Question

f(t) = sin⁴(t) cos⁵(t)

Explanation:

Step1: Rewrite odd cosine term

$\cos^5(t) = \cos^4(t)\cos(t) = (1-\sin^2(t))^2\cos(t)$
So $f'(t) = \sin^4(t)(1-\sin^2(t))^2\cos(t)$

Step2: Substitute $u=\sin(t)$

Let $u = \sin(t)$, then $du = \cos(t)dt$
The integral becomes $\int u^4(1-u^2)^2 du$

Step3: Expand the integrand

$(1-u^2)^2 = 1-2u^2+u^4$
$u^4(1-2u^2+u^4) = u^4 - 2u^6 + u^8$

Step4: Integrate term-by-term

$\int (u^4 - 2u^6 + u^8)du = \frac{u^5}{5} - \frac{2u^7}{7} + \frac{u^9}{9} + C$

Step5: Substitute back $u=\sin(t)$

Replace $u$ with $\sin(t)$

Answer:

$\frac{\sin^5(t)}{5} - \frac{2\sin^7(t)}{7} + \frac{\sin^9(t)}{9} + C$