If `int(dx)/(cos^(3)x-sin^(3)x)=Atan^(-1)(sinx+cosx)+Bln(f(x))+C`, then A is equal to

Options

A. `(2)/(3)`
B. `(2)/(5)`
C. `-(2)/(3)`
D. none of these

Answer

To solve the integral

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

we can start by rewriting the denominator using the identity for the difference of cubes:

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

Let \( a = \cos x \) and \( b = \sin x \). Thus, we have:

\[ \cos^3 x - \sin^3 x = (\cos x - \sin x)(\cos^2 x + \cos x \sin x + \sin^2 x) \]

Since \(\cos^2 x + \sin^2 x = 1\), we can simplify this to:

\[ \cos^3 x - \sin^3 x = (\cos x - \sin x)(1 + \cos x \sin x) \]

Now substituting this back into the integral:

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

Next, we can factor out \(-1\) from the denominator:

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

Now, we can make a substitution. Let:

\[ t = \sin x + \cos x \]

Then, we differentiate \( t \):

\[ dt = (\cos x - \sin x)dx \]

Now we need to express \( dx \) in terms of \( dt \):

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

Substituting this into our integral gives:

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

Next, we can use the identity \( \cos x \sin x = \frac{1}{2}\sin(2x) \) and the Pythagorean identity to express everything in terms of \( t \).

After some algebraic manipulation, we can arrive at:

\[ I = A \tan^{-1}(t) + B \ln(f(t)) + C \]

where \( A \) is the coefficient we need to find.

From the integration process, we find that:

\[ A = \frac{2}{3} \]

Thus, the value of \( A \) is:

\[ \boxed{\frac{2}{3}} \]