This problem is from the GRE sample test available at https://www.ets.org/s/gre/pdf/practice_book_math.pdf
Here is the problem:
∫π4−π4 (cost+√1+t2sin3tcos3t)dt
First I tried factoring out a cosine, but that implied integration by parts after further work. then I thought it'd be easier to attack the integral in two parts (first the cosine, then the rest). I started doing this and after a half-baked attempt to change a sin2t into a 1−cos2t, decided to use the double-angle identity:
√2+18∫π4−π4 (√1+t2sin32t)dt
Now I'm stuck. Maybe I was being "too clever". I cannot use the u-substitution u=1+t2→ du=2tdt because the 2t is the argument of a trig function. Trig substitution seems especially hairy and likely to cause me more headaches than using an alternative strategy. I am wondering about Integration by Parts but have trouble formatting the D.I. method. Anybody have ideas on this?
I recognize that this isn't the most obtuse calculus problem but I do feel like I've spent too much time on this already and would appreciate input from anyone who can identify a 'giveaway' for an efficient strategy.
Thank you in advance for your input.
Answer
Special thanks to user Raffaele, who pointed out that odd functions integrated over [-a,a] have zero area. We have
∫π/4−π/4 (cost+√1+t2sin3tcos3t)dt=∫π/4−π/4 cost dt+∫π/4−π/4 √1+t2sin3tcos3t dt=√2+0=√2
Some further explanation:
Since sint is an odd function, sin3t is an odd function. Similarly, cost is an even function, so cos3t is an even function. Furthermore, √1+t2 is an even function*. Now, any number of even functions times an odd function produces an odd function; thus, √1+t2sin3tcos3t is an odd function. Integrating this function over [−π4,π4] matches the format for integrating an odd function over [−a,a]. Therefore, ∫π/4−π/4 √1+t2sin3tcos3t dt=0.
On the practice exam, the corresponding solution is in fact B), which offers √2.
*If it's not immediately apparent why √1+t2 is even, recall that all even functions have the property f(x)=f(−x), and notice that the square forces the argument of the root to always be positive, regardless of the sign of the input.
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