Step-by-Step Calculus Practice Exam 2 Guidance

Q1(a). Evaluate the integral $\int \cos^3(2x) \sin^5(2x)\,dx$

Background

Topic: Integration Techniques (Trigonometric Integrals)

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

Key Terms and Formulas

• Trigonometric identities: $\sin^2 x = 1 - \cos^2 x$, $\cos^2 x = 1 - \sin^2 x$

• Substitution method

• Reduction formulas for powers of sine and cosine

Step-by-Step Guidance

1. Notice that both sine and cosine have odd powers. Factor one sine out to use substitution.

2. Rewrite $\sin^5(2x)$ as $\sin^4(2x)\sin(2x)$, and $\sin^4(2x)$ as $[\sin^2(2x)]^2$.

3. Express $\sin^2(2x)$ in terms of $\cos^2(2x)$ using the identity $\sin^2(2x) = 1 - \cos^2(2x)$.

4. Let $u = \cos(2x)$, then compute $du$ in terms of $dx$.

5. Substitute all terms in terms of $u$ and $du$, and set up the new integral.

Try solving on your own before revealing the answer!

Final Answer:

$\int \cos^3(2x) \sin^5(2x)\,dx = -\frac{1}{16} \cos^4(2x) + \frac{1}{12} \cos^6(2x) - \frac{1}{8} \cos^8(2x) + C$

This result comes from using substitution and expanding the integrand as described above.

Q1(b). Evaluate the integral $\int x^2 e^{-2x}\,dx$

Background

Topic: Integration by Parts

This question tests your ability to use integration by parts, especially when the integrand is a product of a polynomial and an exponential function.

Key Terms and Formulas

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

• Choose $u$ and $dv$ appropriately (LIATE rule can help: Logarithmic, Inverse trig, Algebraic, Trig, Exponential)

Step-by-Step Guidance

1. Let $u = x^2$ and $dv = e^{-2x} dx$.

2. Compute $du$ and $v$ (integrate $dv$).

3. Apply the integration by parts formula: $\int u\,dv = uv - \int v\,du$.

4. Notice that the new integral will again require integration by parts. Repeat the process for the remaining integral.

5. Continue until the polynomial part is reduced to zero.

Try solving on your own before revealing the answer!

Final Answer:

$\int x^2 e^{-2x}\,dx = -\frac{1}{2} x^2 e^{-2x} - \frac{1}{2} \int 2x e^{-2x}\,dx$

After repeating integration by parts, the full answer is $\left(-\frac{1}{2}x^2 - \frac{1}{2}x - \frac{1}{4}\right)e^{-2x} + C$.

Q1(c). Evaluate the integral $\int \frac{x^2}{x^2 + 1}\,dx$

Background

Topic: Rational Functions and Polynomial Division

This question tests your ability to integrate rational functions, possibly using polynomial division and substitution.

Key Terms and Formulas

• Polynomial division

• Substitution for integrals involving $\frac{1}{x^2 + 1}$

• Standard integral: $\int \frac{1}{x^2 + 1} dx = \arctan x + C$

Step-by-Step Guidance

1. Divide $x^2$ by $x^2 + 1$ to rewrite the integrand as $1 - \frac{1}{x^2 + 1}$.

2. Split the integral into two parts: $\int 1\,dx$ and $-\int \frac{1}{x^2 + 1}\,dx$.

3. Integrate each part separately using standard formulas.

Try solving on your own before revealing the answer!

Final Answer:

$\int \frac{x^2}{x^2 + 1}\,dx = x - \arctan x + C$

We used polynomial division and standard integrals to find the result.

Q1(d). Evaluate the integral $\int \sqrt{49 - x^2}\,dx$

Background

Topic: Trigonometric Substitution

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

Key Terms and Formulas

• Trigonometric substitution: For $\sqrt{a^2 - x^2}$, use $x = a \sin \theta$

• Identity: $1 - \sin^2 \theta = \cos^2 \theta$

Step-by-Step Guidance

1. Let $x = 7 \sin \theta$, so $dx = 7 \cos \theta d\theta$.

2. Substitute $x$ and $dx$ into the integral, and simplify $\sqrt{49 - x^2}$ using the identity.

3. Integrate with respect to $\theta$.

4. Convert your answer back to $x$ using inverse trigonometric functions.

Try solving on your own before revealing the answer!

Final Answer:

$\int \sqrt{49 - x^2}\,dx = \frac{1}{2}x\sqrt{49 - x^2} + \frac{49}{2}\arcsin\left(\frac{x}{7}\right) + C$

This result uses trigonometric substitution and back-substitution to $x$.

Q1(e). Evaluate the integral $\int \frac{dx}{\sqrt{4x^2 - 49}}$

Background

Topic: Trigonometric Substitution

This question tests your ability to integrate expressions involving $\sqrt{a^2x^2 - b^2}$, often using trigonometric or hyperbolic substitution.

Key Terms and Formulas

• For $\sqrt{x^2 - a^2}$, use $x = a \sec \theta$

• Standard integral: $\int \frac{dx}{\sqrt{x^2 - a^2}} = \ln|x + \sqrt{x^2 - a^2}| + C$

Step-by-Step Guidance

1. Rewrite $4x^2 - 49$ as $[2x]^2 - 7^2$.

2. Let $2x = 7 \sec \theta$, so $x = \frac{7}{2} \sec \theta$ and $dx = \frac{7}{2} \sec \theta \tan \theta d\theta$.

3. Substitute into the integral and simplify the square root.

4. Integrate with respect to $\theta$ and convert back to $x$.

Try solving on your own before revealing the answer!

Final Answer:

$\int \frac{dx}{\sqrt{4x^2 - 49}} = \frac{1}{7} \ln\left|2x + \sqrt{4x^2 - 49}\right| + C$

We used trigonometric substitution and the standard logarithmic form for this type of integral.