1

Let $a$ be a real number in the interval $[0,314]$ such that

$$\displaystyle \int^{3\pi+a}_{\pi+a}|x-a-\pi|\sin \frac{x}{2}dx=-16.$$

Determine the number of such real values of $a.$

What I tried:

Put $x-a-\pi=t$. Then,

$$\displaystyle \int^{2\pi}_{0}|t|\sin \bigg(\frac{t+a+\pi}{2}\bigg)dx=-16$$

$$\int^{2\pi}_{0}t\cos\bigg(\frac{t+a}{2}\bigg)dt=-16$$

$$2t\sin \frac{t+a}{2}\bigg|^{2\pi}_{0}+4\int^{2\pi}_{0}\cos\frac{a+t}{2}dt=-16$$

$$-2\pi\sin \frac{a}{2}+4\int^{2\pi}_{0}\cos\frac{a+t}{2}dt=-16.$$

How do I solve this?

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jacky
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1 Answers1

2

You already have $$\int^{2\pi}_{0}t\cos\bigg(\frac{t+a}{2}\bigg)dt=-16$$ From here, we have $$2t\sin\frac{t+a}{2}\bigg|^{2\pi}_{0}-2\int_0^{2\pi}\sin\frac{t+a}{2}dt=-16$$

$$\iff -4\pi\sin\frac a2-2\int_0^{2\pi}\bigg(\sin\frac t2\cos\frac a2+\cos\frac t2\sin\frac a2\bigg)dt=-16$$

$$\iff -4\pi\sin\frac a2-2\cos\frac a2\int_0^{2\pi}\sin\frac t2dt-2\sin\frac a2\int_0^{2\pi}\cos\frac t2dt=-16$$

$$\iff -4\pi\sin\frac a2-8\cos\frac a2=-16$$

$$\iff \pi\sin\frac a2+2\cos\frac a2=4\tag1$$ The LHS of $(1)$ can be written as $$\sqrt{\pi^2+2^2}\sin\bigg(\frac a2+b\bigg)$$ for some $b\in\mathbb R$.

Now, we have $$\sqrt{\pi^2+2^2}\sin\bigg(\frac a2+b\bigg)\le \sqrt{\pi^2+4}\lt\sqrt{(2\sqrt 3)^2+4}=4$$ because of $\pi\lt 2\sqrt 3$. (See here) So, the LHS of $(1)$ is smaller than $4$.

It follows that the number of $a$ is $\color{red}0$.

mathlove
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