If given integration takes the form $\int f(g(x))g’(x)dx$ then it can be converted to a simpler integration that we may be able to evaluate by the substitution $u=g(x)$. In fact, the integration is given in terms of the new variable $u$ as

$$\int f(g(x))g’(x)dx=\int f(u)du.$$

*Example*. Evaluate $\int x\sqrt{1+x^2}dx$.

*Solution*. Let $u=1+x^2$. Then $du=2xdx$. So,

\begin{align*}

\int x\sqrt{1+x^2}dx&=\frac{1}{2}\int\sqrt{u}du\\

&=\frac{1}{3}u^{\frac{3}{2}}+C\\

&=\frac{1}{3}(1+x^2)^{\frac{3}{2}}+C,

\end{align*}

where $C$ is an arbitrary constant.

*Example*. Evaluate $\int\cos (7\theta+5)d\theta$.

*Solution*. Let $u=7\theta+5$. Then $du=7d\theta$. So,

\begin{align*}

\int\cos (7\theta+5)d\theta&=\frac{1}{7}\int\cos udu\\

&=\frac{1}{7}\sin u+C\\

&=\frac{1}{7}\sin(7\theta+5)+C,

\end{align*}

where $C$ is an arbitrary constant.

*Example*. Evaluate $\int x^2\sin(x^3)dx$.

*Solution*. Let $u=x^3$. Then $du=3x^2dx$. So,

\begin{align*}

\int x^2\sin(x^3)dx&=\frac{1}{3}\int \sin udu\\

&=-\frac{1}{3}\cos u+C\\

&=-\frac{1}{3}\cos(x^3)+C,

\end{align*}

where $C$ is an arbitrary constant.

How do we evaluate a definite integral of the form $\int_a^b f(g(x))g’(x)dx$? The following example shows you how.

*Example*. Evaluate $\int_{-1}^1 3x^2\sqrt{x^3+1}dx$.

*Solution*. First let us calculate the indefinite integral $\int 3x^2\sqrt{x^3+1}dx$. Let $u=x^3+1$. Then $du=3x^2dx$. So,

\begin{align*}

\int 3x^2\sqrt{x^3+1}dx&=\int\sqrt{u}du\\

&=\frac{2}{3}u^{\frac{3}{2}}+C\\

&=2(x^3+1)^{\frac{3}{2}}+C,

\end{align*}

where $C$ is an arbitrary constant. Now by Fundamental Theorem of Calculus,

\begin{align*}

\int_{-1}^1 3x^2\sqrt{x^3+1}dx&=\frac{2}{3}[(x^3+1)^{\frac{3}{2}}]_{-1}^1\\

&=\frac{4\sqrt{2}}{3}.

\end{align*}

But there is a better way to do this as shown in the following theorem. Its proof is straightforward.

*Theorem*. If $u’$ is continuous on $[a,b]$ and $f$ is continuous on the range of $u$, then

$$\int_a^bf(u(x))u’(x)dx=\int_{u(a)}^{u(b)}f(u)du.$$

*Example*. Let us replay the previous example using this theorem. Again let $u=x^3+1$. Then $du=3x^2dx$ and $u(-1)=0$, $u(1)=2$. Now, by the above theorem,

\begin{align*}

\int_{-1}^1 3x^2\sqrt{x^3+1}dx&=\int_0^2\sqrt{u}du\\

&=\frac{2}{3}[u^{\frac{3}{2}}]_0^2\\

&=\frac{4\sqrt{2}}{3}.

\end{align*}

I believe you will find this more simple than previous method.

I will finish this lecture with the following nice properties.

*Theorem*. Let $f$ be a continuous function on $[-a,a]$.

(a) If $f$ is an even function, then $\int_{-a}^a f(x)dx=2\int_0^a f(x)dx$.

(b) If $f$ is an odd function, then $\int_{-a}^a f(x)dx=0$.

This can be easily understood from pictures using the symmetries of even and odd functions. But the theorem can be proved using substitution. I will leave it to you.