Substitution Method

Theorem (Substitution Rule).
Let \(g\) be a differentiable function on an interval \(I\), and let \(f\) be continuous on the range of \(g\). Suppose \(F\) is an antiderivative of \(f\), i.e. \(F'(u) = f(u)\). Then:
\[
\int f\bigl(g(x)\bigr)\, g'(x)\,dx
= F\bigl(g(x)\bigr) + C.
\]

Proof.

1. Chain Rule:
   By the Chain Rule, if \(F'(u) = f(u)\), then
   \[
   \frac{d}{dx}\Bigl(F\bigl(g(x)\bigr)\Bigr)
   = F’\bigl(g(x)\bigr)\, g'(x)
   = f\bigl(g(x)\bigr)\, g'(x).
   \]
2. Match the Integrand:
   We see that the integrand \(f(g(x))\,g'(x)\) is exactly the derivative of \(F(g(x))\). Thus,
   \[
   \int f\bigl(g(x)\bigr)\, g'(x)\,dx
   = \int \frac{d}{dx}\Bigl(F\bigl(g(x)\bigr)\Bigr)\, dx.
   \]
3. Fundamental Theorem of Calculus:
   Integrating a derivative recovers the original function (up to a constant):
   \[
   \int \frac{d}{dx}\Bigl(F\bigl(g(x)\bigr)\Bigr)\, dx
   = F\bigl(g(x)\bigr) + C.
   \]
4. Conclusion:
   Hence,
   \[
   \int f\bigl(g(x)\bigr)\, g'(x)\,dx
   = F\bigl(g(x)\bigr) + C,
   \]
   proving the Substitution Rule.

Alternative \(u\)-Substitution Perspective: Let \(u = g(x)\). Then \(du = g'(x)\,dx\), and the integral becomes
\[
\int f\bigl(g(x)\bigr)\, g'(x)\,dx
= \int f(u)\, du
= F(u) + C
= F\bigl(g(x)\bigr) + C,
\]
which is the same result as above.

Problem: Evaluate
\(\displaystyle \int 2x\,\bigl(x^2+1\bigr)^{99}\,dx.\)

1. Set up the substitution:
   Let \(u = x^2 + 1.\) Then:
   \[
   \frac{du}{dx} = 2x
   \quad\Longrightarrow\quad
   du = 2x\,dx.
   \]
2. Rewrite the integral:
   Notice that \(2x\,dx\) becomes \(du\), and \((x^2+1)^{99}\) becomes \(u^{99}\). Thus:
   \[
   \int 2x\,(x^2+1)^{99}\,dx
   = \int u^{99}\,du.
   \]
3. Integrate with respect to \(u\):
   \[
   \int u^{99}\,du
   = \frac{u^{100}}{100} + C.
   \]
4. Back-substitute:
   Replace \(u\) with \(x^2+1\):
   \[
   \frac{u^{100}}{100} + C
   = \frac{(x^2+1)^{100}}{100} + C.
   \]

Final Answer:
\[
\int 2x\,(x^2+1)^{99}\,dx
= \frac{(x^2+1)^{100}}{100} + C.
\]

Problem: Evaluate
\(\displaystyle \int x^2\,\sqrt{x^3 + 1}\,dx.\)

1. Set up the substitution:
   Let \(u = x^3 + 1\). Then
   \[
   \frac{du}{dx} = 3x^2
   \quad\Longrightarrow\quad
   du = 3x^2\,dx
   \quad\Longrightarrow\quad
   x^2\,dx = \frac{1}{3}\,du.
   \]
2. Rewrite the integral:
   Notice that \(\sqrt{x^3 + 1} = \sqrt{u} = u^{1/2}\). Hence:
   \[
   \int x^2\,\sqrt{x^3 + 1}\,dx
   = \int \sqrt{u}\,\bigl(x^2\,dx\bigr)
   = \int u^{1/2} \left(\frac{1}{3}\,du\right)
   = \frac{1}{3}\int u^{1/2}\,du.
   \]
3. Integrate with respect to \(u\):
   \[
   \frac{1}{3}\int u^{1/2}\,du
   = \frac{1}{3} \cdot \frac{u^{3/2}}{\frac{3}{2}}
   = \frac{1}{3} \cdot \frac{2}{3}u^{3/2}
   = \frac{2}{9} \,u^{3/2} + C.
   \]
4. Back-substitute:
   Replace \(u\) with \(x^3+1\):
   \[
   \frac{2}{9}\,(x^3+1)^{3/2} + C.
   \]

Final Answer:
\[
\int x^2\,\sqrt{x^3 + 1}\,dx
= \frac{2}{9}\,(x^3+1)^{3/2} + C.
\]

Problem: Evaluate
\(\displaystyle \int e^{\sin(\theta)} \cos(\theta)\,d\theta.\)

