Hi there!
In my Business English Class, I often give my students a mathematical word problem as an optional question, for the majority of them used to major Not in humanities But in Science in their university.
One time factorization, other time differentiation, and whatnot. Just studying English is boring, so giving them a mathematical word problem can be a good stimulant.
Today, let me review the constant integration of a trigonometric function whose word problem, I remember, would be elicited from one of the foreign blogs. If someone identified its source, I would be more than happy to mention it without delay.
Evaluate the following integral, if possible. If it is impossible, clearly explain why it is not possible to evaluate the integral.
\(\int_{0}^{\frac{π}{4}}\frac{8\cos(2t)}{\sqrt{9-5\sin(2t)}}dt\)
Here is my solution to the problem.
First of all, let’s do the substitution. The substitution for this problem is,
\(u = 9 – 5sin(2t)\)
Here is the actual substitution for this problem.
\(du = ―10cos(2t)dt\), so \(cos(2t)dt = -\frac{1}{10}du\)
\(t = 0：u = 9\)
\(t = \frac{π}{4}：u = 4\)
Now convert the limits to u’s to avoid dealing with the back substitution after doing the integral. Here is the integral after substitution.
\(\int_{0}^{\frac{π}{4}}\frac{8\cos(2t)}{\sqrt{9-5\sin(2t)}}dt\)
\(= -\frac{8}{10}\int_{9}^{4}u^{-\frac{1}{2}}du\)
The integral is then,
\(\int_{0}^{\frac{π}{4}}\frac{8\cos(2t)}{\sqrt{9-5\sin(2t)}}dt\)
\(= [-\frac{8}{5}u^{\frac{1}{2}}]_{9}^{4}\)
\(= – \frac{16}{5} – (-\frac{24}{5}) = \frac{8}{5}\)
That’s great! I’ve managed to review what I had studied before. Studying integration is becoming more and more intriguing.
Stay tuned, and expect to see my next post.
Keep well.
Frank Yoshida
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【グローバルビジネスで役立つ数学】でもっと学習する b＾＾)
【参考図書】『もう一度高校数学』（著者：高橋一雄氏）株式会社日本実業出版社
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