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Evaluate

$$\int\sec^4(u) \operatorname d \!u$$

I don't know what to substitute: I've tried $1+\tan(u)$ and integration by parts. I know the general formula for $\sec^n(u)$, but I want to be able to do this integral on my own.

Martin Sleziak
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ahorn
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2 Answers2

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$$\int \sec^4(u) du = \int \sec^2(u)\cdot \sec^2(u) du = \int \sec^2(u)(1+\tan^2(u))du$$

Now let $x=\tan(u)$ so that $dx = \sec^2(u)du$ thus transforming the integral to: $$ \int (1+x^2)dx $$

Rustyn
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    Note that this technique works just as well for $\int \tan^k u \sec^l u ,du$, for integers $k, l$, $k \geq 0$, $l > 0$ and even. – Travis Willse Oct 23 '14 at 09:53
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Note that $\sec^4(x) = \sec^2(x)\sec^2(x)= \sec^2(x)[\tan^2(x)+1]$.($\leftarrow$ from math)

So we have,

$$\int\sec^2(x)[\tan^2(x)+1] dx=\int[\sec^2(x)\tan^2(x)+\sec^2(x)]dx.$$

Now use $u=\tan^2(x)$ and see where that leads you.

H_B
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  • Would you care to make it look it nice –  Oct 23 '14 at 08:11
  • If you surround an equation with $$ instead of $ it becomes a larger function: Here's a difference: $\int x+5 dx$ becomes $\int x+5 dx$ while $$\int x+5 dx$$ becomes $$\int x+5 dx$$ – Alice Ryhl Oct 23 '14 at 08:25
  • Is this better? I just noticed the other person pretty much said the same thing as I did. But I could not see that until I got off my tablet. Would it better edicate to just delete my answer? @DigitalBrain – H_B Oct 23 '14 at 08:25
  • oh cool, ill do that. – H_B Oct 23 '14 at 08:25