Q.Integrate the function $\frac{\sin^8 x-\cos^8 x}{1-2\sin^2 x\cos^2 x}$

Concept: Trigonometric Simplification — factorise the numerator using difference of squares, then simplify using identities.

First, factorise the numerator:

Since $\sin^2 x + \cos^2 x = 1$, this becomes $(\sin^2 x - \cos^2 x)(\sin^4 x + \cos^4 x)$.

Now rewrite $\sin^4 x + \cos^4 x$:

$\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x$ …

Why this approach works

When you see high powers of sine and cosine together, your first instinct should be to look for factorisation. The numerator $\sin^8 x - \cos^8 x$ is a difference of fourth powers, which itself is a difference of squares. The denominator $1 - 2\sin^2 x \cos^2 x$ looks suspiciously like something that might cancel with part of that factorisation — and indeed it does.

The key insight: $\sin^8 x - \cos^8 x = (\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)$. And $\sin^4 x - \cos^4 x$ is itself $(\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x) = (\sin^2 x - \cos^2 x) \cdot 1$. So the numerator contains a factor $\sin^2 x - \cos^2 x = -\cos 2x$.

Meanwhile, the denominator $1 - 2\sin^2 x \cos^2 x$ turns out to equal $\sin^4 x + \cos^4 x$ — a neat identity worth remembering.

This is derived from $(\sin^2 x + \cos^2 x)^2 = 1$, expanding to $\sin^4 x + \cos^4 x + 2\sin^2 x \cos^2 x = 1$, then rearranging.

So the denominator exactly cancels the $\sin^4 x + \cos^4 x$ factor from the numerator, leaving only $-\cos 2x$.

Step-by-step solution

1. Factor the numerator

$\sin^8 x - \cos^8 x = (\sin^4 x)^2 - (\cos^4 x)^2 = (\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)$

Now factor the first bracket again:

$\sin^4 x - \cos^4 x = (\sin^2 x)^2 - (\cos^2 x)^2 = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)$

Since $\sin^2 x + \cos^2 x = 1$, this simplifies to $\sin^2 x - \cos^2 x$.

So the numerator becomes $(\sin^2 x - \cos^2 x)(\sin^4 x + \cos^4 x)$.

2. Recognise the double-angle form

$\sin^2 x - \cos^2 x = -(\cos^2 x - \sin^2 x) = -\cos 2x$ …

Method: Algebraically simplify high-power trig before integrating

Use this when sines and cosines appear in high even powers: repeated difference-of-squares factoring plus the Pythagorean identity almost always collapses the integrand to something with an elementary antiderivative.

Steps

Step 1: Factor the numerator as a difference of squares, repeatedly.

$A^4-B^4=(A^2-B^2)(A^2+B^2)$ and $A^2-B^2=(A-B)(A+B)$. Use $\sin^2x+\cos^2x=1$ to kill any factor that becomes $1$.

Step 2: Recognise the standard identity for the denominator.

From $(\sin^2x+\cos^2x)^2=1$, expand to get …

Common Mistakes

Mistake 1: Trying to integrate the original expression directly.

Why it's wrong: $\dfrac{\sin^8x-\cos^8x}{1-2\sin^2x\cos^2x}$ has no obvious antiderivative until it is simplified. Correct approach: factor and cancel first — it reduces to $-\cos 2x$.

Mistake 2: Not recognising $1-2\sin^2x\cos^2x=\sin^4x+\cos^4x$. …