Q.Integrate the following function: $\cos^4 2x$

The key idea is to use the sine/cosine power reduction formula repeatedly to eliminate the high power.

First, recall the identity:

Here $\theta = 2x$, so $\cos^4 2x = (\cos^2 2x)^2$.

Step 1: Apply the identity once:

Step 2: Square the result:

Step 3: Reduce $\cos^2 4x$ using the same identity with $\theta = 4x$:

Substitute back:

Simplify: …

Why power reduction?

When you see $\cos^4 2x$, your first instinct might be to expand using $\cos^2\theta = \frac{1+\cos 2\theta}{2}$. That’s exactly the right move — but we need to apply it carefully because the angle inside the cosine is already $2x$, and the power is 4. The trick is to rewrite $\cos^4 2x$ as $(\cos^2 2x)^2$, then reduce the square inside, then square the result. This avoids messy expansions and keeps everything in terms of simple cosines that integrate cleanly.

Step-by-step solution

1. Write the fourth power as a square of a square.

We have:

2. Apply the power-reduction formula to $\cos^2 2x$.

Recall:

Here $\theta = 2x$, so:

3. Square the result.

Now:

4. Reduce $\cos^2 4x$ again.

Apply the same formula to $\cos^2 4x$ (with $\theta = 4x$):

Substitute back:

5. Simplify the expression.

Combine terms inside the parentheses:

Multiply by $\frac{1}{4}$:

$\cos^4 2x = \frac{3}{8} + \frac{1}{2}\cos 4x + \frac{1}{8}\cos 8x$ …

Method: Repeated power reduction for an even power of cosine with a scaled argument

$\cos^4(2x)$ is handled exactly like $\cos^4 x$, but every angle is doubled — keep the scaled argument consistent through both reduction passes and mind the $\dfrac1k$ factors when integrating.

Steps

Step 1: Reduce the outer square with the scaled argument.

Step 2: Reduce the leftover $\cos^2 4x$.

so all terms become constants or first-degree cosines in $4x$ and $8x$.

Step 3: Integrate, dividing each cosine by its own angle. …

Common Mistakes

Mistake 1: Keeping the argument as $x$ instead of $2x$ during reduction.

Why it's wrong: the integrand is $\cos^4(2x)$, so the identity gives $\cos^2 2x=\tfrac{1+\cos 4x}{2}$ and then $\cos^2 4x=\tfrac{1+\cos 8x}{2}$ — the angles are $4x$ and $8x$, not $2x$ and $4x$. Correct approach: apply power reduction with $\theta=2x$ consistently.

Mistake 2: Forgetting the $\dfrac1k$ factors, especially for $\cos 8x$. …