Step-by-Step Guidance for Calculus Integration Techniques

Q1. Evaluate the indefinite integral: $\int x \arctan(x) \, dx$

Background

Topic: Integration by Parts

This question tests your ability to use integration by parts, a technique for integrating products of functions.

Key Terms and Formulas

• Integration by Parts formula: $\int u \, dv = u v - \int v \, du$

• Choose $u$ and $dv$ from the integrand. Typically, $u$ is a function that becomes simpler when differentiated.

Step-by-Step Guidance

1. Identify $u$ and $dv$ in the integral. For $\int x \arctan(x) \, dx$, let $u = \arctan(x)$ and $dv = x \, dx$.

2. Compute $du$ and $v$: $du = \frac{1}{1 + x^2} \, dx$, and $v = \frac{x^2}{2}$ (by integrating $x$).

3. Apply the integration by parts formula: $\int x \arctan(x) \, dx = u v - \int v \, du$.

4. Set up the remaining integral: $\int \frac{x^2}{2} \cdot \frac{1}{1 + x^2} \, dx$.

Try solving on your own before revealing the answer!

Final Answer: $\frac{(x^2 + 1)}{2} \arctan(x) - \frac{x}{2} + C$

After applying integration by parts and simplifying, you arrive at this result. The arbitrary constant $C$ is included for indefinite integrals.

Q2. Evaluate the indefinite integral: $\int x^2 \sin(2x) \, dx$

Background

Topic: Integration by Parts (Repeated)

This question tests your ability to use integration by parts, possibly more than once, to integrate a product of a polynomial and a trigonometric function.

Key Terms and Formulas

• Integration by Parts: $\int u \, dv = u v - \int v \, du$

• For repeated integration by parts, choose $u$ as the polynomial ($x^2$) and $dv$ as the trigonometric part ($\sin(2x) \, dx$).

Step-by-Step Guidance

1. Let $u = x^2$, $dv = \sin(2x) \, dx$.

2. Compute $du = 2x \, dx$, and $v = -\frac{1}{2} \cos(2x)$ (integrate $\sin(2x)$).

3. Apply integration by parts: $\int x^2 \sin(2x) \, dx = u v - \int v \, du$.

4. Set up the new integral: $\int -\frac{1}{2} \cos(2x) \cdot 2x \, dx$ and simplify.

5. Notice that you will need to use integration by parts again for the new integral.

Try solving on your own before revealing the answer!

Final Answer: $-\frac{x^2}{2} \cos(2x) + \frac{x}{2} \sin(2x) + \frac{1}{4} \cos(2x) + C$

Integration by parts is applied twice, and the result is simplified. Don't forget the constant $C$.

Q3. Evaluate the indefinite integral: $\int \sin^3(x) \cos^4(x) \, dx$

Background

Topic: Trigonometric Integrals

This question tests your ability to integrate powers of sine and cosine, often using reduction formulas or substitution.

Key Terms and Formulas

• Reduction formulas for powers of sine and cosine.

• Substitution: $u = \sin(x)$ or $u = \cos(x)$, depending on which has an odd power.

• Use identities: $\sin^2(x) = 1 - \cos^2(x)$, $\cos^2(x) = 1 - \sin^2(x)$.

Step-by-Step Guidance

1. Since $\sin^3(x)$ has an odd power, save one $\sin(x)$ and rewrite the rest in terms of $\cos(x)$.

2. Express $\sin^3(x) = (1 - \cos^2(x)) \sin(x)$, so the integral becomes $\int (1 - \cos^2(x)) \sin(x) \cos^4(x) \, dx$.

3. Expand and split the integral: $\int \sin(x) \cos^4(x) \, dx - \int \sin(x) \cos^6(x) \, dx$.

4. Use substitution $u = \cos(x)$, $du = -\sin(x) \, dx$ for each term.

Try solving on your own before revealing the answer!

Final Answer: $\frac{1}{7} \cos^7(x) - \frac{1}{5} \cos^5(x) + C$

After substitution and integrating, you get the result in terms of powers of cosine. The constant $C$ is included.

Q4. Evaluate the indefinite integral: $\int \tan^3(x) \sec^3(x) \, dx$

Background

Topic: Trigonometric Integrals

This question tests your ability to integrate products of powers of tangent and secant, often using substitution or reduction formulas.

