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Examples In this video, we work three examples, one with $x=a
\tan(\theta)$, one with $x = a\sin(\theta)$, and one with $x = a
\sec(\theta)$. Worked out solutions are written below the
video. Examples from the video.

Consider our answer above. In order to convert back into terms of $x$, we must figure out what $\sin(\theta)$ is in terms of $x$. By rewriting our original substitution we see that $\tfrac x 2=\tan\theta$. Use this to draw a right triangle, with opposite side $x$ and adjacent side $a=2$. The hypotenuse is then $\sqrt{a^2+x^2}=\sqrt{4+x^2}$. We need to find $\sin\theta$ in terms of $x$, and we see from the triangle that $\sin\theta=\frac{x}{\sqrt{x^2+4}}$. 

Here we have used the methods of the last
learning module to evaluate the trig integral, including the handy trig identities for
$\cos^2\theta$ and $\sin(2\theta)$. (You need to
know these by heart). We look at the
terms in our final answer above. We use the triangle
to convert $\sin\theta\cos\theta$ back into terms of
$x$. Finally, we must write $\theta$ in terms of
$x$. We use our original substitution:
$\frac{x}{3}=\sin x$ gives us $\sin^{1}(\tfrac x
3)=\theta$. 

We now convert to terms of $x$:
$\sec\theta=2x$, so $\cos\theta=\frac{1}{2x}$. The
triangle could therefore have adjacent side of 1, and
hypotenuse of $2x$, and the opposite side
$\sqrt{4x^21}$. We can also have adjacent of
$\frac{1}{2}$ and hypotenuse of $x$. (Why?) This gives the
opposite side the value of $\sqrt{x^2\frac{1}{4}}$, as in
the diagram. Either way is fine. Looking at our
values above, we only need deal with $\tan \theta$, since we
know $\sec\theta=2x$. From the triangle,
$\tan\theta=\frac{\sqrt{x^21/4}}{1/2}$. If we had
used the other triangle, we would get
$\tan\theta=\frac{\sqrt{4x^21}}{1}$  are these the same values? 
