Almost every method this course teaches for solving a differential equation ends the same way: an integral you actually have to evaluate. Separating variables, applying an integrating factor, inverting a Laplace transform — each one reduces the differential equation to a plain antiderivative problem, at which point every technique from integral calculus is back on the table.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
Find \displaystyle\int \dfrac{x}{x^2+4}\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\int \tfrac12\cdot\tfrac{1}{u}\,du = \tfrac12\ln|u| = \tfrac12\ln(x^2+4), plus C.
Correct answer
Also accepted:
Why
Let u=x^2+4, so du=2x\,dx.
From Calculus II — Integral Calculus & Infinite Series
Find \displaystyle\int (2x+1)^5\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
\int u^5 \cdot \tfrac12\,du = \tfrac{1}{12}u^6 = \tfrac{(2x+1)^6}{12}, plus C.
Correct answer
Also accepted:
Why
Let u=2x+1, so du=2\,dx.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
Evaluate \displaystyle\int_0^{\pi/2} x\sin x\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Why
[-x\cos x+\sin x]_0^{\pi/2} = (0+1)-(0+0) = 1.
Correct answer
Also accepted:
Why
Let u=x, dv=\sin x\,dx, then evaluate [-x\cos x+\sin x] at the bounds.
From Calculus II — Integral Calculus & Infinite Series
Find \displaystyle\int x\sin(2x)\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
-\tfrac x2\cos2x - \int\left(-\tfrac12\cos2x\right)dx = -\tfrac x2\cos2x+\tfrac14\sin2x, plus C.
Correct answer
Also accepted:
Why
Let u=x, dv=\sin(2x)\,dx, so v=-\tfrac12\cos(2x).
Think of every integration technique you know as a specific wrench in a toolbox. A differential equation doesn't hand you a new toolbox. It hands you a slightly stranger bolt. You still reach for the same wrench you'd have used on an ordinary integral.
Two habits are worth tightening before they resurface under new names. First: an antiderivative is a whole family, F(x)+C, not one function — forgetting the constant is invisible in a pure integration exercise but becomes a genuine wrong answer once an initial condition is applied to pin down C. Second: recognizing which technique a given integral calls for — a substitution hiding in a chain-rule shape, a product crying out for integration by parts — is exactly the same recognition skill "choosing a solution method" for a differential equation will ask of you two chapters from now, aimed at equations instead of integrals.
Same segments, different C
Drag the slider and watch F(x)+C stay tangent to every slope-field segment, no matter which member of the family it is.
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Worked example
A substitution hiding in a chain-rule shape
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1
Evaluate \displaystyle\int x\,e^{x^2}\,dx.
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2
Let u=x^2, so du=2x\,dx and x\,dx = \tfrac12\,du.
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3\int x e^{x^2}\,dx = \tfrac12\int e^u\,du = \tfrac12 e^{x^2} + C.\boxed{\tfrac12 e^{x^2}+C}
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Worked example
A product that needs integration by parts
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1
Evaluate \displaystyle\int x\sin x\,dx.
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2
Let u=x, dv=\sin x\,dx, so du=dx, v=-\cos x.
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3\int x\sin x\,dx = -x\cos x + \int\cos x\,dx = -x\cos x + \sin x + C.\boxed{-x\cos x+\sin x+C}
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Where this goes next
Every solving technique ahead
Separation of variables, the integrating-factor method, and inverting a Laplace transform all bottom out in an antiderivative. A shaky grip on substitution or integration by parts here becomes a wrong differential equation solution later, for a reason that has nothing to do with differential equations at all.
Check your understanding (counts toward your progress)
Evaluate \displaystyle\int \dfrac{x}{x^2+1}\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
With u=x^2+1, du=2x\,dx, so the integral becomes \tfrac12\int\frac{du}{u}=\tfrac12\ln|x^2+1|+C.
Correct answer
Also accepted:
Why
Let u=x^2+1.
Evaluate \displaystyle\int x\cos x\,dx.
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
u=x, dv=\cos x\,dx gives du=dx, v=\sin x: \int x\cos x\,dx = x\sin x - \int\sin x\,dx = x\sin x+\cos x+C.
Correct answer
Also accepted:
Why
Let u=x, dv=\cos x\,dx.
To evaluate \displaystyle\int 2x\cos(x^2)\,dx, which substitution actually simplifies the integral?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Setting u equal to the outer function leaves du with no match for the 2x sitting outside.
This drops the argument of cosine entirely -- the substitution has to absorb the WHOLE inner function.
This leaves an extra factor of 2 unaccounted for, and du=dx doesn't match 2x\,dx.
Why
With u=x^2, du=2x\,dx matches the 2x\,dx sitting outside cosine exactly, leaving \int\cos u\,du.
A student integrates \int x e^x\,dx and gets x e^x - e^x + C. Checking by differentiating this answer, is it correct?
Right on try — credit. Right — the answer had already been shown, so no credit this time. Next try is worth Answer shown after three tries — this one no longer earns credit. You chose to see the answer — this one earns no credit.
Differentiate xe^x-e^x: the product rule on xe^x gives e^x+xe^x, and the derivative of -e^x is -e^x -- they cancel, leaving exactly xe^x.
An indefinite integral always carries an arbitrary constant; its presence is correct, not a flaw.
Differentiating the proposed answer is always a valid, self-contained way to check an antiderivative.
Why
Differentiating xe^x-e^x via the product rule gives (e^x+xe^x)-e^x=xe^x, confirming the antiderivative is correct.
Differentiate xe^x - e^x using the product rule on the first term.