Key Terms and Formulas

• Reduction formulas for $\tan^n(x) \sec^m(x)$ integrals.

• Recall $\frac{d}{dx} \tan(x) = \sec^2(x)$ and $\frac{d}{dx} \sec(x) = \sec(x) \tan(x)$.

• Substitution: $u = \tan(x)$ or $u = \sec(x)$, depending on the powers.

Step-by-Step Guidance

1. Rewrite $\tan^3(x) \sec^3(x)$ as $\tan^2(x) \tan(x) \sec^3(x)$.

2. Recall $\tan^2(x) = \sec^2(x) - 1$ and substitute.

3. Expand: $\int (\sec^2(x) - 1) \tan(x) \sec^3(x) \, dx$.

4. Split into two integrals and consider substitution $u = \sec(x)$ for each.

Try solving on your own before revealing the answer!

Final Answer: $\frac{1}{5} \sec^5(x) - \frac{1}{3} \sec^3(x) + C$

After expanding and integrating, you get the result in terms of powers of secant. The constant $C$ is included.

Q5. Evaluate the indefinite integral: $\int \frac{x^3}{\sqrt{x^2 + 9}} \, dx$

Background

Topic: Trigonometric Substitution

This question tests your ability to use trigonometric substitution to integrate expressions involving square roots of quadratic terms.

Key Terms and Formulas

• Trigonometric substitution: For $\sqrt{x^2 + a^2}$, use $x = a \tan(\theta)$.

• Recall $\tan^2(\theta) + 1 = \sec^2(\theta)$.

• Substitute and rewrite the integral in terms of $\theta$.

Step-by-Step Guidance

1. Let $x = 3 \tan(\theta)$, so $dx = 3 \sec^2(\theta) \, d\theta$.

2. Rewrite $\sqrt{x^2 + 9}$ as $\sqrt{9 \tan^2(\theta) + 9} = 3 \sec(\theta)$.

3. Substitute all terms into the integral and simplify.

4. Expand and integrate in terms of $\theta$, then convert back to $x$ using the original substitution.

Try solving on your own before revealing the answer!

Final Answer: $\frac{1}{3}(x^2 - 18) \sqrt{x^2 + 9} + C$

Alternatively, $\frac{1}{3}(x^2 + 9)^{3/2} - 9 \sqrt{x^2 + 9} + C$. The constant $C$ is included.

Q6. Evaluate the indefinite integral: $\int \frac{1}{(4 - x^2)^{3/2}} \, dx$

Background

Topic: Trigonometric Substitution

This question tests your ability to use trigonometric substitution for integrals involving expressions like $a^2 - x^2$.

Key Terms and Formulas

• For $\sqrt{a^2 - x^2}$, use $x = a \sin(\theta)$.

• Recall $\sin^2(\theta) + \cos^2(\theta) = 1$.

• Rewrite the integral in terms of $\theta$ and simplify.

Step-by-Step Guidance

1. Let $x = 2 \sin(\theta)$, so $dx = 2 \cos(\theta) \, d\theta$.

2. Rewrite $4 - x^2$ as $4 - 4 \sin^2(\theta) = 4 \cos^2(\theta)$.

3. Substitute into the integral and simplify the denominator.

4. Integrate in terms of $\theta$, then convert back to $x$ using the original substitution.

Try solving on your own before revealing the answer!

Final Answer: $\frac{x}{4 \sqrt{4 - x^2}} + C$

After substitution and simplification, you arrive at this result. The constant $C$ is included.

Q7. Evaluate the indefinite integral: $\int \frac{2x^2 + 5x - 1}{(x - 1)(x^2 + 1)} \, dx$

Background

Topic: Partial Fraction Decomposition

This question tests your ability to decompose a rational function into partial fractions and integrate each term.

Key Terms and Formulas

• Partial fraction decomposition: Express $\frac{2x^2 + 5x - 1}{(x - 1)(x^2 + 1)}$ as $\frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}$.

• Integrate each term separately.

Step-by-Step Guidance

1. Set up the partial fraction decomposition: $\frac{2x^2 + 5x - 1}{(x - 1)(x^2 + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + 1}$.

