Question

\( \int_{0}^{\pi/4} \frac{\cos^2 x \sin^2 x}{\left( \cos^3 x + \sin^3 x \right)^2} \, dx \) is equal to:

• A. \( \frac{1}{12} \)
• B. \( \frac{1}{9} \)
• C. \( \frac{1}{6} \)
• D. \( \frac{1}{3} \)

Answer 1

The Correct Option is C

Divide the numerator and denominator by \(\cos x\):

\[ \int_0^{\pi/4} \frac{\tan^2 x \sec^2 x \, dx}{(1 + \tan^3 x)^2}. \]

Let \(1 + \tan^3 x = t\). Then:

\[ \tan^2 x \sec^2 x \, dx = \frac{dt}{3}. \]

The limits transform as:

• When \(x = 0\), \(t = 1\),
• and when \(x = \frac{\pi}{4}\), \(t = 2\).

Substitute into the integral:

\[ \int_0^{\pi/4} \frac{\tan^2 x \sec^2 x \, dx}{(1 + \tan^3 x)^2} = \frac{1}{3} \int_1^2 \frac{dt}{t^2}. \]

Solve the integral:

\[ \frac{1}{3} \int_1^2 t^{-2} \, dt = \frac{1}{3} \left[ -\frac{1}{t} \right]_1^2. \]

Simplify:

\[ \frac{1}{3} \left[ -\frac{1}{2} - (-1) \right] = \frac{1}{3} \left[ -\frac{1}{2} + 1 \right] = \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}. \]

Answer 2

The problem is to evaluate the definite integral \( I = \int_{0}^{\pi/4} \frac{\cos^2 x \sin^2 x}{\left( \cos^3 x + \sin^3 x \right)^2} \, dx \).

Concept Used:

This integral can be solved using the method of substitution. The key steps are:

1. Algebraically manipulate the integrand. A common technique for trigonometric functions is to divide the numerator and denominator by a high power of \( \cos x \) to convert the expression into terms of \( \tan x \) and \( \sec^2 x \).

2. Perform a u-substitution, typically by setting \( u \) equal to an expression involving \( \tan x \).

3. Change the limits of integration to correspond to the new variable \( u \).

4. Evaluate the transformed integral, usually with the power rule for integration: \( \int u^n \, du = \frac{u^{n+1}}{n+1} \).

Step-by-Step Solution:

Step 1: To simplify the integrand, divide both the numerator and the denominator by \( \cos^6 x \). This is chosen because the denominator contains a term \( (\cos^3 x)^2 = \cos^6 x \).

\[ I = \int_{0}^{\pi/4} \frac{\frac{\cos^2 x \sin^2 x}{\cos^6 x}}{\frac{\left( \cos^3 x + \sin^3 x \right)^2}{\cos^6 x}} \, dx \]

Step 2: Simplify the numerator and denominator separately.

Numerator:

\[ \frac{\cos^2 x \sin^2 x}{\cos^6 x} = \frac{\sin^2 x}{\cos^4 x} = \frac{\sin^2 x}{\cos^2 x} \cdot \frac{1}{\cos^2 x} = \tan^2 x \sec^2 x \]

Denominator:

\[ \frac{\left( \cos^3 x + \sin^3 x \right)^2}{\cos^6 x} = \left( \frac{\cos^3 x + \sin^3 x}{\cos^3 x} \right)^2 = \left( 1 + \frac{\sin^3 x}{\cos^3 x} \right)^2 = (1 + \tan^3 x)^2 \]

Step 3: Rewrite the integral with the simplified terms.

\[ I = \int_{0}^{\pi/4} \frac{\tan^2 x \sec^2 x}{(1 + \tan^3 x)^2} \, dx \]

Step 4: Perform a u-substitution. Let \( u = 1 + \tan^3 x \).

Step 5: Differentiate \( u \) with respect to \( x \) to find \( du \).

\[ \frac{du}{dx} = \frac{d}{dx}(1 + \tan^3 x) = 3 \tan^2 x \cdot \frac{d}{dx}(\tan x) = 3 \tan^2 x \sec^2 x \]

This gives us \( du = 3 \tan^2 x \sec^2 x \, dx \), which means \( \tan^2 x \sec^2 x \, dx = \frac{du}{3} \).

Step 6: Change the limits of integration based on the substitution for \( u \).

When \( x = 0 \) (lower limit):

\[ u = 1 + \tan^3(0) = 1 + 0 = 1 \]

When \( x = \pi/4 \) (upper limit):

\[ u = 1 + \tan^3(\pi/4) = 1 + (1)^3 = 2 \]

Step 7: Substitute \( u \), \( du \), and the new limits into the integral.

\[ I = \int_{1}^{2} \frac{1}{u^2} \left( \frac{du}{3} \right) = \frac{1}{3} \int_{1}^{2} u^{-2} \, du \]

Final Computation & Result:

Step 8: Integrate with respect to \( u \) using the power rule.

\[ I = \frac{1}{3} \left[ \frac{u^{-1}}{-1} \right]_{1}^{2} = -\frac{1}{3} \left[ \frac{1}{u} \right]_{1}^{2} \]

Step 9: Apply the new limits of integration to find the final value.

\[ I = -\frac{1}{3} \left( \frac{1}{2} - \frac{1}{1} \right) = -\frac{1}{3} \left( \frac{1 - 2}{2} \right) = -\frac{1}{3} \left( -\frac{1}{2} \right) \] \[ I = \frac{1}{6} \]

Thus, the value of the integral is 1/6.