b) ∫sin⁴(x)cos⁵(x) dx solve by trigonometric integration strategies.

Explanation:

Step1: Rewrite $\cos^5(x)$

We know that $\cos^5(x)=\cos^4(x)\cos(x)=(1 - \sin^2(x))^2\cos(x)$. So the integral becomes $\int\sin^4(x)(1 - \sin^2(x))^2\cos(x)dx$.

Step2: Use substitution

Let $u = \sin(x)$, then $du=\cos(x)dx$. The integral is now $\int u^4(1 - u^2)^2du$.

Step3: Expand $(1 - u^2)^2$

$(1 - u^2)^2=1 - 2u^2+u^4$. So the integral is $\int u^4(1 - 2u^2+u^4)du=\int(u^4 - 2u^6+u^8)du$.

Step4: Integrate term - by - term

$\int(u^4 - 2u^6+u^8)du=\int u^4du-2\int u^6du+\int u^8du$. Using the power rule $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we get $\frac{u^{5}}{5}-2\times\frac{u^{7}}{7}+\frac{u^{9}}{9}+C$.

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

The result is $\frac{\sin^{5}(x)}{5}-\frac{2\sin^{7}(x)}{7}+\frac{\sin^{9}(x)}{9}+C$.

Answer:

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