please explain how to solve the indefinite integral for tan(sqrt(x))dx If the indefinite integral is not solvable, please explain how to solve it as a definite integral. I am not interested in numerical solutions.

Gemini 4 Argon

Answer: The indefinite integral of tan(sqrt(x))dx cannot be expressed in terms of elementary functions, requiring either the use of polylogarithm special functions, an infinite power series, or complex contour integration for specific definite cases.

To begin analyzing the integral I = ∫ tan(√x) dx, we first simplify the radical by applying the substitution u = √x. Squaring both sides gives x = u², which means dx = 2u du. Substituting these into the original expression transforms the integral into I = ∫ 2u tan(u) du. We can then apply integration by parts by setting w = 2u and dv = tan(u) du, which yields dw = 2 du and v = -ln|cos(u)|. This leaves us with I = -2u ln|cos(u)| + 2 ∫ ln|cos(u)| du. At this point, we hit a roadblock regarding standard calculus techniques. The remaining integral, ∫ ln|cos(u)| du, is a well-known non-elementary integral. According to Liouville's theorem and the Risch algorithm, its antiderivative cannot be expressed as a finite combination of elementary functions (polynomials, exponentials, logarithms, or trigonometric functions). However, if we expand our toolkit to include special functions, we can find an exact closed-form antiderivative using the dilogarithm function, Li₂(z). By rewriting cos(u) in its complex exponential form, cos(u) = (1/2)e^(-iu)(1 + e^(2iu)), we can integrate term-by-term. Substituting √x back in for u, the exact indefinite integral is: I = -2√x ln|cos(√x)| + i x - 2√x ln(2) - i Li₂(-e^(2i√x)) + C (taking the real part for real √x). Alternatively, we can express the indefinite integral analytically as an infinite series without using complex numbers. The Taylor series for tan(u) is ∑ [(-1)^(n-1) 2^(2n) (2^(2n) - 1) B_(2n) / (2n)!] u^(2n-1) for |u| < π/2, where B_(2n) are the Bernoulli numbers. Multiplying by 2u and integrating term-by-term with respect to u gives an exact power series representation. Substituting u = √x yields: I = ∑_(n=1 to ∞) [(-1)^(n-1) 2^(2n+1) (2^(2n) - 1) B_(2n) / ((2n+1)!)] x^(n + 1/2) + C, valid for 0 ≤ x < π²/4. When evaluating this as a definite integral, the approach depends on the limits of integration because tan(√x) has vertical asymptotes wherever √x = π/2 + kπ (i.e., at x = (π/2 + kπ)²). For an interval without singularities, such as from x = 0 to x = π²/16 (where u goes from 0 to π/4), we use the integration by parts formula directly: ∫_(0)^(π²/16) tan(√x) dx = [-2u ln(cos(u))]_(0)^(π/4) + 2 ∫_(0)^(π/4) ln(cos(u)) du. Using the Fourier series of ln(cos(u)), the remaining definite integral evaluates exactly to -(π/4)ln(2) - G/2, where G is Catalan's constant. Thus, the exact definite integral on [0, π²/16] is (π/4)ln(2) - G. If you wish to evaluate the definite integral across a singularity—for example, from x = 0 to x = π² (so u goes from 0 to π, crossing the pole at u = π/2)—the standard Riemann integral diverges, and we must use the Cauchy Principal Value (PV). Using the integration by parts setup, PV ∫_(0)^(π) 2u tan(u) du reduces to 2 ∫_(0)^(π) ln|cos(u)| du because the boundary terms at u = 0 and u = π vanish, and the symmetric logarithmic singularities at u = π/2 cancel out in the principal value limit. Using the symmetry of|cos(u)| and Euler's famous log-sine integral result (∫_(0)^(π/2) ln(cos(u)) du = -(π/2)ln(2)), we find that PV ∫_(0)^(π²) tan(√x) dx = -2π ln(2).

