webwork / ma114f25 - participation / e\n1\nprevious problem problem list\nexam 1 review day\n(2…

Explanation:

Step1: Rewrite $\sin^{5}(x)$

Use $\sin^{2}(x)=1 - \cos^{2}(x)$, so $\sin^{5}(x)=\sin^{4}(x)\cdot\sin(x)=(1 - \cos^{2}(x))^{2}\sin(x)$. The integral becomes $\int(1 - \cos^{2}(x))^{2}\sin(x)\cos^{6}(x)dx$.

Step2: Use substitution

Let $u = \cos(x)$, then $du=-\sin(x)dx$. The integral is $-\int(1 - u^{2})^{2}u^{6}du$.

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

$(1 - u^{2})^{2}=1 - 2u^{2}+u^{4}$. So the integral is $-\int(1 - 2u^{2}+u^{4})u^{6}du=-\int(u^{6}-2u^{8}+u^{10})du$.

Step4: Integrate term - by - term

$\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ ($n\neq - 1$). $-\int(u^{6}-2u^{8}+u^{10})du=-\left(\frac{u^{7}}{7}-\frac{2u^{9}}{9}+\frac{u^{11}}{11}\right)+C$.

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

$=-\frac{\cos^{7}(x)}{7}+\frac{2\cos^{9}(x)}{9}-\frac{\cos^{11}(x)}{11}+C$.

Answer:

$-\frac{\cos^{7}(x)}{7}+\frac{2\cos^{9}(x)}{9}-\frac{\cos^{11}(x)}{11}+C$