Q.Evaluate: $\int \dfrac{\sin^6 x+\cos^6 x}{\sin^2 x\,\cos^2 x}\,dx$

Concept: Sine Double Angle Integration — rewrite the numerator using the identity $a^3+b^3=(a+b)(a^2-ab+b^2)$.

First, note $\sin^6 x+\cos^6 x = (\sin^2 x)^3+(\cos^2 x)^3$.

Factor: $(\sin^2 x+\cos^2 x)(\sin^4 x-\sin^2 x\cos^2 x+\cos^4 x) = 1\cdot[(\sin^2 x+\cos^2 x)^2-3\sin^2 x\cos^2 x] = 1-3\sin^2 x\cos^2 x$.

Thus the integrand becomes:

$\frac{1-3\sin^2 x\cos^2 x}{\sin^2 x\cos^2 x} = \frac{1}{\sin^2 x\cos^2 x} - 3.$ …

This problem looks messy at first — sixth powers of sine and cosine in the numerator, and only $\sin^2 x \cos^2 x$ in the denominator. But there's a beautiful simplification hiding in plain sight.

The core idea: $\sin^6 x + \cos^6 x$ is a sum of cubes. Recall that $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. Here $a = \sin^2 x$ and $b = \cos^2 x$, so:

Since $\sin^2 x + \cos^2 x = 1$, we get:

Now $\sin^4 x + \cos^4 x$ itself can be simplified. Write it as $(\sin^2 x)^2 + (\cos^2 x)^2$ and use $a^2 + b^2 = (a+b)^2 - 2ab$:

Substitute this back:

This is the master key. Now the integrand becomes:

Split it into two fractions:

Now we need to handle $\frac{1}{\sin^2 x \cos^2 x}$. Write it as:

That is:

So the entire integrand simplifies beautifully to: …

Method: Collapsing high trig powers with algebraic identities

Use this for integrands like $\dfrac{\sin^6 x+\cos^6 x}{\sin^2 x\cos^2 x}$, where an algebraic identity reduces the numerator and the fraction splits into standard $\sec^2$/$\csc^2$ pieces.

Steps

Step 1: Reduce the numerator with a sum-of-powers identity.

using $a^3+b^3=(a+b)(a^2-ab+b^2)$ with $a=\sin^2x,\,b=\cos^2x$.

Step 2: Divide by $\sin^2 x\cos^2 x$.

$\frac{1-3\sin^2x\cos^2x}{\sin^2x\cos^2x}=\frac{1}{\sin^2x\cos^2x}-3.$ …

Common Mistakes

Mistake 1: Not reducing $\sin^6+\cos^6$.

Why it's wrong: expanding sixth powers directly is error-prone; the identity $\sin^6+\cos^6=1-3\sin^2\cos^2$ is the clean route. Correct approach: use the sum-of-cubes factorisation.

Mistake 2: Mishandling $\tfrac{1}{\sin^2x\cos^2x}$.

Why it's wrong: it equals $\sec^2x+\csc^2x$ (via $\sin^2+\cos^2=1$ on top), not $\csc^2x\sec^2x$ left as-is. Correct approach: split using the Pythagorean identity in the numerator. …