Integrate inverse circular kernels and change variables

Connect recognition with substitution

Here log means the natural logarithm (base e), and all angles are in radians. Reciprocal, logarithmic and supplied-derivative recognition give exact primitives; retain their constants. The chain rule explains a change of variable: if F′(u)=f(u), then the derivative of F(g(x)) is f(g(x))g′(x). Substitution names the inner function and rewrites that entire derivative pattern

Recover the inverse-sine kernel

For a>0, differentiate arcsin(x/a). Its derivative is (1/a)/√(1−x²/a²)=1/√(a²−x²), because the positive scale a leaves the square root as √(a²−x²)/a. The identity holds on −a<x<a. The primitive has finite endpoint values, but the original integrand is undefined there

Recover the inverse-tangent kernel

For a>0, differentiating arctan(x/a) gives a/(a²+x²). Thus a numerator1 requires an extra factor1/a; a numerator a does not. The positive denominator makes the derivative identity valid for all real x.

Which primitive has derivative 1/(1+x²) on all real inputs?

For ∫6/(9+x²) dx = K arctan(x/3)+C, find K.

Keep domains and constants separate

A primitive family is defined on an interval where its derivative equals the original integrand. The derivative of log|x| is1/x on each component of x≠0; the constants on the negative and positive components may differ. Absolute values do not remove a zero denominator. On a valid connected interval, all primitives differ by a constant

Is log|1|−log|−1| a valid evaluation of the ordinary finite integral of 1/x from −1 to1?

Include the affine chain scale

For u=bx+c with b≠0, the differential is du=b dx. The arcsine-kernel coefficient is k/b in ∫k/√(a²−(bx+c)²) dx. The arctangent-kernel coefficient is k/(ab) in ∫k/(a²+(bx+c)²) dx. A negative b changes the primitive scale and may reverse transformed bounds

Work an affine inverse-sine primitive

For ∫3/√(4−(1−2x)²) dx, use u=1−2x and du=−2 dx. The result is −(3/2)arcsin((1−2x)/2)+C. The real integrand requires |1−2x|<2, giving −1/2<x<3/2. Differentiation restores both the negative inner derivative and the positive numerator

For the integral of three over the square root of [four minus (one minus two x) squared], give the coefficient multiplying arcsin in a primitive

Complete the square before choosing a kernel

For ∫6/(x²−6x+13) dx, rewrite the denominator as (x−3)²+4. Its scale is a=2, so the primitive is3 arctan((x−3)/2)+C on R. The positive residual4 matters; a difference of squares is not the same arctangent kernel

For the integral of six over x squared minus six x plus thirteen, what coefficient multiplies arctan in a primitive?

Take a positive factor out of a square root

For ∫4/√(27+18x−9x²) dx, complete the square to get9[4−(x−1)²]. The square root contributes3, so the primitive is(4/3)arcsin((x−1)/2)+C, valid on −1<x<3. Extracting9 instead of3 from the root gives the wrong scale

Rewrite the entire integral in one variable

For ∫x√(x²+5) dx, set u=x²+5 and du=2x dx. Then the integral is(1/2)∫u^(1/2) du=(1/3)u^(3/2)+C. Return to x to obtain(x²+5)^(3/2)/3+C. Do not transform the square root while leaving an unrelated x dx behind

A substitution need not be polynomial

For ∫e^(2x)(1+e^(2x))³ dx, set u=1+e^(2x), du=2e^(2x)dx. The result is u⁴/8+C=(1+e^(2x))⁴/8+C. For ∫(log x)²/x dx on x>0, u=log x gives u³/3+C. The stated domain is part of each result

Use a nonlinear inverse-tangent pattern

For ∫6x/(9+x⁴) dx, set u=x², du=2x dx. This gives3∫du/(9+u²)=arctan(u/3)+C. Therefore the primitive is arctan(x²/3)+C on R. At x=0 the original numerator is zero and the denominator is nonzero; matching2x dx to du remains valid there

Give K in ∫6x/(9+x⁴) dx = K arctan(x²/3)+C.

Check the domain after an exponential substitution

For ∫4e^(2x)/√(9−e^(4x)) dx, let u=e^(2x). Since du=2e^(2x)dx, the primitive is2 arcsin(e^(2x)/3)+C. Require e^(2x)<3, hence x<(log3)/2. Positivity of e^(2x) does not remove the upper restriction

Use a condition to choose one primitive

Suppose F′(x)=3/√(4−(2x−1)²) and F(1)=π/2. The primitive scale is3/2. At x=1 its inverse input is1/2 and arcsin(1/2)=π/6, so the inverse term contributesπ/4. Therefore F(x)=(3/2)arcsin((2x−1)/2)+π/4 on −1/2<x<3/2.

