Different answers for definite integration

Question

Me and my friend were trying to solve this integration question for our upcoming exams and we came up with 2 different ways, both are shown in pictures below, I don't know why we got two different answers. I think it's related to domain/range of inverse functions. Please let me know.

1^st Method.^{(original image)}

\begin{align*} 2I &= \pi \int_{0}^{\pi} \frac{\sec x \tan x}{1 + \sec^2 x} \, \mathrm{d}x \\ &= \pi \int_{0}^{\frac{\pi}{2}} \left( \frac{\sec x \tan x}{1 + \sec^2 x} + \frac{\sec(\pi-x) \tan(\pi-x)}{1 + \sec^2 x} \right) \, \mathrm{d}x \\ &= 2\pi \int_{0}^{\frac{\pi}{2}} \frac{\sec x \tan x}{1 + \sec^2 x} \, \mathrm{d}x \\ &= 2\pi \left[ \tan^{-1}(\sec x)\right]_{0}^{\frac{\pi}{2}} \\ &= 2\pi \left[ \tan^{-1}(\infty) - \tan^{-1}(1) \right] \\ &= 2\pi \left[ \frac{\pi}{2} - \frac{\pi}{4} \right] = \frac{\pi^2}{2}. \\[1em] \implies & \quad I = \frac{\pi^2}{4}. \end{align*}

2^nd Method.^{(original image)}

\begin{align*} 2I &= \int_{0}^{\pi} \frac{\pi \sec x \tan x}{1 + \sec^2 x} \, \mathrm{d}x. \end{align*}

Let $t = \sec x$. Then

$$ \mathrm{d}t = \sec x \tan x \mathrm{d}x, \qquad \text{Limits:} \quad \begin{array}{|c|c|c|} \hline x & 0 & \pi \\ \hline t & 1 & -1 \\ \hline \end{array} $$

So,

\begin{gather*} 2I = \int_{1}^{-1} \frac{\pi}{1+t^2} \, \mathrm{d}t = \left[ \pi \tan^{-1}(t) \right]_{1}^{-1} = \pi \left[ -\frac{\pi}{4} - \frac{\pi}{4} \right] \\[1em] \implies \quad -\frac{\pi^2}{4}. \end{gather*}

Answer 1

2^nd Method. \begin{align*} 2I &= \int_{0}^{\pi} \frac{\pi \sec x \tan x}{1 + \sec^2 x} \, \mathrm{d}x. \end{align*} Let $t = \sec x$.

@RamanujanXXV But the first method involves $\sec x$ too and the upper limit is $\frac{\pi}2.$ Is that not incorrect? Besides, if the function that we're integrating is defined in the given interval, why would $\sec x$ matter? Is it because we substitute $t=\sec(x)?$

Yes, the substitution technique that you use in Method 2 (but not Method 1) is legitimate only if certain specific conditions are satisfied:

• From: Justification for Integration by Substitution

  If $g'$ is integrable on $[a,b]$ and $f$ is integrable, and has an antiderivative, on $g[a,b],$ then, substituting $t=g(x)$ into the LHS, $$\int_a^bf\big(g(x)\big)\, g'(x)\,\mathrm{d}x=\int_{g(a)}^{g(b)}f(t)\,\mathrm{d}t.$$

$g(x)=\sec x$ not being bounded on $[0,π]$ invalidates your Method 2 because

• $g$ is not bound on $[0,π]\implies g' $ is not bound on $[0,π]\implies g'$ is not integrable on $[0,π].$

Generally, if $g$ is continuously differentiable on $[a,b],$ then $g'$ is integrable on $[a,b],$ which means that it is valid to proceed with the change of integration variable (and that condition on $f$ is automatically satisfied on succesful integration with the new variable $t$).