Let

Note that we can write cos⁵x = cos⁴x × cos x

⇒ cos⁵x = (cos²x)² × cos x

We also have sin²x + cos²x = 1

⇒ cos⁵x = (1 – sin²x)²cos x

So,

Put sin x = t

⇒ cos x dx = dt (Differentiating both sides)

When x = 0, t = sin 0 = 0

So, the new limits are 0 and 1.

Substituting this in the original integral,

Recall