$\int {\cos }^3\left(x\right){\sin }^2\left(x\right) \mathit{\ⅆ}x$

$\= -\frac{{\cos }^4\left(x\right)\sin \left(x\right)}{3\+2}\+\frac{2-1}{3\+2}\int {\cos }^3\left(x\right) \mathit{\ⅆ}x$

$\= -\frac{{\cos }^4\left(x\right)\sin \left(x\right)}{5}\+\frac{1}{5}\left(\frac{{\cos }^2\left(x\right)\sin \left(x\right)}{3}\+\frac{3-1}{3}\int \cos \left(x\right) \mathit{\ⅆ}x\right)$

$\= -\frac{{\cos }^4\left(x\right)\sin \left(x\right)}{5}\+\frac{{\cos }^2\left(x\right)\sin \left(x\right)}{15}\+\frac{2}{15}\sin \left(x\right)$

$\int {\cos }^3\left(x\right){\sin }^2\left(x\right) \mathit{\ⅆ}x$

$\= \int \cos \left(x\right)\left(1- {\sin }^2\left(x\right)\right){\sin }^2\left(x\right) \mathrm{dx}$

$\= \int {\sin }^2\left(x\right)\cos \left(x\right)\mathrm{dx}- \int {\sin }^4\left(x\right)\cos \left(x\right) \mathrm{dx}$

$\= \int u^2 \mathrm{du}-\int u^4 \mathrm{du}$

$\{\begin{array}{c}u\=\sin \left(x\right)\\ \mathrm{du}\=\cos \left(x\right)\mathrm{dx}\end{array}$

$\= \frac{u^3}{3}-\frac{u^5}{5}$

$\= \frac{{\sin }^3\left(x\right)}{3}-\frac{{\sin }^5\left(x\right)}{5}$

Evaluation of the integral

$\int {\cos }^3\left(x\right){\sin }^2\left(x\right) \mathit{\ⅆ}x$ = $-\frac{1}{5}{\cos \left(x\right)}^4\sin \left(x\right)\+\frac{1}{15}{\cos \left(x\right)}^2\sin \left(x\right)\+\frac{2}{15}\sin \left(x\right)$$$

