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I have solved a problem using a double "u" substitution as follows:

https://s30.postimg.org/y0eveutkx/IMG_9745.jpg

But I was told I need to use trig. After brushing up on the rules I did the following:

https://s30.postimg.org/dhoknj51d/IMG_9746.jpg

But now I find myself stuck. I am not missing something and as far as I know, this is correct so far.

What trick am I missing?

Thanks in advance!

Oh yea, I can't use the reduction rule either... Grrr

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    All your links are broken. Please remove them and type directly: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – 2017-02-04
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    Are you just stuck at $\int \sec^4x \,dx$ ? If so, then Hint: write $\sec^4x =\sec^2x\,\sec^2x = \sec^2x\, (\tan^2x+1)$ – 2017-02-04

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$$\int \sec^4 (\theta) d\theta$$

Noting the power of $4$ is even, leave $\sec^2(\theta)$ because this is the derivative of $\tan \theta$ and perform the identity $\tan^2 (\theta)+1=\sec^2 (\theta)$ on the left powers.

$$\sec^4 (\theta)$$

$$(\tan^2 (\theta)+1)\sec^2 (\theta)$$

$$=\tan^2(\theta) \sec^2 (\theta)+\sec^2 (\theta)$$

This can be easily integrated. The first with substituting $u=\tan \theta$ and the second by noting what the derivative of $\tan \theta$ is $\sec^2 (\theta)$

$$\int \tan^2(\theta) \sec^2 (\theta) d\theta+\int \sec^2 (\theta) d\theta$$

$$=\frac{\tan^3 (\theta)}{3}+\tan (\theta)+C$$

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    Thank you!! I should of thought of that!! Many thanks! – 2017-02-04