \sin x \) in terms of \( f(x) \):

\[ \cos x \sin x = \frac{1}{2}(\sin(2x)) = \frac{1}{2}(f^2 - 1) \]

Thus, we can rewrite the integral as:

\[ I = \int \frac{2 \, df}{(1 - (f^2 - 1))(1 + \frac{1}{2}(f^2 - 1))} \]

This simplifies to:

\[ I = \int \frac{2 \, df}{(2 - f^2)(\frac{3}{2} + \frac{1}{2}f^2)} \]

Now, we can separate the integral into two parts and integrate:

\[ I = \frac{2}{3} \tan^{-1}(f(x)) + \frac{1}{2\sqrt{2}} \ln \left| \frac{\sqrt{2} + f(x)}{\sqrt{2} - f(x)} \right| + C \]

From the given equation \( I = A \tan^{-1}(f(x)) + B \ln \left| \frac{\sqrt{2} + f(x)}{\sqrt{2} - f(x)} \right| + C \), we can identify:

\[ A = \frac{2}{3}, \quad B = \frac{1}{2\sqrt{2}} \]

Now, we need to find the value of \( 12A + 9\sqrt{2}B - 3 \):

Calculating \( 12A \):

\[ 12A = 12 \cdot \frac{2}{3} = 8 \]

Calculating \( 9\sqrt{2}B \):

\[ 9\sqrt{2}B = 9\sqrt{2} \cdot \frac{1}{2\sqrt{2}} = \frac{9}{2} \]

Now substituting these values into the expression:

\[ 12A + 9\sqrt{2}B - 3 = 8 + \frac{9}{2} - 3 \]

Converting \( 8 \) and \( -3 \) to halves:

\[ = \frac{16}{2} + \frac{9}{2} - \frac{6}{2} = \frac{16 + 9 - 6}{2} = \frac{19}{2} \]

Thus, the final answer is:

\[ \frac{19}{2} \]