2. Multiply both sides by $(x - 1)(x^2 + 1)$ and solve for $A$, $B$, and $C$ by equating coefficients.

3. Once you have $A$, $B$, and $C$, rewrite the integral as a sum of simpler integrals.

4. Integrate each term: $\int \frac{A}{x - 1} \, dx$, $\int \frac{Bx}{x^2 + 1} \, dx$, $\int \frac{C}{x^2 + 1} \, dx$.

Try solving on your own before revealing the answer!

Final Answer: $3 \ln|x - 1| - \frac{1}{2} \ln(x^2 + 1) + 4 \arctan(x) + C$

After decomposing and integrating, you get the result. The constant $C$ is included.

Q8. Evaluate the indefinite integral: $\int \frac{x^2 + 1}{x^3 - x^2} \, dx$

Background

Topic: Partial Fraction Decomposition

This question tests your ability to decompose a rational function and integrate each term.

Key Terms and Formulas

• Factor the denominator: $x^3 - x^2 = x^2(x - 1)$.

• Set up partial fractions: $\frac{x^2 + 1}{x^3 - x^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 1}$.

• Integrate each term separately.

Step-by-Step Guidance

1. Set up the partial fraction decomposition: $\frac{x^2 + 1}{x^3 - x^2} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x - 1}$.

2. Multiply both sides by $x^3 - x^2$ and solve for $A$, $B$, and $C$.

3. Rewrite the integral as a sum of simpler integrals.

4. Integrate each term: $\int \frac{A}{x} \, dx$, $\int \frac{B}{x^2} \, dx$, $\int \frac{C}{x - 1} \, dx$.

Try solving on your own before revealing the answer!

Final Answer: $\frac{1}{x} + 2 \ln|x - 1| - \ln|x| + C$

After decomposing and integrating, you get the result. The constant $C$ is included.

Q9. Evaluate the improper integral (or state if it diverges): $\int_1^{\infty} \frac{\ln(x)}{x^2} \, dx$

Background

Topic: Improper Integrals

This question tests your ability to evaluate improper integrals, which involve infinite limits or unbounded integrands.

Key Terms and Formulas

• Improper integral: $\int_a^{\infty} f(x) \, dx = \lim_{b \to \infty} \int_a^b f(x) \, dx$

• Integration by parts may be needed for $\int \frac{\ln(x)}{x^2} \, dx$.

Step-by-Step Guidance

1. Rewrite the improper integral as a limit: $\lim_{b \to \infty} \int_1^b \frac{\ln(x)}{x^2} \, dx$.

2. Use integration by parts: Let $u = \ln(x)$, $dv = x^{-2} \, dx$.

3. Compute $du = \frac{1}{x} \, dx$, $v = -x^{-1}$.

4. Apply the integration by parts formula and set up the resulting expression.

5. Evaluate the limit as $b \to \infty$ after plugging in the bounds.

Try solving on your own before revealing the answer!

Final Answer: $1$ (converges)

The improper integral converges to $1$ after evaluating the limit.

Q10. Evaluate the improper integral (or state if it diverges): $\int_0^2 \frac{1}{\sqrt{2 - x}} \, dx$

Background

Topic: Improper Integrals

This question tests your ability to evaluate improper integrals where the integrand becomes unbounded at one endpoint.

Key Terms and Formulas

• Improper integral: $\int_a^b f(x) \, dx$ where $f(x)$ is unbounded at $b$.

• Rewrite as a limit: $\lim_{c \to b^-} \int_a^c f(x) \, dx$.

• Substitution may help: Let $u = 2 - x$.

Step-by-Step Guidance

1. Rewrite the integral as a limit: $\lim_{c \to 2^-} \int_0^c \frac{1}{\sqrt{2 - x}} \, dx$.

2. Use substitution $u = 2 - x$, $du = -dx$.

3. Change the bounds accordingly and rewrite the integral in terms of $u$.

4. Integrate $\int u^{-1/2} \, du$ and set up the expression for the limit.

5. Evaluate the limit as $c \to 2$.

Try solving on your own before revealing the answer!

Final Answer: $2 \sqrt{2}$ (converges)

The improper integral converges to $2 \sqrt{2}$ after evaluating the limit.