evaluate the indefinite integral given below.\n int (sin^{3}(x)-3sin^{2}(x))(sin^{4}(x)-4sin^{3}(x))^{8}cos(x…

Explanation:

Step1: Set substitution

Let $u = \sin^{4}(x)-4\sin^{3}(x)$. Then $du=(4\sin^{3}(x)\cos(x)- 12\sin^{2}(x)\cos(x))dx = 4(\sin^{3}(x)-3\sin^{2}(x))\cos(x)dx$. So, $(\sin^{3}(x)-3\sin^{2}(x))\cos(x)dx=\frac{1}{4}du$.

Step2: Rewrite the integral

The original integral $\int(\sin^{3}(x)-3\sin^{2}(x))(\sin^{4}(x)-4\sin^{3}(x))^{8}\cos(x)dx$ becomes $\frac{1}{4}\int u^{8}du$.

Step3: Integrate with power - rule

Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\frac{1}{4}\times\frac{u^{9}}{9}+C=\frac{u^{9}}{36}+C$.

Step4: Substitute back

Substitute $u=\sin^{4}(x)-4\sin^{3}(x)$ back into the result. We get $\frac{(\sin^{4}(x)-4\sin^{3}(x))^{9}}{36}+C$.

Answer:

$\frac{(\sin^{4}(x)-4\sin^{3}(x))^{9}}{36}+C$