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I know the answer but I don't know how to find the proper u-sub for it. I'm told I need to use U-sub for all the integrals. Here is where I end up

$$\int (1-\sec^2(\theta))\,d\theta -\int \tan^2(\theta)\sec^2(\theta)\,d\theta$$

My natural u-sub should be $\tan x$ since $dx$ would be $\sec^2$ but I don't know what to do about the other parts. Can I just u sub one section of the integral?

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  • $\begingroup$ your done. Note, $\sec^2 \theta$ is elementary and $u=\tan \theta$ has $du = \sec^2 \theta d\theta$. $\endgroup$ – James S. Cook Jul 5 '14 at 14:24
  • $\begingroup$ I'm really asking if I can just use u-sub for one part of the integral. $\endgroup$ – Joshhw Jul 5 '14 at 14:26
  • $\begingroup$ Yes, evaluate the two integrals separately. The first is easy, and the second is done with substitution. $\endgroup$ – StrangerLoop Jul 5 '14 at 14:38
  • $\begingroup$ Notice that there should be a + sign in front of the second term; see Michael Hardy's answer. $\endgroup$ – user84413 Jul 5 '14 at 21:27
  • $\begingroup$ @Joshhw you can use as many different substitutions as you wish. Remember the integral has the linearity properties $\int f+g = \int f+\int g$ and $\int cf = c \int f$ so we can divide and conquer by disparate methods. I'm sure amWhy's answer cleared it up for you. $\endgroup$ – James S. Cook Jul 6 '14 at 1:06
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Yes. You can use u-substitution on the second integral without using it on the first. You'll just need to back-substitute on the second, once integrated.

On the first, note that $\frac{d}{d\theta}(\tan \theta + C) = \sec^2\theta \iff \int \sec^2\theta\,d\theta = \tan \theta + C$.

$$\int \underbrace{(1-\sec^2(\theta))d\theta}_{\large = \theta - \tan \theta} -\int \tan^2(\theta)\sec^2(\theta)d\theta$$

Using $u = \tan \theta \implies du = \sec^2 \theta$ on the second integral: $$= \theta - \tan\theta - \int u^2\,du \tag{$u = \tan\theta$}$$

$$= \theta - \tan \theta - \frac{u^3}3 + C\tag{$u = \tan\theta$}$$ $$ = \theta - \tan \theta - \frac 13 \tan^3 \theta + C$$

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\begin{align} & \int \tan^4\theta\,d\theta = \int (\tan^2\theta)(\sec^2\theta-1)\,d\theta \\[10pt] = {} & \int\tan^2\theta\Big(\sec^2\theta\,d\theta\Big) - \int\tan^2\theta\,d\theta \\[10pt] = {} & \int\tan^2\theta\Big(\sec^2\theta\,d\theta\Big) - \int(\sec^2\theta-1)\,d\theta \\[10pt] = {} & \int u^2\,du - \tan\theta + \theta+\text{constant} \end{align}

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  • $\begingroup$ Why the down-vote here? $\endgroup$ – Michael Hardy Jul 6 '14 at 14:59
  • $\begingroup$ I see no dv here.Maybe it's gone. :) +1 $\endgroup$ – mrs Aug 11 '14 at 19:19
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It is good to know that while working with $$\int\sin^m(x)\cos^n(x)dx$$ where $m,~n$ are integers then if $m+n$ is even number so $t=\tan(x)$ is a good substitution.

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Here's another substitution option. Substitute $u=\tan{(\theta)}$. Then, $\theta=\arctan{(u)}$, and $\mathrm{d}\theta=\frac{\mathrm{d}u}{1+u^2}$, and:

$$\int\tan^4{(\theta)}\,\mathrm{d}\theta=\int\frac{u^4\,\mathrm{d}u}{1+u^2}.$$

Then apply partial fractions.

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  • $\begingroup$ I think this makes the problem much harder, because you either have to do a nontrivial integration by parts or partial fractions with complex numbers. In fact, if presented with the right hand side, I would do the reverse of your substitution to work with the left hand side instead. $\endgroup$ – Ian Jul 5 '14 at 14:57
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    $\begingroup$ However, one nice thing about this approach is $\int \tan^4(\theta) d\theta = \int \frac{u^4}{1+u^2} du = \int \left ( u^2 - \frac{u^2}{1+u^2} \right ) du$. The second term yields an easier trigonometric integral. This gives a systematic way to integrate $\tan^{2n}(\theta)$ for any $n$. $\endgroup$ – Ian Jul 5 '14 at 15:04
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    $\begingroup$ @Ian I completely agree this way is much harder. That's part of the reason I suggested trying to solve it this way. $\endgroup$ – David H Jul 5 '14 at 15:06
  • $\begingroup$ @David H I think this is the most efficient way to integrate positive integral powers of the tangent. $\endgroup$ – user84413 Jul 5 '14 at 21:33

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