Annotated stepwise solution

Loading Student:-Calculus1

$\int {\cos }^3\left(x\right){\sin }^2\left(x\right) \mathit{\ⅆ}x$$\stackrel{\text{show solution steps}}{\to }$$\begin{array}{lll}& & \text{Integration Steps}\\ & & \int {\cos \left(x\right)}^3{\sin \left(x\right)}^2\ⅆx\\ \text{▫}& & \text{1. Rewrite}\\ & \text{◦}& \text{Equivalent expression}\\ & & {\cos \left(x\right)}^3=\cos \left(x\right)-\cos \left(x\right){\sin \left(x\right)}^2\\ & & \text{This gives:}\\ & & \int -{\sin \left(x\right)}^2\left({\sin \left(x\right)}^2-1\right)\cos \left(x\right)\ⅆx\\ \text{▫}& & \text{2. Rewrite}\\ & \text{◦}& \text{Equivalent expression}\\ & & -{\sin \left(x\right)}^2\left({\sin \left(x\right)}^2-1\right)\cos \left(x\right)=-\cos \left(x\right){\sin \left(x\right)}^4+\cos \left(x\right){\sin \left(x\right)}^2\\ & & \text{This gives:}\\ & & \int \left(-\cos \left(x\right){\sin \left(x\right)}^4+\cos \left(x\right){\sin \left(x\right)}^2\right)\ⅆx\\ \text{▫}& & \text{3. Apply the}\mathbf{\text{sum}}\text{rule}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{sum}}\text{rule}\\ & & \int \left(f\left(x\right)+g\left(x\right)\right)\ⅆx=\int f\left(x\right)\ⅆx+\int g\left(x\right)\ⅆx\\ & & f\left(x\right)=-\cos \left(x\right){\sin \left(x\right)}^4\\ & & g\left(x\right)=\cos \left(x\right){\sin \left(x\right)}^2\\ & & \text{This gives:}\\ & & \int -\cos \left(x\right){\sin \left(x\right)}^4\ⅆx+\int \cos \left(x\right){\sin \left(x\right)}^2\ⅆx\\ \text{▫}& & \text{4. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int -\cos \left(x\right){\sin \left(x\right)}^4\ⅆx\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int \left[\right]f\left(x\right)\ⅆx=\left[\right]\left(\int f\left(x\right)\ⅆx\right)\\ & \text{◦}& \text{This means:}\\ & & \int -\cos \left(x\right){\sin \left(x\right)}^4\ⅆx=-\left(\int \cos \left(x\right){\sin \left(x\right)}^4\ⅆx\right)\\ & & \text{We can rewrite the integral as:}\\ & & -\left(\int \cos \left(x\right){\sin \left(x\right)}^4\ⅆx\right)+\int \cos \left(x\right){\sin \left(x\right)}^2\ⅆx\\ \text{▫}& & \text{5. Apply a change of variables to rewrite the integral in terms of}u\\ & \text{◦}& \text{Let}u\text{be}\\ & & u=\sin \left(x\right)\\ & \text{◦}& \text{Isolate equation for}x\\ & & x=\arcsin \left(u\right)\\ & \text{◦}& \text{Differentiate both sides}\\ & & \left[\right]=\frac{\left[\right]}{\sqrt{-u^2+1}}\\ & \text{◦}& \text{Substitute the values for x and dx back into the original}\\ & & \int \cos \left(x\right){\sin \left(x\right)}^4\ⅆx=\int u^4\ⅆu\\ & & \text{This gives:}\\ & & -\left(\int u^4\ⅆu\right)+\int \cos \left(x\right){\sin \left(x\right)}^2\ⅆx\\ \text{▫}& & \text{6. Apply the}\mathbf{\text{power}}\text{rule to the term}\int u^4\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule, for n}\text{≠}\text{-1}\\ & & \int u^{\left[\right]}\ⅆu=\left[\right]\\ & \text{◦}& \text{This means:}\\ & & \int u^4\ⅆu=\left[\right]\\ & \text{◦}& \text{So,}\\ & & \int u^4\ⅆu=\frac{u^5}{5}\\ & & \text{We can rewrite the integral as:}\\ & & -\frac{u^5}{5}+\int \cos \left(x\right){\sin \left(x\right)}^2\ⅆx\\ \text{▫}& & \text{7. Revert change of variable}\\ & \text{◦}& \text{Variable we defined in step}5\\ & & u=\sin \left(x\right)\\ & & \text{This gives:}\\ & & -\frac{{\sin \left(x\right)}^5}{5}+\int \cos \left(x\right){\sin \left(x\right)}^2\ⅆx\\ \text{▫}& & \text{8. Apply a change of variables to rewrite the integral in terms of}u\\ & \text{◦}& \text{Let}u\text{be}\\ & & u=\sin \left(x\right)\\ & \text{◦}& \text{Isolate equation for}x\\ & & x=\arcsin \left(u\right)\\ & \text{◦}& \text{Differentiate both sides}\\ & & \left[\right]=\frac{\left[\right]}{\sqrt{-u^2+1}}\\ & \text{◦}& \text{Substitute the values for x and dx back into the original}\\ & & \int \cos \left(x\right){\sin \left(x\right)}^2\ⅆx=\int u^2\ⅆu\\ & & \text{This gives:}\\ & & -\frac{{\sin \left(x\right)}^5}{5}+\int u^2\ⅆu\\ \text{▫}& & \text{9. Apply the}\mathbf{\text{power}}\text{rule to the term}\int u^2\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule, for n}\text{≠}\text{-1}\\ & & \int u^{\left[\right]}\ⅆu=\left[\right]\\ & \text{◦}& \text{This means:}\\ & & \int u^2\ⅆu=\left[\right]\\ & \text{◦}& \text{So,}\\ & & \int u^2\ⅆu=\frac{u^3}{3}\\ & & \text{We can rewrite the integral as:}\\ & & -\frac{{\sin \left(x\right)}^5}{5}+\frac{u^3}{3}\\ \text{▫}& & \text{10. Revert change of variable}\\ & \text{◦}& \text{Variable we defined in step}8\\ & & u=\sin \left(x\right)\\ & & \text{This gives:}\\ & & \frac{{\sin \left(x\right)}^3}{3}-\frac{{\sin \left(x\right)}^5}{5}\end{array}$