Change the variable

Why substitution works

Use one variable throughout the transformed integral:

1. Differentiate the stated substitution to obtain the new differential.
2. Replace the integrand and differential completely, keeping any sign or interval restriction.
3. Integrate in the new variable.
4. Return to the original variable and differentiate to check.

A square root is nonnegative. A trigonometric substitution must use an angle interval that justifies the sign when simplifying it.

The source contrasts $\int1/\sqrt{1-4x^2}\,dx$, which matches the composite arcsine formula after supplying the inner derivative, with the motivating question $\int\sqrt{1-x^2}\,dx$, which does not match that reciprocal-radical form directly. A suitable change of variable can turn it into standard integrals.

Let $x=g(u)$, $dx/du=g^{\prime}(u)$ and $y=f(x)$. If $\int y\,dx=\int f(x)\,dx=F(x)+C$, then $F^{\prime}(x)=f(x)$. The chain rule gives $\frac{d}{du}F(g(u))=F^{\prime}(g(u))g^{\prime}(u)=f(g(u))g^{\prime}(u)$. Integrating with respect to $u$ therefore gives

$$\int f(x)\,dx=\int f(g(u))g^{\prime}(u)\,du.$$

An additive constant is absorbed in the resulting indefinite integral. Every occurrence of $x$ in the transformed integrand must be rewritten in terms of $u$, including the differential. The Teacher highlights this change of variable and corrects an incomplete handwritten $\int f(g(\theta))\,d\theta$ by inserting $dx/d\theta$; using $x=g(\theta)$ requires $dx=(dx/d\theta)\,d\theta$. The margin chain-rule reminder is $d(y^2)/dx=2y\,dy/dx$. Apply this derivation in Example 16 and Self-Practice 9 below.

Explain every factor in a given substitution

Use the given substitution $u=\sqrt{x+1}$ to evaluate $\int x/\sqrt{x+1}\,dx$. State the original real domain and explain why simply replacing the denominator by $u$ is not enough. Differentiate your final expression to check it.

After your attempt: compare the reasoning

The integrand requires $x>-1$, hence $u>0$. Squaring gives $x=u^2-1$ and $dx=2u\,du$. Therefore

$\int\frac{x}{\sqrt{x+1}}\,dx=\int2(u^2-1)\,du=\frac23u^3-2u+C.$

Returning to x gives $\frac23(x+1)^{3/2}-2\sqrt{x+1}+C$. Its derivative is $\sqrt{x+1}-1/\sqrt{x+1}=x/\sqrt{x+1}$. Omitting the differential factor $2u$ changes the integral; leaving x in the transformed expression mixes variables.