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Find $$\int_{0}^{\frac{\pi}{2}} sin^5\theta \ \ d\theta$$ Hence, with reference to graphs of circular functions, find $$\int_{0}^{\frac{\pi}{2}} cos^5\theta \ \ d\theta$$ ... explaining your reasoning.

The first integral is not too difficult to compute and I was successfully able to get $\frac{8}{15}$. However, the issue is with the next part. This is a non-calculator question, and, as becomes obvious by graphing the two, the second integral is identical in value due to the symmetry of the two graphs.

As someone who has no idea what either of ${sin}^5\theta$ or ${cos}^5\theta$ look like, how could I answer the second part without having to compute it manually?

Note: The time given for the second part of the question (as this was a two part questions instead of one, as opposed to how I've shown it) was less than that of the first part, implying that the knowledge of such symmetry of graph was expected, and that was what was being tested.

StopReadingThisUsername
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  • The powers are irrelevant; what matters is that the cosine is a translate of sine. Nevertheless, it's not hard to sketch these graphs if you keep in mind the values of sine and cosine are between $-1$ and $1$. What happens when you take the fifth power of such a number? The values get closer to zero, except in the limiting cases of $-1$ and $1$, which stay fixed. – symplectomorphic Jun 17 '16 at 04:07

2 Answers2

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I was stuck on this question as well, however I had to go back to basics and realise a function of the integral is to show you the area under the curve. Therefore if you sketch the curves for both sin x and cos x at the given range you find the areas are the same ( as cos and sin are translations across the x-axis of each other) and they are reflected across x/4. Hence, the areas are the same despite the exponent. That is why the mark scheme asks for an appropriate reference to symmetry and graphs.

Stephanie
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Using the identity $\cos \theta = \sin(\pi/2 - \theta)$ lets you use $u$-substitution to relate the second integral to $$\int_0^{\pi/2} \sin^5 u\,du = \int_0^{\pi/2} \sin^5 \theta\,d\theta, $$ which you already know.

pjs36
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  • Graphically, the main identity is visible: We can transform $\sin \theta \mapsto \sin(\pi/2 - \theta)$ by first reflecting across the $y$-axis ($\theta \mapsto - \theta$) then shifting $\pi/2$ units left ($-\theta \mapsto -\theta + \pi/2$). I have crudely drawn pictures as proof, if there is any interest. – pjs36 Jun 17 '16 at 02:57