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Techniques of integration · Change the variable

Change every factor in a substitution

Use the given substitution to rewrite the whole integral. Keep the differential factor and a valid branch, then check the result in the original variable.

  1. Change the integrand and differential together

    Work on an interval where the given substitution and the required inverse branch are valid. If u is given in terms of x, rearrange or differentiate that relationship as needed.

    x=g(u).x=g(u).

    dx=g′(u) du.dx=g^{\prime}(u)\,du.

    1 · DifferentialDifferentiate the relationship

    For x=g(u)x=g(u), use dx=g′(u) dudx=g^{\prime}(u)\,du. If you instead differentiate u=h(x)u=h(x), include the corresponding factor when rewriting dxdx. Replacing dxdx by dudu alone is generally wrong.

    2 · One variableReplace every occurrence of x

    Rewrite both the integrand and differential, so the transformed integral contains only the new variable and its differential:

    ∫f(g(u))g′(u) du.\int f(g(u))g^{\prime}(u)\,du.

    A mixture of x and u means the change is incomplete.

    3 · Domain and branchJustify square-root signs

    For real v, v2=∣v∣\sqrt{v^2}=|v|. An angle interval may justify replacing ∣cos⁡θ∣|\cos\theta| by cos⁡θ\cos\theta; the identity alone does not. Keep every original denominator restriction.

    A correct-looking new variable is not enough: check the multiplier from the differential separately. The supplied substitution determines the change; this map does not require inventing one.

  2. Return to the original variable and verify

    These steps concern indefinite integrals. After integrating in the new variable, express the answer in x on the original valid interval.

    F(x)+C.F(x)+C.

    F′(x)=f(x).F^{\prime}(x)=f(x).

    1 · IntegrateUse the transformed expression

    Choose a standard form or simplify the new integrand. Carry its coefficient and sign through the integration, and include an arbitrary constant.

    2 · Restore xUse the stated substitution to return

    Replace the new variable in the final indefinite answer. For a trigonometric substitution, use signed identities or a triangle with its stated range; a positive-side triangle alone does not justify negative inputs.

    3 · CheckDifferentiate on the original domain

    The derivative must equal the original integrand, including its coefficient and sign. Retain the interval restrictions: a substitution does not remove a pole. Antiderivatives can differ by a constant on a connected interval, and constants can be chosen independently on disconnected intervals.

    If the derivative does not match, revisit the differential, algebra and integration. For definite integrals, the bounds must also be handled consistently with the chosen variable.

Try a worked example

Read a case, make a prediction, then reveal the reasoning. The examples keep their stated values.

Example 16(a): change both factors

For the existing part (a), evaluate ∫x(2−3x)11 dx\int x(2-3x)^{11}\,dx using u=2−3xu=2-3x. Before integrating, predict the sign of dx/dudx/du and rewrite the remaining x factor.

Show reasoning

  1. 1. Change every factor

    The supplied linear substitution gives a negative differential factor. The separate x factor also needs rewriting.

    x=2−u3,dx=−13 du,∫x(2−3x)11 dx=−19∫(2u11−u12) du.\begin{aligned}x&=\frac{2-u}{3},\\dx&=-\frac13\,du,\\\int x(2-3x)^{11}\,dx&=-\frac19\int(2u^{11}-u^{12})\,du.\end{aligned}

  2. 2. Integrate and return to x

    Keep the coefficient through both power integrals, then replace u by the original expression. The indefinite answer is F(x)+CF(x)+C, using F below. This polynomial integrand is defined for every real x.

    F(x)=−(2−3x)1254+(2−3x)13117.\begin{aligned}F(x)&=-\frac{(2-3x)^{12}}{54}\\&\quad+\frac{(2-3x)^{13}}{117}.\end{aligned}

  3. 3. Check the original integrand

    Both chain-rule factors are needed. Combining the terms recovers the original x multiplier.

    F′(x)=23(2−3x)11−13(2−3x)12=x(2−3x)11.\begin{aligned}F^{\prime}(x)&=\frac23(2-3x)^{11}\\&\quad-\frac13(2-3x)^{12}\\&=x(2-3x)^{11}.\end{aligned}
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Self-Practice 9: the square-root branch

Use the existing substitution x=sin⁡θx=\sin\theta for ∫1−x2 dx\int\sqrt{1-x^2}\,dx. On −1<x<1-1<x<1, explain which angle interval makes 1−sin⁡2θ=cos⁡θ\sqrt{1-\sin^2\theta}=\cos\theta valid. Then compare the source’s cosine-substitution answer.

Show reasoning

  1. 1. Choose a valid branch and change dx

    On the open principal sine interval, cosine is positive. The root and the differential each contribute a cosine factor.

    −π2<θ<π2,dx=cos⁡θ dθ,1−sin⁡2θ=∣cos⁡θ∣=cos⁡θ.\begin{aligned}-\frac\pi2&<\theta<\frac\pi2,\\dx&=\cos\theta\,d\theta,\\\sqrt{1-\sin^2\theta}&=|\cos\theta|=\cos\theta.\end{aligned}

  2. 2. Integrate the transformed square

    Use double angle after the complete variable change. Integrating only cos θ would lose the differential factor.

    ∫1−x2 dx=∫cos⁡2θ dθ=θ2+sin⁡θcos⁡θ2+C.\begin{aligned}\int\sqrt{1-x^2}\,dx&=\int\cos^2\theta\,d\theta\\&=\frac\theta2+\frac{\sin\theta\cos\theta}{2}+C.\end{aligned}

  3. 3. Return and differentiate

    The result applies on the stated open interval. Differentiate the product and inverse sine, then combine terms to recover the original radical.

    Fs(x)=x1−x22+sin⁡−1x2,Fs′(x)=1−x22+1−x221−x2=1−x2.\begin{aligned}F_s(x)&=\frac{x\sqrt{1-x^2}}2\\&\quad+\frac{\sin^{-1}x}{2},\\F_s^{\prime}(x)&=\frac{\sqrt{1-x^2}}2\\&\quad+\frac{1-x^2}{2\sqrt{1-x^2}}\\&=\sqrt{1-x^2}.\end{aligned}

  4. 4. Compare the source alternative

    For the source’s cosine substitution take 0<θ<π0<\theta<\pi, so sine is positive. The principal identity sin⁡−1x+cos⁡−1x=π/2\sin^{-1}x+\cos^{-1}x=\pi/2 explains why the two answers differ by a constant. Their arbitrary constants need not have the same value.

    Fc(x)=x1−x22−cos⁡−1x2,Fc(x)−Fs(x)=−π4.\begin{aligned}F_c(x)&=\frac{x\sqrt{1-x^2}}2\\&\quad-\frac{\cos^{-1}x}{2},\\F_c(x)-F_s(x)&=-\frac\pi4.\end{aligned}
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