Monday Math 143

Find .


The presence of the term leads us to attempt a trigonometric substitution of the form or . For reasons that will become clear in a moment, we will choose the latter. The limits x=-1 and x=1 become θ=π and θ=0, respectively, while , so we have
.

Now, the key is to note the trigonometric identity , so we see
.
Now, since we are integrating from zero to π, is positive for the entire interval, and so we may remove the absolute values:
.
Next, we can use again via θθ/2:
,
and since

then
.
Plugging these in,
,
and since and are both positive for our entire region of integration, we have
,
and since the half-angle formula for tangent says
,
,
and via ,
,

We can find an exact value for the secant above using half-angle identities along with and , we can find that
,
so
.

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