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The following is from page 142 in chapter 7 of Barron's E-Z Calculus (formerly Calculus the Easy Way), fifth edition:

Evaluate:

43. $y = \int \sec^{2}{x} \tan^{3}{x} \, \mathrm{d}x$

WolframAlpha and the Maxima CAS both agree that the correct answer is $\frac{\tan^{4}{x}}{4} + C$. However, I don't understand how this answer was reached.

I tried the substitution $u = \tan{x}, u' = \sec^{2}{x}$, but that still leaves $\tan^{2}{x}$ in the integral.

What is the correct substitution, and how does it lead to the answer? Where did the 4 in the denominator come from?

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    $\begingroup$ $u^3=\tan^3 x$. $\endgroup$ – T.J. Gaffney Jan 20 '14 at 21:14
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We can set $z = \tan x$, and we get that $dz = \sec^2 x dx$. We can then substitute them in to get that $$\int \sec^2 x \tan^3 x dx = \int z^3 dz = \frac{z^4}{4} + C$$ Substitute back in, and we're done.

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  • $\begingroup$ $\frac{\mathrm{d}}{\mathrm{d}x} \tan^{3}{x} = 3 \sec^{2}{x} \tan^{2}{x}$, so how does this fit into the integral? $\endgroup$ – bwDraco Jan 20 '14 at 21:24
  • $\begingroup$ We don't take the derivative of $\tan^3 x$, we take the derivative of $\tan x$. Alternatively, you could take the derivative of $\tan^4 x$, which would give $4 \sec^2 x \tan^3 x$. $\endgroup$ – 2012ssohn Jan 20 '14 at 21:27
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    $\begingroup$ Also, could whoever downvoted this please tell me WHY you downvoted it so that I can improve this answer? $\endgroup$ – 2012ssohn Jan 20 '14 at 21:27
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    $\begingroup$ Okay, I get it. I didn't realize that the $u$ could be under an exponent outside the $\mathrm{d}u$ in the integral. $\endgroup$ – bwDraco Jan 20 '14 at 21:37
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Take $\displaystyle\int{\sec^2{x} \tan^3 {x}} dx$ and note that this is the same as $\displaystyle\int{\tan^3 {x}\sec^2{x} } dx$.

Knowing that $\displaystyle\frac{d}{dx}{\tan{x}}=\sec^2{x}$, then $\displaystyle\frac{d}{dx}{f(\tan{x})}=f'(\tan{x})\sec^2{x}$ using the chain rule.

Finding the integral of $\sec^2{x} \tan^3 {x}$ is finding such a class of functions whose derivatives are all $\sec^2{x} \tan^3 {x}$.

Knowing that the power rule is given by $\displaystyle\frac{d}{dx}x^n=nx^{n-1}$, then the power of $\tan x$ in the integral indicates the function from which it came. In this case, $\tan$ takes power 3, so the integral will be power 4. This applies to any composite function.

Using the law of constants (constant multiplier in = constant multiplier out), we can write

$\displaystyle\int{\tan^3 {x}\sec^2{x} } dx$ = $\displaystyle\frac{1}{4}\displaystyle\int{4\tan^3 {x}\sec^2{x} } dx$

and then seeing that the expression inside the integral is the derivative of an $f(\tan{x})$ we have

$\displaystyle\frac{1}{4}\displaystyle\int{4\tan^3 {x}\sec^2{x} } dx$ = $\displaystyle\frac{1}{4}(\tan^4 {x} + k)=\displaystyle\frac{1}{4}\tan^4 {x} + C$

recognising in our head that $\displaystyle\frac{d}{dx}(\tan^4{x}+k)=4\tan^{4-1}{x}\sec^2{x}$.

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Here is a slightly different approach :

Make the following substitution :

$$t=\tan(x) \,\,\,(♣)$$.

$$\implies t^3=\tan^3(x) \,\,\,(♦)$$

Differentiate both sides of $(♦)$ to get :

$$3t^2dt =3\tan^2(x) . \sec^2(x)$$

Cancel $3$ from both sides to get :

$$t^2dt=\sec^2(x).\tan^2(x) dx\,\,\,(♠)$$

Now take the integral :

$$y=\int \sec^2(x)\tan^3(x)dx=\int tan(x) \times \left(\sec^2(x)tan^2(x)\right)dx \,\,\,(♥)$$

Using $(♣)$ and $(♠) , \,\,\,(♥)$ becomes :

$$\int (t) \times (t^2 dt) = \int t^3 dt =\dfrac{t^4}{4}+C\,\,\,\,(♫)$$

Plugging $(♣)$ into $(♫)$, we get :

$$y=\dfrac{\tan^4(x)}{4} + C$$

Hope this helps. $:)$

NOTE : My approach is slightly different from @2012ssohn . He differentiates $u=tan(x)$ whereas I differentiate $\tan^3(x)$. It makes almost no difference. It is just a slightly different way of looking at the same problem.

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