1. Set up the substitution:
   Let \(u = \sin(\theta)\). Then:
   \[
   \frac{du}{d\theta} = \cos(\theta)
   \quad\Longrightarrow\quad
   du = \cos(\theta)\,d\theta.
   \]
2. Rewrite the integral:
   Substituting \(u = \sin(\theta)\) and \(\cos(\theta)\,d\theta = du\), we get:
   \[
   \int e^{\sin(\theta)} \cos(\theta)\,d\theta
   = \int e^{u}\,du.
   \]
3. Integrate with respect to \(u\):
   \[
   \int e^{u}\,du = e^{u} + C.
   \]
4. Back-substitute:
   Replace \(u\) with \(\sin(\theta)\):
   \[
   e^{\sin(\theta)} + C.
   \]

Final Answer:
\[
\int e^{\sin(\theta)} \cos(\theta)\,d\theta
= e^{\sin(\theta)} + C.
\]

Problem: Evaluate
\(\displaystyle \int \frac{e^x}{e^x + 1}\,dx.\)

1. Set up the substitution:
   Let \(u = e^x + 1\). Then:
   \[
   \frac{du}{dx} = e^x
   \quad\Longrightarrow\quad
   du = e^x\,dx.
   \]
2. Rewrite the integral:
   Substituting \(u = e^x + 1\) and \(e^x\,dx = du\), the integral becomes:
   \[
   \int \frac{e^x}{e^x + 1}\,dx
   = \int \frac{du}{u}.
   \]
3. Integrate with respect to \(u\):
   \[
   \int \frac{du}{u} = \ln|u| + C.
   \]
4. Back-substitute:
   Replace \(u\) with \(e^x + 1\):
   \[
   \ln|u| + C = \ln|e^x + 1| + C.
   \]

Final Answer:
\[
\int \frac{e^x}{e^x + 1}\,dx
= \ln|e^x + 1| + C.
\]

Solution:
Set \(u = 8x^2\). Then:
\[
\frac{du}{dx} = 16x \quad \Longrightarrow \quad du = 16x\,dx \quad \Longrightarrow \quad \frac{1}{16} du = x\,dx.
\]

Substituting into the integral:
\[
\int x \cos(8x^2)\,dx = \int \cos(u) \frac{1}{16} du = \frac{1}{16} \int \cos(u)\,du.
\]

The integral of \(\cos(u)\) is \(\sin(u)\), so:
\[
\frac{1}{16} \int \cos(u)\,du = \frac{1}{16} \sin(u) + C.
\]

Back-substitute \(u = 8x^2\):
\[
\int x \cos(8x^2)\,dx = \frac{1}{16} \sin(8x^2) + C.
\]

Final Answer:
\[
\frac{1}{16} \sin(8x^2) + C.
\]

Solution:
Set \(u = 3x – 7\). Then:
\[
\frac{du}{dx} = 3 \quad \Longrightarrow \quad du = 3\,dx \quad \Longrightarrow \quad \frac{1}{3} du = dx.
\]

Substituting into the integral:
\[
\int 9 (3x – 7)^{-5}\,dx = 9 \int u^{-5} \frac{1}{3}\,du = 3 \int u^{-5}\,du.
\]

The integral of \(u^{-5}\) is \(\frac{u^{-4}}{-4}\), so:
\[
3 \int u^{-5}\,du = 3 \cdot \frac{u^{-4}}{-4} = -\frac{3}{4} u^{-4} + C.
\]

Back-substitute \(u = 3x – 7\):
\[
\int 9 (3x – 7)^{-5}\,dx = -\frac{3}{4} (3x – 7)^{-4} + C.
\]

Final Answer:
\[
-\frac{3}{4} (3x – 7)^{-4} + C.
\]

Solution:
Set \(u = 7x^2 + 3\). Then:
\[
\frac{du}{dx} = 14x \quad \Longrightarrow \quad du = 14x\,dx \quad \Longrightarrow \quad x\,dx = \frac{1}{14} du.
\]

Substituting into the integral:
\[
\int \frac{x}{(7x^2 + 3)^5} \, dx = \int \frac{1}{u^5} \cdot \frac{1}{14}\,du = \frac{1}{14} \int u^{-5}\,du.
\]

The integral of \(u^{-5}\) is:
\[
\int u^{-5}\,du = \frac{u^{-4}}{-4} = -\frac{1}{4}u^{-4}.
\]

Substituting back:
\[
\frac{1}{14} \int u^{-5}\,du = \frac{1}{14} \cdot -\frac{1}{4} u^{-4} = -\frac{1}{56} u^{-4} + C.
\]

Back-substitute \(u = 7x^2 + 3\):
\[
-\frac{1}{56} (7x^2 + 3)^{-4} + C.
\]