Given that F prime is three over the square root of [four minus (two x minus one) squared] and F(1) is π/2, give the coefficient of π in the additive constant of an arcsin primitive

Transform bounds at the same time as the variable

For a proper definite integral, either return to the original variable before applying the original bounds, or keep the substituted variable and transform both bounds. Never apply x-endpoints directly to a primitive written in u unless their images happen to agree. The arbitrary constant cancels

Keep a negative inner slope and descending bounds consistent

For ∫ from0 to1 of6/[1+(1−2x)²] dx, set u=1−2x. The bounds become1 and−1 and dx=−du/2. Thus the integral is−3[arctan u] from1 to−1=3π/2. Reversing the u-bounds instead gives+3∫ from−1 to1 of1/(1+u²) du, the same result

What is the transformed upper endpoint when x=1 and u=1−2x?

A valid primitive can handle a turning substitution

For ∫ from−1 to1 of2x/(1+x⁴) dx, use the primitive arctan(x²). It is differentiable throughout the interval, including0. The endpoint values are bothπ/4, so the signed integral is0. The chain-rule justification does not require inverting x² or dividing by2x at0.

For the integrand two x over one plus x to the fourth power, between x equals −1 and x equals 1, what do the two lobes imply about its signed integral and geometric area?

Derive the kernels and their alternatives, ungraded

Starting from the inverse-circular derivatives, derive both scaled integral kernels for a>0. Explain the different reciprocal scale factors, the real domains and why −arccos(x/a) is also a primitive of1/√(a²−x²). State how that alternative differs from arcsin(x/a)

Model: differentiate and use a constant identity

The derivative of arcsin(x/a) is1/√(a²−x²) on(−a,a), while the derivative of arctan(x/a) is a/(a²+x²) on R. Therefore a numerator1 needs coefficient1/a only in the second kernel. Since arcsin t+arccos t=π/2 for−1≤t≤1, −arccos(x/a)=arcsin(x/a)−π/2 on the shared interval. The two primitives differ by a constant. Finite endpoint values do not make the original square-root denominator nonzero

Design and justify substitutions, ungraded

Choose substitutions and find complete primitives for ∫(log x)⁴/x dx on x>0, ∫cos x/√(4−sin²x) dx, and ∫x(2x²+3)^(5/2) dx. Show how the full differential transforms, return to x, state domains and differentiate each answer to check it

Model: match each inner derivative

Use u=log x to obtain(log x)⁵/5+C on x>0. Use u=sin x to obtain arcsin(sin x/2)+C on R; the denominator is always at least√3, and the chain identity remains valid where cos x=0. Use u=2x²+3 with du=4x dx to obtain(2x²+3)^(7/2)/14+C on R. In each case differentiating recovers the whole integrand, including its scale

Justify the definite substitution identity, ungraded

Suppose f is continuous on an interval containing the image of a continuously differentiable g on[L,R]. Derive ∫ fromL toR of f(g(x))g′(x) dx = ∫ fromg(L) tog(R) of f(u) du using an antiderivative of f. Explain why this does not license endpoint subtraction across a singularity

Model: use the chain rule and the fundamental theorem

Choose F with F′=f on the image interval. Then(F∘g)′=f(g)g′ throughout[L,R], so the original integral is F(g(R))−F(g(L)), equal to the transformed integral. This proof permits a turning g because it never solves for an inverse or divides by g′. Continuity and a valid primitive throughout the interval are essential. A pole in1/x breaks those hypotheses; log|x| cannot be used as a single interval primitive across0.

Explain signed integral and area, ungraded

Use the integrand2x/(1+x⁴) to find its signed integral and geometric area between−1 and1. Explain why equal transformed endpoint values do not imply that the integrand is zero, and identify the point at which the area calculation must split

Model: split at the sign change

The primitive is arctan(x²). The signed integral isπ/4−π/4=0. On[−1,0] the integral is−π/4, while on[0,1] it isπ/4. Geometric area is the sum of their absolute contributions, π/2. Split at0 because the integrand changes sign, not because it is singular. The substitution runs from1 down to0 and back to1.

Verify with technology and report domains, ungraded

Use radian-mode symbolic or numerical technology to check the affine inverse-sine initial-value example and the negative-slope definite integral above. Differentiate the proposed primitive, substitute the initial value and compare the exact definite result with a decimal evaluation. Record the domain restrictions and your own calculations

Model: check three separate claims

Differentiating(3/2)arcsin((2x−1)/2)+π/4 gives3/√(4−(2x−1)²) on−1/2<x<3/2, and its value at1 isπ/2. The separate negative-slope integral equals3π/2, approximately4.71239. Agreement at selected points is a check, not a derivation. A tool’s real domain and angle mode must match the mathematics

Exact values and complete substitution explanations

Give exact inverse primitives, coefficients, conditions and transformed endpoints in the stated forms. Use reciprocal, logarithmic and supplied-derivative recognition where appropriate. General proof, unrestricted symbolic-expression entry, independent substitution design, full graph explanation and technology execution remain ungraded