<s>[INST] Unusual Constant appearing for definite integral I’m learning integration and have come across a question in a textbook. I arrived to the same answer as the textbook itself, but I know it is not correct. It starts as part a) show that $$\frac{d}{dx} tan^{-1} \left(\frac{3tan(x)}{2} \right)= \frac{6}{5sin^2(x)+4}$$ Hence, find area bounded by curve from $ \dfrac{1}{5sin^2(x) + 4}$ from $x=0$ to $x=7$. Clearly using part A to integrate and substituting 7 and 0, I arrive to 0.153 square units, in line with the textbook. Yet, by using desmos to graph the curve, using lower rectangles, minimum area has to be $0.111111 \times 7 = 0.77777$ units squared. I put the integral on integral calculator and it yielded $0.153 + \dfrac{\pi}{3}$, without explaining where this constant came from. I am unfamiliar with constants in definite integrals, and struggled to find anything online to answer this. Thanks.
(Note: I am not asking for homework help, Im just asking about how to find constants in these definite integrals, particularly in a question like this, which the textbook itself overlooked. Thank you. Any help is appreciated.)
 [/INST] This happens because $\tan x$ is not defined at the points $x = \frac{\pi}{2},  \frac{3\pi}{2} $ which lie in $(0,7)$ so $\tan^{-1} \left(\frac{3\tan(x)}{2} \right)$ becomes discontinuous at these points.
This is why you have to split the interval into sub-intervals $(0,\frac{\pi}{2}), (\frac{\pi}{2}, \frac{3\pi}{2}), (\frac{3\pi}{2},7)  $ while substituting the limits.
$$ \Delta = \dfrac16 \left[ \tan^{-1} \left(\frac{3\tan(x)}{2} \right)_0^\frac{\pi^-}{2} + \tan^{-1} \left(\frac{3\tan(x)}{2} \right)_\frac{\pi^+}{2}^\frac{3\pi^-}{2} + \tan^{-1} \left(\frac{3\tan(x)}{2} \right)_\frac{3\pi^+}{2}^7 \right] $$
$$ \Delta = \dfrac16 \left[ \dfrac{\pi}{2} - 0+ \dfrac{\pi}{2} - \left(-  \dfrac{\pi}{2} \right) + \tan^{-1}\left(\frac{3\tan7}{2} \right)  - \left(-  \dfrac{\pi}{2} \right) \right] $$
$$ \Delta = \dfrac{\pi}{3}+\dfrac16 \tan^{-1}\left(\frac{3\tan7}{2} \right) $$
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