$\int {\tan }^3\left(x\right) \mathit{\ⅆ}x$

$\= \int \tan \left(x\right)\cdot {\tan }^2\left(x\right) \mathit{\ⅆ}x$

$\= \int \tan \left(x\right)\left({\sec }^2\left(x\right)-1\right) \mathit{\ⅆ}x$

$\= \int \tan \left(x\right){\sec }^2\left(x\right) \mathit{\ⅆ}x-\int \tan \left(x\right) \mathit{\ⅆ}x$

$\{\begin{array}{c}u\=\tan \left(x\right)\\ \mathrm{du}\={\sec }^2\left(x\right)\mathrm{dx}\end{array}$

$\= \int u \mathit{\ⅆ}u-\int \tan \left(x\right) \mathit{\ⅆ}x$

$\=\frac{u^2}{2}-\ln \(\left|\sec \left(x\right)\right|\)$

$\=\frac{{\tan }^2\left(x\right)}{2}-\ln \(\left|\sec \left(x\right)\right|\)$

Table 6.2.4(a)   Indefinite integral of ${\tan }^3\left(x\right)$ 

Evaluation in Maple

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

$\begin{array}{c}\int {\tan \left(x\right)}^3\ⅆx\hfill \\ & \text{=}& \int \left({\sec \left(x\right)}^2-1\right)\tan \left(x\right)\ⅆx\hfill & \left[\mathrm{rewrite}\,{\tan \left(x\right)}^2={\sec \left(x\right)}^2-1\right]\\ & \text{=}& \int {\sec \left(x\right)}^2\tan \left(x\right)\ⅆx+\ln \left(\cos \left(x\right)\right)\hfill & \left[\mathrm{rewrite}\,\left({\sec \left(x\right)}^2-1\right)\tan \left(x\right)={\sec \left(x\right)}^2\tan \left(x\right)-\tan \left(x\right)\right]\\ & \text{=}& \int u\ⅆu+\ln \left(\cos \left(x\right)\right)\hfill & \left[\mathrm{change}\,u=\sec \left(x\right)\,u\right]\\ & \text{=}& \frac{u^2}{2}+\ln \left(\cos \left(x\right)\right)\hfill & \left[\mathrm{power}\right]\\ & \text{=}& \frac{{\sec \left(x\right)}^2}{2}+\ln \left(\cos \left(x\right)\right)\hfill & \[\mathrm{re}\mathrm{vert}\end{array}$

Table 6.2.4(b)   Evaluation of $\int {\tan }^3\left(x\right) \mathrm{dx}$ via the Integration Methods tutor