Claude Opus 5.5

Answer: The integral ∫tan(√x)dx has no elementary antiderivative, but it does have an exact closed form in terms of the Clausen function (equivalently, the dilogarithm): ∫tan(√x)dx = −2√x·ln|2cos√x| + Cl₂(π−2√x) + C, valid on any interval avoiding the poles x = (π/2+kπ)².

Step 1: Substitution. Let x = t², with t = √x ≥ 0, so dx = 2t dt. The integral becomes ∫2t·tan t dt. Step 2: Integration by parts. Take u = t and dv = tan t dt, so du = dt and v = −ln|cos t|. This gives ∫2t tan t dt = −2t ln|cos t| + 2∫ln|cos t| dt. Step 3: The non-elementary part. ∫ln|cos t| dt has no elementary antiderivative, and this is why the original integral cannot be written with elementary functions alone. It can still be handled exactly, without numerics, using the Fourier series ln|2cos t| = −Σ_{k≥1} (−1)^k cos(2kt)/k. Integrating term by term gives ∫ln|cos t| dt = −t ln2 − ½Σ(−1)^k sin(2kt)/k² + C. The series that appears is the Clausen function Cl₂(θ) = Σ sin(kθ)/k² = Im Li₂(e^{iθ}), evaluated at θ = 2t+π. Because Cl₂ is odd and 2π-periodic, Cl₂(2t+π) = −Cl₂(π−2t). Therefore ∫ln|cos t| dt = −t ln2 + ½Cl₂(π−2t) + C. Check: Cl₂′(θ) = −ln|2sin(θ/2)|. So d/dt[½Cl₂(π−2t)] = ln|2cos t|. Adding the derivative of −t ln2 leaves ln|cos t|, as required. Step 4: Assemble the result and substitute back t = √x: ∫tan(√x)dx = −2√x·ln|2cos√x| + Cl₂(π−2√x) + C. An equivalent dilogarithm form, which you can verify directly by differentiation, is: ∫tan(√x)dx = i·x − 2√x·ln(1+e^{2i√x}) + i·Li₂(−e^{2i√x}) + C. Its derivative with respect to t is 2it(1−e^{2it})/(1+e^{2it}) = 2t tan t. Take care with branches, and use the antiderivative only on intervals between poles. Step 5: Definite integrals. tan(√x) has non-integrable singularities at x = (π/2+kπ)². Near each one, tan behaves like 1/(π/2−t), which produces a logarithmic divergence. So a definite integral exists only if [a, b] avoids these points. On such an interval, simply evaluate F(b) − F(a) using the Clausen form above. Example: on [0, π²/16], the upper limit corresponds to t = π/4. The lower limit contributes Cl₂(π) = 0. The upper limit contributes Cl₂(π/2) = G (Catalan's constant) and −2(π/4)ln(√2) = −(π/4)ln2. Hence ∫₀^{π²/16} tan(√x)dx = G − (π/4)·ln2, an exact closed form. Its value is about 0.372, which serves as a sanity check only. Summary: substitute x = t², integrate by parts, and recognize ∫ln cos t dt as a Clausen/dilogarithm function. The answer is exact but not elementary.

GPT-6 Astra

Answer: The integral \(\int \tan(\sqrt{x})\,dx\) has no elementary antiderivative, but it has an exact expression using the Clausen function, which also gives exact definite integrals.

