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 will use the identity for the difference of cubes:

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

where \( a = \cos x \) and \( b = \sin x \). Therefore, we can rewrite the denominator:

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

Since \( \cos^2 x + \sin^2 x = 1 \), we have:

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

Now substituting this back into the integral, we get:

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

Next, we will perform a substitution. Let:

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

Then, differentiating \( t \):

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

We can also express \( \cos x \sin x \) in terms of \( t \). We know:

\[ t^2 = \sin^2 x + \cos^2 x + 2\sin x \cos x = 1 + 2\sin x \cos x \]

This gives:

\[ \sin x \cos x = \frac{t^2 - 1}{2} \]

Now substitute \( \sin x \cos x \) back into the integral:

\[ I = \int \frac{dt}{t (1 + \frac{t^2 - 1}{2})} \]

Simplifying the denominator:

\[ 1 + \frac{t^2 - 1}{2} = \frac{2 + t^2 - 1}{2} = \frac{t^2 + 1}{2} \]

Thus, we have:

\[ I = \int \frac{2 \, dt}{t (t^2 + 1)} \]

Now we can separate this into partial fractions:

\[ \frac{2}{t(t^2 + 1)} = \frac{A}{t} + \frac{Bt + C}{t^2 + 1} \]

Multiplying through by the denominator \( t(t^2 + 1) \):

\[ 2 = A(t^2 + 1) + (Bt + C)t \]

Expanding and collecting like terms, we can solve for \( A \), \( B \), and \( C \).

Setting \( t = 0 \):

\[ 2 = A(0^2 + 1) \implies A = 2 \]

Now, we can find \( B \) and \( C \) by substituting other convenient values for \( t \) or by comparing coefficients.

After finding \( A \), \( B \), and \( C \), we can integrate each term separately. The integral of \( \frac{2}{t} \) gives \( 2 \ln |t| \), and the integral of \( \frac{Bt + C}{t^2 + 1} \) gives \( B \tan^{-1}(t) + C \ln(t^2 + 1) \).

Finally, we can express the result in the form:

\[ I = 2 \ln |\sin x + \cos x| + B \tan^{-1}(\sin x + \cos x) + C \]

From the problem statement, we know that:

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

Thus, we find that \( A = 2 \).

### Final Answer:
\[ A = 2 \]