Key Concepts and Formulas

• Substitution Method for Definite Integrals: If $I = \int_a^b f(g(x)) g'(x) dx$, then by substituting $u = g(x)$, we get $du = g'(x) dx$. The new limits of integration will be $g(a)$ and $g(b)$, so $I = \int_{g(a)}^{g(b)} f(u) du$.
• Trigonometric Identities and Manipulations: Useful for transforming trigonometric expressions, such as $\tan x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$.
• Power Rule of Integration: $\int u^n du = \frac{u^{n+1}}{n+1} + C$ for $n \neq -1$.

---

Step-by-Step Solution

Let the given integral be $I$. I = \int_\limits0^{\pi / 4} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x

Step 1: Transform the integrand into terms of $\tan x$ and $\sec^2 x$.

• Why this step? The integrand involves powers of $\sin x$ and $\cos x$. Dividing the numerator and denominator by a suitable power of $\cos x$ can convert the expression into a form involving $\tan x$ and $\sec x$, which are derivatives of each other or related to each other's derivatives. This prepares the integral for a substitution.
• How to do it: Divide the numerator and the denominator by $\cos^6 x$. This choice is motivated by the $(\cos^3 x + \sin^3 x)^2$ term in the denominator, which upon dividing by $\cos^3 x$ inside the parenthesis becomes $\cos^6 x$ when squared. I = \int_\limits0^{\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}} d x Simplify the numerator: $\frac{\cos ^2 x \sin ^2 x}{\cos^6 x} = \frac{\sin ^2 x}{\cos^4 x} = \left(\frac{\sin x}{\cos x}\right)^2 \left(\frac{1}{\cos x}\right)^2 = \tan^2 x \sec^2 x$ Simplify the denominator: $\frac{\left(\cos ^3 x+\sin ^3 x\right)^2}{\cos^6 x} = \left(\frac{\cos ^3 x+\sin ^3 x}{\cos^3 x}\right)^2 = \left(\frac{\cos^3 x}{\cos^3 x} + \frac{\sin^3 x}{\cos^3 x}\right)^2 = (1 + \tan^3 x)^2$ Substitute these back into the integral: I = \int_\limits0^{\pi / 4} \frac{\tan^2 x \sec^2 x}{(1+\tan^3 x)^2} d x

Step 2: Perform the first substitution: $t = \tan x$.

• Why this step? The integrand now has $\tan x$ and $\sec^2 x$, and we know that the derivative of $\tan x$ is $\sec^2 x$. This makes $t = \tan x$ a suitable substitution, with $dt = \sec^2 x \, dx$.
• How to do it:
  • Let $t = \tan x$.
  • Then, $dt = \sec^2 x \, dx$.
  • Change the limits of integration:
    • When $x = 0$, $t = \tan(0) = 0$.
    • When $x = \pi/4$, $t = \tan(\pi/4) = 1$. Substitute these into the integral: I = \int_\limits0^1 \frac{t^2}{(1+t^3)^2} dt

Step 3: Perform the second substitution: $z = 1+t^3$.

• Why this step? The integral is now in the form $\int \frac{t^2}{(1+t^3)^2} dt$. The derivative of $1+t^3$ is $3t^2$, which is proportional to the $t^2$ term in the numerator. This suggests another substitution to simplify the denominator.
• How to do it:
  • Let $z = 1+t^3$.
  • Then, $dz = 3t^2 \, dt$, which implies $t^2 \, dt = \frac{1}{3} dz$.
  • Change the limits of integration:
    • When $t = 0$, $z = 1+(0)^3 = 1$.
    • When $t = 1$, $z = 1+(1)^3 = 2$. Substitute these into the integral: I = \int_\limits1^2 \frac{1}{z^2} \left(\frac{1}{3} dz\right) I = \frac{1}{3} \int_\limits1^2 z^{-2} dz

Step 4: Integrate and evaluate the definite integral.

• Why this step? The integral is now in a standard form that can be solved using the power rule of integration.
• How to do it: Apply the power rule $\int z^n dz = \frac{z^{n+1}}{n+1}$. \begin{aligned} I &= \frac{1}{3} \left[ \frac{z^{-2+1}}{-2+1} \right]_\limits1^2 \\ &= \frac{1}{3} \left[ \frac{z^{-1}}{-1} \right]_\limits1^2 \\ &= \frac{1}{3} \left[ -\frac{1}{z} \right]_\limits1^2 \end{aligned} Now, evaluate using the limits of integration: $\begin{aligned} I &= -\frac{1}{3} \left( \frac{1}{2} - \frac{1}{1} \right) \\ &= -\frac{1}{3} \left( \frac{1}{2} - 1 \right) \\ &= -\frac{1}{3} \left( -\frac{1}{2} \right) \\ &= \frac{1}{6}\end{aligned}$

The final answer is $\frac{1}{6}$.

---

Common Mistakes & Tips

• Forgetting to Change Limits: When performing a substitution in a definite integral, it is imperative to change the limits of integration to match the new variable. Failure to do so will result in an incorrect answer.
• Algebraic Errors in Trigonometric Manipulation: Be careful when dividing by powers of trigonometric functions and simplifying terms like $\frac{\sin^2 x}{\cos^4 x}$.
• Missing Constant Factors: When $dz = k \cdot du$, remember to include the factor $\frac{1}{k}$ when substituting $du$.

---

Summary

The given definite integral was solved by first transforming the integrand into a form suitable for substitution, using trigonometric manipulations to introduce $\tan x$ and $\sec^2 x$. A sequence of two substitutions, first $t = \tan x$ and then $z = 1+t^3$, simplified the integral to a basic power function. Crucially, the limits of integration were adjusted for each substitution. Finally, the resulting integral was evaluated using the power rule of integration.

The final answer is $\boxed{1/6}$.