**1. Substitute and integrate by parts.** Set \(t=\sqrt{x}\), so \(x=t^2\) and \(dx=2t\,dt\). Then \[ \int\tan(\sqrt{x})\,dx=2\int t\tan t\,dt. \] Using integration by parts with \(u=t\) and \(dv=\tan t\,dt\), hence \(v=-\ln|\cos t|\), gives \[ 2\int t\tan t\,dt =-2t\ln|\cos t|+2\int\ln|\cos t|\,dt. \] The remaining integral \(\int\ln|\cos t|\,dt\) is a standard non-elementary integral: it cannot be expressed using a finite combination of elementary functions. Thus the original integral is not elementary either. This does **not** mean it cannot be expressed exactly. **2. Write an exact antiderivative.** Define the Clausen function by the convergent series \[ \operatorname{Cl}_2(\theta)=\sum_{n=1}^{\infty}\frac{\sin(n\theta)}{n^2}. \] Its derivative, away from multiples of \(2\pi\), is \[ \operatorname{Cl}_2'(\theta)=-\ln\left|2\sin\frac{\theta}{2}\right|. \] Consequently, \[ \int\ln|\cos t|\,dt =\frac12\operatorname{Cl}_2(\pi-2t)-t\ln2+C. \] Substituting back yields \[ \boxed{\int\tan(\sqrt{x})\,dx =-2\sqrt{x}\ln|2\cos(\sqrt{x})| +\operatorname{Cl}_2(\pi-2\sqrt{x})+C.} \] This holds on each interval where the integrand is defined. **3. Evaluate definite integrals exactly.** For \(0\le a<b\), provided there are no poles on \([a,b]\), \[ \boxed{\int_a^b\tan(\sqrt{x})\,dx =\left[-2\sqrt{x}\ln|2\cos(\sqrt{x})| +\operatorname{Cl}_2(\pi-2\sqrt{x})\right]_a^b.} \] Without specified bounds, this is the general exact answer. Definite integration does not automatically eliminate the need for special functions, although particular bounds can simplify the result. For example, choose \(a=0\) and \(b=\pi^2/16\). Since \(\operatorname{Cl}_2(\pi)=0\) and \(\operatorname{Cl}_2(\pi/2)=G\), we obtain \[ \boxed{\int_0^{\pi^2/16}\tan(\sqrt{x})\,dx =G-\frac{\pi}{4}\ln2,} \] where \(G\) is Catalan’s constant, defined exactly by \[ G=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)^2}. \] No numerical approximation is involved. **Watch for singularities:** the real integrand has poles at \(x=(\pi/2+k\pi)^2\), for \(k=0,1,2,\ldots\). An integral reaching or crossing one of these poles does not converge as an ordinary improper integral; for example, \(\int_0^{\pi^2/4}\tan(\sqrt{x})\,dx=+\infty\).

Grok 4.7

Answer: The indefinite integral ∫ tan(√x) dx equals -2√x ln|2 cos(√x)| - Im[Li_{2}(-e^{2i √x})] + C (equivalently via its Fourier series).

Substitute u = √x so that x = u^{2} and dx = 2u du. The integral becomes 2 ∫ u tan u du. Integration by parts with the choice p = u, dq = tan u du yields -u ln|cos u| + ∫ ln|cos u| du (the constant factors are restored at the end). The Fourier series ln(cos u) = -ln 2 + ∑_{k=1}^∞ (-1)^{k+1} cos(2ku)/k (valid for |u| < π/2) integrates term by term to -u ln 2 + (1/2) ∑_{k=1}^∞ (-1)^{k+1} sin(2ku)/k^{2}. Multiplying by the outer factor of 2 produces the explicit antiderivative -2u ln|2 cos u| + ∑_{k=1}^∞ (-1)^{k+1} sin(2ku)/k^{2}. The infinite sum is precisely -Im[Li_{2}(-e^{2iu})], giving the closed form in the dilogarithm. Restoring u = √x supplies the claimed antiderivative; direct differentiation confirms that it recovers tan(√x) wherever the expression is defined (i.e., away from the poles of tan). The same formula, evaluated at the endpoints and continued analytically across branches, therefore also yields any definite integral whose path avoids those poles.

Final Summary

The AI Roundtable unanimously agreed that integrating tan(√x) dx yields a non-elementary antiderivative, but it can still be solved exactly without numerical approximation. By substituting u = √x and applying integration by parts, Gemini 4 Argon, Claude Opus 5.5, GPT-6 Astra, and Grok 4.7 all isolated the non-elementary term ∫ ln|cos(u)| du. The models then mapped this term to closed-form special functions—specifically the Clausen function Cl₂(π - 2√x) or the dilogarithm Li₂—and showed how bounds like [0, π²/16] evaluate to exact expressions involving Catalan's constant.

All 4 models agreed