If $\int_{\pi/2}^\theta\sin x\,dx=\sin2\theta$, then the value of $\theta$ satisfying $0<\theta<\pi$, is
(a) $3\pi/2$
(b) $\pi/6$
(c) $5\pi/6$
(d) $\pi/2$

Method 1:

I apply Leibniz rule and differentiate both the sides with respect to $\theta$. No option seems to be correct.

Method 2:

I integrate the left hand side and get to a conclusion of option d).

Now which method to follow?

  • 13
    $\begingroup$ Suppose you are given "If $x^2=x$, then find $x$". Would you differentiate both sides to get the solution? $\endgroup$
    – leonbloy
    Commented May 26, 2017 at 13:49
  • 2
    $\begingroup$ @TheGreatDuck Actually, no. $2x = 1 \implies x=1/2$. $\endgroup$
    – leonbloy
    Commented May 26, 2017 at 20:52
  • $\begingroup$ @TheGreatDuck Then $2x=0 \Rightarrow x=0$? $\endgroup$ Commented May 27, 2017 at 7:27

3 Answers 3


The left and right side aren't the same function, they just happen to intersect at a specific angle $\theta$. Because of that, you cannot expect both to have the same derivative, so Method 1 isn't correct.

  • 4
    $\begingroup$ In other words, $\theta$ is not a variable, so you cannot differentiate with respect to $\theta$. $\endgroup$
    – 5xum
    Commented May 26, 2017 at 8:22
  • 1
    $\begingroup$ I was being very dumb. Thanks for making me realize that. $\endgroup$ Commented May 26, 2017 at 8:26

Method $(2)$ is correct, and as you found, the correct choice is $(\text{d})$.

Method $(1)$ is seriously flawed.

The given equation does not assert that two functions of $\theta$ are identically equal. Rather it asks for a value of $\theta$ for which the two functions of $\theta$ are equal.

To dramatize the error, suppose we want a value of $\theta$ with $0 < \theta < \pi$ such that $$\sin(\theta) = \frac{\theta}{2}$$ It makes no sense to differentiate both sides. If you differentiated anyway, ignoring the fact that equation does not assert that the LHS and RHS are the same function, you would get $$\cos(\theta) = \frac{1}{2}$$ yielding $\theta = \pi/3$, which doesn't satisfy the original equation.

  • 1
    $\begingroup$ Even more noticeable if you then try to apply the differentiation technique again to get $$-\sin(\theta) = 0$$ $\endgroup$
    – Chris
    Commented May 26, 2017 at 10:49

Careless differentiation is dangerous. Imagine you are trying to solve

$$ 2\theta-1 = \theta$$

The obvious and only answer is $\theta = 1$. If you differentiate, you get $2 = 1$, which might induce the faulty idea that there is no solution.

So, let's check again, without too much computations, only basic assumptions. Interestingly, the question resorts to different arguments.

  • (a) 3π/2 : since it is $\ge \pi$, ruled out,
  • (b) π/6: $\sin x > 0$ in $]0,\pi/2[$, and $0< π/6< \pi/2$, hence $∫^{π/6}_{π/2}\sin x dx <0$, ruled out!
  • (c) 5π/6: $\sin (5\pi/3) = $\sin ((6-1)\pi/3) = $\sin ((-1)\pi/3) $ is negative. But $\sin x > 0$ in $]\pi/2,\pi[$, and $\pi/2< 5π/6< \pi$, hence $∫^{5π/6}_{π/2}\sin x dx >0$, ruled out!
  • (d) π/2: the only solution left, and easily verified.

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