MA005: Calculus I

Once an antiderivative in terms of $u$ is found, we have a choice of methods. We can

(a) rewrite our antiderivative in terms of the original variable $x$ , and then evaluate the antiderivative at the integration endpoints and subtract, or

(b) change the integration endpoints to values of $u$, and evaluate the antiderivative in terms of $u$ before subtracting.

If the original integral had endpoints $x=\mathrm{a}$ and $x=\mathrm{b}$, and we make the substitution $u=\mathrm{g}(x)$ and $\mathrm{du}=g^{\prime}(x) \mathrm{d} \mathrm{x}$, then the new integral will have endpoints $u=\mathrm{g}(\mathrm{a})$ and $u=\mathrm{g}(\mathrm{b})$ and

$\int_{x=\mathrm{a}}^{x=\mathrm{b}}$ (original integrand) $dx$ becomes $\int_{u=g(a)}^{u=g(b)}$ (new integrand) $du$.

To evaluate $\int_{0}^{1}(3 x-1)^{4} \mathrm{~d} \mathrm{x}$, we can put $u=3 x-1$. Then $\mathbf{d u}=\mathrm{d}(3 x-1)=\mathbf{3} \mathbf{d} \mathbf{x}$ so the integral becomes $\int u^{4} \frac{\mathbf{1}}{\mathbf{3}} \mathrm{du}=\frac{1}{15} u^{5}+\mathrm{C}$.

(a) Converting our antiderivative back to the variable $x$ and evaluating with the original endpoints:

$\int_{0}^{1}(3 x-1)^{4} \mathrm{dx}=\left.\frac{1}{15}(3 x-1)^{5}\right|_{0} ^{1}=\frac{1}{15}(3 \cdot 1-1)^{5}-\frac{1}{15}(3 \cdot 0-1)^{5}=\frac{1}{15}(2)^{5}-\frac{1}{15}(-1)^{5}=\frac{32}{15}-\frac{-1}{15}=\frac{33}{15}$.

(b) Converting the integration endpoints to values of $u$: when $\mathrm{x}=0$, then $u=3 x-1=3 \cdot 0-1=-1$, and when $\mathrm{x}=1$, then $u=3 x-1=3 \cdot 1-1=2$ so

$\int_{x=0}^{x=1}(3 x-1)^{4} \mathrm{dx}=\int_{u=-1}^{u=2} u^{4} \frac{1}{3} \mathrm{du}=\left.\frac{1}{15} \mathrm{u}^{5}\right|_{-1} ^{2}=\frac{1}{15}(2)^{5}-\frac{1}{15}(-1)^{5}=\frac{32}{15}-\frac{-1}{15}=\frac{33}{15}$.

Both approaches typically involve about the same amount of work and calculation.

Practice 4: If the original integrals in Example 4 had endpoints (a) $\mathrm{x}=0$ to $\mathrm{x}=\pi$, (b) $\mathrm{x}=0$ to $\mathrm{x}=2$, and (c) and $\mathrm{x}=0$ to $x=\ln (3)$, then the new integrals should have what endpoints?

Special Transformations for $\int \sin ^{2}(x) \mathrm{dx} \text { and } \int \cos ^{2}(x) \mathrm{dx}$

The integrals of $\sin ^{2}(x)$ and $\cos ^{2}(x)$ occur relatively often, and we can find their antiderivatives with the help of two trigonometric identities for $\cos (2 x)$:

$\cos (2 x)=1-2 \sin ^{2}(x) \text { and } \cos (2 x)=2 \cos ^{2}(x)-1$

Solving the identies for $\sin ^{2}(x)$ and $\cos ^{2}(x)$, we get $\sin ^{2}(x)=\frac{1-\cos (2 x)}{2}$ and $\cos ^{2}(x)=\frac{1+\cos (2 x)}{2}$

Then $\begin{aligned} \int \sin ^{2}(x) \mathrm{dx} &=\int \frac{1-\cos (2 x)}{2} \mathrm{dx}=\int \frac{1}{2} \mathrm{~d} x-\int \frac{1}{2} \cos (2 x) \mathrm{d} x \\ &=\frac{x}{2}-\frac{1}{2} \frac{\sin (2 x)}{2}+\mathrm{C}=\frac{x}{2}-\frac{\sin (2 x)}{4}+\mathrm{C} \text { or } \frac{1}{2}\{x-\sin (x) \cos (x)\}+\mathrm{C} \end{aligned}$

Similarly, $\begin{aligned} \int \cos ^{2}(x) \mathrm{dx} &=\int \frac{1+\cos (2 x)}{2} \mathrm{dx}=\int \frac{1}{2} \mathrm{~d} x+\int \frac{1}{2} \cos (2 x) \mathrm{d} x \\ &=\frac{x}{2}+\frac{1}{2} \frac{\sin (2 x)}{2}+\mathrm{C}=\frac{x}{2}+\frac{\sin (2 x)}{4}+\mathrm{C} \text { or } \frac{1}{2}\{x+\sin (x) \cos (x)\}+\mathrm{C} \end{aligned}$