Loading Student:-Calculus1

$\int {\tan }^3\left(x\right) \mathit{\ⅆ}x$$\stackrel{\text{show solution steps}}{\to }$$\begin{array}{lll}& & \text{Integration Steps}\\ & & \int {\tan \left(x\right)}^3\ⅆx\\ \text{▫}& & \text{1. Rewrite}\\ & \text{◦}& \text{Equivalent expression}\\ & & {\tan \left(x\right)}^2={\sec \left(x\right)}^2-1\\ & & \text{This gives:}\\ & & \int \left({\sec \left(x\right)}^2-1\right)\tan \left(x\right)\ⅆx\\ \text{▫}& & \text{2. Rewrite}\\ & \text{◦}& \text{Equivalent expression}\\ & & \left({\sec \left(x\right)}^2-1\right)\tan \left(x\right)={\sec \left(x\right)}^2\tan \left(x\right)-\tan \left(x\right)\\ & & \text{This gives:}\\ & & \int \left({\sec \left(x\right)}^2\tan \left(x\right)-\tan \left(x\right)\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)={\sec \left(x\right)}^2\tan \left(x\right)\\ & & g\left(x\right)=-\tan \left(x\right)\\ & & \text{This gives:}\\ & & \int {\sec \left(x\right)}^2\tan \left(x\right)\ⅆx+\int -\tan \left(x\right)\ⅆx\\ \text{▫}& & \text{4. Apply a change of variables to rewrite the integral in terms of}u\\ & \text{◦}& \text{Let}u\text{be}\\ & & u=\sec \left(x\right)\\ & \text{◦}& \text{Isolate equation for}x\\ & & x=\mathrm{arcsec}\left(u\right)\\ & \text{◦}& \text{Differentiate both sides}\\ & & \mathrm{dx}=\frac{\mathrm{du}}{u^2\sqrt{1-\frac{1}{u^2}}}\\ & \text{◦}& \text{Substitute the values for x and dx back into the original}\\ & & \int {\sec \left(x\right)}^2\tan \left(x\right)\ⅆx=\int u\ⅆu\\ & & \text{This gives:}\\ & & \int u\ⅆu+\int -\tan \left(x\right)\ⅆx\\ \text{▫}& & \text{5. Apply the}\mathbf{\text{power}}\text{rule to the term}\int u\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule, for n}\text{≠}\text{-1}\\ & & \int u^n\ⅆu=\frac{u^{n+1}}{n+1}\\ & \text{◦}& \text{This means:}\\ & & \int u\ⅆu=\frac{u^{1+1}}{1+1}\\ & \text{◦}& \text{So,}\\ & & \int u\ⅆu=\frac{u^2}{2}\\ & & \text{We can rewrite the integral as:}\\ & & \frac{u^2}{2}+\int -\tan \left(x\right)\ⅆx\\ \text{▫}& & \text{6. Revert change of variable}\\ & \text{◦}& \text{Variable we defined in step}4\\ & & u=\sec \left(x\right)\\ & & \text{This gives:}\\ & & \frac{{\sec \left(x\right)}^2}{2}+\int -\tan \left(x\right)\ⅆx\\ \text{▫}& & \text{7. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int -\tan \left(x\right)\ⅆx\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int Cf\left(x\right)\ⅆx=C\left(\int f\left(x\right)\ⅆx\right)\\ & \text{◦}& \text{This means:}\\ & & \int -\tan \left(x\right)\ⅆx=-\left(\int \tan \left(x\right)\ⅆx\right)\\ & & \text{We can rewrite the integral as:}\\ & & \frac{{\sec \left(x\right)}^2}{2}-\left(\int \tan \left(x\right)\ⅆx\right)\\ \text{▫}& & \text{8. Rewrite}\\ & \text{◦}& \text{Equivalent expression}\\ & & \tan \left(x\right)=\frac{\sin \left(x\right)}{\cos \left(x\right)}\\ & & \text{This gives:}\\ & & \frac{{\sec \left(x\right)}^2}{2}-\left(\int \frac{\sin \left(x\right)}{\cos \left(x\right)}\ⅆx\right)\\ \text{▫}& & \text{9. Apply a change of variables to rewrite the integral in terms of}u\\ & \text{◦}& \text{Let}u\text{be}\\ & & u=\cos \left(x\right)\\ & \text{◦}& \text{Isolate equation for}x\\ & & x=\arccos \left(u\right)\\ & \text{◦}& \text{Differentiate both sides}\\ & & \mathrm{dx}=-\frac{\mathrm{du}}{\sqrt{-u^2+1}}\\ & \text{◦}& \text{Substitute the values for x and dx back into the original}\\ & & \int \frac{\sin \left(x\right)}{\cos \left(x\right)}\ⅆx=\int -\frac{1}{u}\ⅆu\\ & & \text{This gives:}\\ & & \frac{{\sec \left(x\right)}^2}{2}-\left(\int -\frac{1}{u}\ⅆu\right)\\ \text{▫}& & \text{10. Apply the}\mathbf{\text{constant multiple}}\text{rule to the term}\int -\frac{1}{u}\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant multiple}}\text{rule}\\ & & \int Cf\left(u\right)\ⅆu=C\left(\int f\left(u\right)\ⅆu\right)\\ & \text{◦}& \text{This means:}\\ & & \int -\frac{1}{u}\ⅆu=-\left(\int \frac{1}{u}\ⅆu\right)\\ & & \text{We can rewrite the integral as:}\\ & & \frac{{\sec \left(x\right)}^2}{2}+\int \frac{1}{u}\ⅆu\\ \text{▫}& & \text{11. Apply the}\mathbf{\text{reciprocal}}\text{rule to the term}\int \frac{1}{u}\ⅆu\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{reciprocal}}\text{rule}\\ & & \int \frac{1}{u}\ⅆu=\ln \left(u\right)\\ & & \text{We can rewrite the integral as:}\\ & & \frac{{\sec \left(x\right)}^2}{2}+\ln \left(u\right)\\ \text{▫}& & \text{12. Revert change of variable}\\ & \text{◦}& \text{Variable we defined in step}9\\ & & u=\cos \left(x\right)\\ & & \text{This gives:}\\ & & \frac{{\sec \left(x\right)}^2}{2}+\ln \left(\cos \left(x\right)\right)\end{array}$

Table 6.2.4(c)   Detailed annotated stepwise evaluation of $\int {\tan }^3\left(x\right) \mathit{\ⅆ}x$