Visibly Unequal Answers in Using Different Methods of Integration

Question

Question: Evaluate the Integral:$\displaystyle\int\sec(x)\,\mathrm dx $

Approach 1 :

Write $\sec(x)$ as $\ \frac{1}{\cos(x)}$,

$$\ \int \frac{1}{\cos(x)}\,\mathrm dx $$

Substitute $\sin^2(x) + \cos^2(x) $ for $1$,

$$\int \frac{\sin^2(x) + \cos^2(x)}{\cos(x)}\,\mathrm dx $$

Expand the fraction and simplify to get a linear integrand,

$$\int\tan(x)\sec(x)\,\mathrm dx + \int\cos(x)\,\mathrm dx $$

Evaluating the integrand using standard integrals,

$$\ (\sec(x)+c_1) + (\sin(x)+c_2)$$

Merging the constants and simplifying the integrals we get,

$$\sec(x) + \sin(x) + C $$

Approach 2 :

Multiplying and dividing by $\sec(x)+\tan(x)$ we get,

$$\int \frac{\sec(x)(\sec(x)+\tan(x))}{\sec(x)+\tan(x)}\,\mathrm dx $$

Let y=$\sec(x)+\tan(x)$ and finding its derivative,

$$\frac{\mathrm dy}{\mathrm dx} = \sec(x)\bigl(\sec(x)+\tan(x)\bigr) $$

$$\,\mathrm dx = \frac{\mathrm dy}{\sec(x)\bigl(\sec(x)+\tan(x)\bigr)} $$

Substituting $y$ and $\mathrm dy$,

$$\ \int \frac{\mathrm dy}y $$

Evaluating the integral,

$$\ln(|y|) $$

Substituting for $y$,

$$\ln\left(\bigl|\sec(x)+\tan(x)\bigr|\right) $$

The answers don't appear to be the same. My question is, since they are the results of the same integral the answers must be the same yet appear different how can this be proved to be the same?

I have already seen this and this questions but I have tried hard to reach a proof showing equivalence at no avail, therefore I am writing a different question.

Edit: I hadn't noticed the error in the Approach 1, which on correction would very well resolve the conflicts. Sorry for an erroneous question.

Seems to be a mistake in your first approach. You equate $\frac{\sin^2 x}{\cos x}$ with $\tan x ~\sec x.$ — user2661923, Aug 25 '21 at 12:35
\sec x, not sec x and \ln x, not ln x. See how to write MathJax. — Toby Mak, Aug 25 '21 at 12:36
There is no reason to delete your well written question, with nice work shown. The fact that you had an analytical error is irrelevant to judging the quality of your question. — user2661923, Aug 25 '21 at 12:37
Related: https://math.stackexchange.com/questions/6695/ways-to-evaluate-int-sec-theta-mathrm-d-theta — Integreek, Dec 06 '24 at 16:19

score 3 · Accepted Answer · answered Aug 25 '21 at 12:34

3

It happens that$$\frac{\sin^2x}{\cos x}=\tan(x)\sin(x)\ne\tan(x)\sec(x).$$

answered Aug 25 '21 at 12:34

José Carlos Santos

440,053

score 2 · Answer 2 · edited Aug 25 '21 at 12:40

2

In "Expand the fraction and simplify to get a linear integrand", it would be $$\int \tan(x) \sin(x)\ dx+\int \cos(x)\ dx$$, and not $$\int \tan(x) \sec(x) \ dx+ \int \cos(x) \ dx$$.

edited Aug 25 '21 at 12:40

Toby Mak

17,073

answered Aug 25 '21 at 12:34

Mano Prakash P

56

score 0 · Answer 3 · answered Dec 06 '24 at 15:59

Let me complete the solution as per Approach 1 for reference:

$$\require{cancel}\begin{align}\int\sec x\mathrm dx&=\int\frac{\sin^2x+\cos^2x}{\cos x}\mathrm dx\\&=\int\frac{\sin^2x}{1-\sin^2x}\mathrm d(\sin x)+\int\cos x\mathrm dx\\&=\cancel{\int-\mathrm d(\sin x)}+\int\frac{\mathrm d(\sin x)}{1-\sin^2x}\cancel{+\sin x}+C\\&=\frac12\ln\left|\frac{1-\sin x}{1+\sin x}\right|+C\\&=\frac12\ln\left|\frac{(1-\sin x)^2}{\cos^2x}\right|+C\\&=\ln|\sec x+\tan x|+C\end{align}$$

Visibly Unequal Answers in Using Different Methods of Integration

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