Integration by Substitution
The Substitution Rule
If $u=g(x)$ is a differentiable function whose range is an interval $I$ and $f$ is continuous on $I$, then
$
\int f(g(x)) g^{\prime}(x) d x=\int f(u) d u
$
Evaluating the integral $\int f(g(x)) g^{\prime}(x) d x$, when $f$ and $g^{\prime}$ are continuous functions
1. Substitute $u=g(x)$ and $d u=g^{\prime}(x) d x$ to obtain the integral $\int f(u) d u$.
2. Integrate with respect to $u$.
3. Replace $u$ by $g(x)$ in the result.
The Substitution Rule for Definite Integrals
If $g^{\prime}$ is continuous on $[a, b]$ and $f$ is continuous on the range of $u=g(x)$, then
$
\int_a^b f(g(x)) \cdot g^{\prime}(x) d x=\int_{g(a)}^{g(b)} f(u) d u
$
Example1
- Find $\int \cos (5 \theta-3) d \theta$.
▶️Answer/Explanation
Solution
$
\begin{array}{ll}
\int \cos (5 \theta-3) d \theta=\int \cos u \cdot \frac{1}{5} d u & \text { Let } u=5 \theta-3, d u=5 d \theta, \frac{1}{5} d u=d \theta . \\
=\frac{1}{5} \int \cos u d u & \text { The integral is now in standard form. } \\
=\frac{1}{5} \sin u+C & \text { Integrate with respect to } u . \\
=\frac{1}{5} \sin (5 \theta-3)+C & \text { Replace } u \text { by } 5 \theta-3 .
\end{array}
$
Example 2
- Evaluate $\int \frac{x}{\sqrt{1-x^2}} d x$
▶️Answer/Explanation
Solution
$\int \frac{x}{\sqrt{1-x^2}} d x=\int \frac{-1 / 2 d u}{\sqrt{u}}$ Let $u=1-x^2, d u=-2 x d x,-\frac{1}{2} d u=x d x$.
$=-\frac{1}{2} \int u^{-1 / 2} d u$ In the form $\int u^n d u$
$=-\frac{1}{2}(2 \sqrt{u})+C$ Integrate with respect to $u$.
$=-\sqrt{u}+C$
$=-\sqrt{1-x^2}+C$ Replace $u$ with $1-x^2$.
Example 3
- Evaluate $\int_0^{\pi / 2} \frac{3 \cos x}{\sqrt{1+3 \sin x}} d x$
▶️Answer/Explanation
Solution
We have two choices.
Method 1: Transform the integral and evaluate the integral with the transformed limits.
$\int_0^{\pi / 2} \frac{3 \cos x}{\sqrt{1+3 \sin x}} d x$ Let $u=1+3 \sin x, d u=3 \cos x d x$.
When $x=0, u=1+3 \sin 0=1$.
When $x=\pi / 2$,
$\begin{aligned} & =\int_1^4 \frac{d u}{\sqrt{u}} \\ & =\left[2 u^{1 / 2}\right]_1^4 \\ & =2\left[4^{1 / 2}-1\right] \\ & =2[2-1]=2\end{aligned}$ Evaluate the new integral.
Method 2: Transform the integral to an indefinite integral, integrate, change back to $x$, and use the original $x$-limits.
$\int \frac{3 \cos x}{\sqrt{1+3 \sin x}} d x=\int \frac{d u}{\sqrt{u}}$ Let $u=1+3 \sin x, d u=3 \cos x d x$.
$=2 u^{\frac{1}{2}}+C$ Integrate with respect to $u$.
$=2(1+3 \sin x)^{\frac{1}{2}}+C$ Replace $u$ with $1+3 \sin x$.
$\int_0^{\pi / 2} \frac{3 \cos x}{\sqrt{1+3 \sin x}} d x$
$=2\left[(1+3 \sin x)^{\frac{1}{2}}\right]_0^{\pi / 2}$ Use the integral found, with limits of integration for $x$.
$\begin{aligned} & =2\left[\left(1+3 \sin \frac{\pi}{2}\right)^{\frac{1}{2}}-(1+3 \sin 0)^{\frac{1}{2}}\right] \\ & =2\left[4^{\frac{1}{2}}-1^{\frac{1}{2}}\right]=2\end{aligned}$
Example 4
- $
\int \sqrt{x} \sin \left(x^{3 / 2}\right) d x=
$
(A) $\frac{2}{3} \cos \left(x^{3 / 2}\right)+C$
(B) $\frac{2}{3} \sqrt{x} \cos \left(x^{3 / 2}\right)+C$
(C) $-\frac{2}{3} x^{3 / 2} \cos \left(x^{3 / 2}\right)+C$
(D) $-\frac{2}{3} \cos \left(x^{3 / 2}\right)+C$
▶️Answer/Explanation
Ans:D
Example 5
- If the substitution $u=2-x$ is made, $\int_1^3 x \sqrt{2-x} d x=$
(A) $\int_{-1}^1 u \sqrt{u} d u$
(B) $-\int_1^3 u \sqrt{u} d u$
(C) $\int_1^3(2-u) \sqrt{u} d u$
(D) $\int_{-1}^1(u-2) \sqrt{u} d u$
▶️Answer/Explanation
Ans:D
Example 6
- If $\int_{-1}^3 f(x+k) d x=8$, where $k$ is a constant, then $\int_{k-1}^{k+3} f(x) d x=$
(A) $8-k$
(B) $8+k$
(C) 8
(D) $k-8$
▶️Answer/Explanation
Ans:C