Final Answer:
\[
\int \frac{x}{(7x^2 + 3)^5} \, dx = -\frac{1}{56} (7x^2 + 3)^{-4} + C.
\]

Solution:
Set \(u = 8x – 2\). Then:
\[
\frac{du}{dx} = 8 \quad \Longrightarrow \quad du = 8\,dx \quad \Longrightarrow \quad dx = \frac{1}{8} du.
\]

Substituting into the integral:
\[
\int \sin(8x – 2) \, dx = \int \sin(u) \cdot \frac{1}{8}\,du = \frac{1}{8} \int \sin(u)\,du.
\]

The integral of \(\sin(u)\) is:
\[
\int \sin(u)\,du = -\cos(u).
\]

Substituting back:
\[
\frac{1}{8} \int \sin(u)\,du = \frac{1}{8} \cdot -\cos(u) = -\frac{1}{8} \cos(u) + C.
\]

Back-substitute \(u = 8x – 2\):
\[
-\frac{1}{8} \cos(8x – 2) + C.
\]

Final Answer:
\[
\int \sin(8x – 2) \, dx = -\frac{1}{8} \cos(8x – 2) + C.
\]

Solution:
Simplify the integrand:
\[
\frac{\ln(x^4)}{x} = \frac{4 \ln(x)}{x}.
\]
Thus:
\[
\int \frac{\ln(x^4)}{x} \, dx = 4 \int \frac{\ln(x)}{x} \, dx.
\]

Set \(u = \ln(x)\). Then:
\[
\frac{du}{dx} = \frac{1}{x} \quad \Longrightarrow \quad du = \frac{1}{x}\,dx.
\]

Substituting into the integral:
\[
4 \int \frac{\ln(x)}{x} \, dx = 4 \int u\,du.
\]

The integral of \(u\) is:
\[
\int u\,du = \frac{u^2}{2}.
\]

Substituting back:
\[
4 \int u\,du = 4 \cdot \frac{u^2}{2} = 2u^2 + C.
\]

Back-substitute \(u = \ln(x)\):
\[
2 (\ln(x))^2 + C.
\]

To simplify further:
\[
2 (\ln(x))^2 = \frac{1}{8} \big[4 \ln(x)\big]^2.
\]
Hence:
\[
\int \frac{\ln(x^4)}{x} \, dx = \frac{1}{8} \big[\ln(x^4)\big]^2 + C.
\]

Final Answer:
\[
\int \frac{\ln(x^4)}{x} \, dx = \frac{1}{8} \big[\ln(x^4)\big]^2 + C.
\]

Solution:
Set \(u = 1 + \cos(at)\). Then:
\[
\frac{du}{dt} = -\sin(at) \cdot a \quad \Longrightarrow \quad du = -a \sin(at) \, dt \quad \Longrightarrow \quad \sin(at) \, dt = -\frac{1}{a} \, du.
\]

Substituting into the integral:
\[
\int (1 + \cos(at))^2 \sin(at) \, dt = \int u^2 \left(-\frac{1}{a}\right) \, du = -\frac{1}{a} \int u^2 \, du.
\]

The integral of \(u^2\) is:
\[
\int u^2 \, du = \frac{u^3}{3}.
\]
So:
\[
-\frac{1}{a} \int u^2 \, du = -\frac{1}{a} \cdot \frac{u^3}{3} = -\frac{1}{3a} u^3 + C.
\]

Back-substitute \(u = 1 + \cos(at)\):
\[
-\frac{1}{3a} u^3 + C = -\frac{1}{3a} (1 + \cos(at))^3 + C.
\]

Final Answer:
\[
\int (1 + \cos(at))^2 \sin(at) \, dt = -\frac{1}{3a} (1 + \cos(at))^3 + C.
\]

Definite Integral: Evaluate \(\int_0^\pi (1 + \cos(at))^2 \sin(at) \, dt\).

Using the result of the indefinite integral:
\[
\int_0^\pi (1 + \cos(at))^2 \sin(at) \, dt = -\frac{1}{3a} \left[(1 + \cos(a\pi))^3 – (1 + \cos(0))^3\right].
\]

Simplify the values of \(\cos(a\pi)\) and \(\cos(0)\):
\[
\cos(a\pi) = -1, \quad \cos(0) = 1.
\]
Substituting:
\[
= -\frac{1}{3a} \left[(1 + (-1))^3 – (1 + 1)^3\right].
\]

Simplify further:
\[
= -\frac{1}{3a} \left[0^3 – 2^3\right] = -\frac{1}{3a} \left[0 – 8\right] = -\frac{1}{3a} \cdot (-8) = \frac{8}{3a}.
\]

Final Answer for the Definite Integral:
\[
\int_0^\pi (1 + \cos(at))^2 \sin(at) \, dt = \frac{8}{3a}.
\]