This TSA checks H2 integration techniques including exponential and logarithmic forms, rational algebraic integration, trigonometric integration, higher powers of trigonometric functions, integration by parts, and integration by substitution.
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Exponential, Logarithmic and Reverse Chain Forms
1. Which expression gives \[ \int 4x\cdot 2^{2x^2+3}\,dx? \]
Tested fundamental: Integrating \(a^{f(x)}f'(x)\).
Explanation: Let \(u=2x^2+3\). Then \(u’=4x\), so \(\int 4x\cdot 2^{2x^2+3}\,dx=\frac{2^{2x^2+3}}{\ln2}+c\).
Common wrong choice: C forgets that \(\int a^x\,dx=\frac{a^x}{\ln a}+c\), not \(a^x+c\).
2. Which expression gives \[ \int \frac{2x}{x^2-5}\,dx? \]
Tested fundamental: Recognising the \(\frac{f'(x)}{f(x)}\) form.
Explanation: This is in the form \(\int \frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+c\). The modulus is important.
Common wrong choice: A misses the modulus signs in \(\ln|x^2-5|\).
3. Which technique is most appropriate for \[ \int x^3e^{-x^2}\,dx? \]
Tested fundamental: Choosing integration by parts when part of the integrand can be integrated by reverse chain rule.
Explanation: Write \(x^3e^{-x^2}=x^2(xe^{-x^2})\). The factor \(xe^{-x^2}\) can be integrated using reverse chain rule, so integration by parts is suitable.
Common wrong choice: D is for differentiation, not integration.
Rational Algebraic Integration
4. Which standard form is most relevant for \[ \int \frac{2}{x^2+9}\,dx? \]
Tested fundamental: Identifying the \(x^2+a^2\) inverse tangent form.
Explanation: The denominator has the form \(x^2+a^2\), which leads to an inverse tangent result.
Common wrong choice: C is used for forms such as \(\frac{f'(x)}{f(x)}\) or certain \(a^2-x^2\) and \(x^2-a^2\) forms, not this one.
5. Which standard form is most relevant for \[ \int \frac{1}{\sqrt{9-x^2}}\,dx? \]
Tested fundamental: Identifying the inverse sine form.
Explanation: The integrand has the standard form \(\frac{1}{\sqrt{a^2-x^2}}\), which leads directly to a \(\sin^{-1}\) result.
Common wrong choice: B is tempting, but the usual standard formula gives \(\sin^{-1}\left(\frac{x}{a}\right)+c\).
6. Which expression gives \[ \int \frac{1}{4-x^2}\,dx? \]
Tested fundamental: Integrating the \(a^2-x^2\) logarithmic form.
Explanation: This matches \(\int \frac{1}{a^2-x^2}\,dx=\frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right|+c\). Here \(a=2\), so the coefficient is \(\frac14\).
Common wrong choice: C wrongly treats the numerator as the derivative of the denominator.
7. To integrate \[ \int \frac{3}{x^2-6x+13}\,dx, \] what should be done first?
Tested fundamental: Recognising when completing the square is needed for rational algebraic integration.
Explanation: Complete the square: \(x^2-6x+13=(x-3)^2+4\). This then leads to an inverse tangent form.
Common wrong choice: D is not valid because the numerator is not the derivative of the denominator.
8. Which rewrite is most useful for integrating \[ \int \frac{2x+5}{x^2+4x+6}\,dx? \]
Tested fundamental: Splitting a rational integrand using the derivative of the denominator.
Explanation: The derivative of the denominator is \(2x+4\). Splitting \(2x+5=(2x+4)+1\) creates a logarithmic part and a remaining rational part.
Common wrong choice: C misses the need to create a multiple of the derivative of the denominator.
9. A rational integrand is improper. What should usually be done before applying standard integration forms?
Tested fundamental: Recognising the first step for improper rational integrands.
Explanation: If the numerator has degree greater than or equal to the denominator, perform long division first.
Common wrong choice: A may be useful later for a quadratic denominator, but long division must be dealt with first when the rational integrand is improper.
10. Which technique is most appropriate for \[ \int \frac{x+3}{\sqrt{x^2-2x+10}}\,dx? \]
Tested fundamental: Splitting a numerator against the derivative of a quadratic under a square root.
Explanation: The derivative of \(x^2-2x+10\) is \(2x-2\). The numerator can be rewritten using a multiple of \(2x-2\), leaving a standard remaining part.
Common wrong choice: B is usually for products of functions where one part simplifies after differentiation or repeated parts is needed.
Trigonometric Integration and Higher Powers
11. Which expression gives \[ \int \mathrm{cosec}^2(3x)\,dx? \]
Tested fundamental: Accounting for the inner derivative in trigonometric integration.
Explanation: Since \(\frac{d}{dx}\cot(3x)=-3\mathrm{cosec}^2(3x)\), we have \(\int \mathrm{cosec}^2(3x)\,dx=-\frac13\cot(3x)+c\).
Common wrong choice: B forgets to divide by the coefficient \(3\) from the inner derivative.
12. Which first step is most suitable for \[ \int \cot^2 3x\,dx? \]
Tested fundamental: Rewriting \(\cot^2x\) using \(1+\cot^2x=\mathrm{cosec}^2x\).
Explanation: Since \(1+\cot^2u=\mathrm{cosec}^2u\), we have \(\cot^2u=\mathrm{cosec}^2u-1\).
Common wrong choice: B reverses the identity and gives the wrong sign.
13. For \[ \int \sin^4x\,dx, \] which technique is most appropriate?
Tested fundamental: Choosing the correct method for even powers of sine or cosine.
Explanation: For even powers of sine or cosine, use double-angle formulae such as \(\sin^2x=\frac12-\frac12\cos2x\).
Common wrong choice: A is unnecessary because the main issue is reducing the even power first.
14. For \[ \int \cos^5x\,dx, \] which first step is most appropriate?
Tested fundamental: Handling odd powers of cosine.
Explanation: For odd powers of cosine, keep one \(\cos x\) aside and rewrite the remaining even power using \(1-\sin^2x\). Then use \(u=\sin x\).
Common wrong choice: C is more suitable for even powers, not odd powers.
15. For \[ \int \tan^5x\,dx, \] which first step is most appropriate?
Tested fundamental: Handling odd powers of tangent.
Explanation: For odd powers of tangent, separate one \(\tan x\), then rewrite the remaining even power using \(\tan^2x=\sec^2x-1\).
Common wrong choice: D treats the power as though it were the standard integral of \(\tan x\).
Integration by Parts
16. The integration by parts formula is \[ \int u\,dv=uv+\int v\,du. \]
Tested fundamental: Knowing the sign in the integration by parts formula.
Explanation: The correct formula is \(\int u\,dv=uv-\int v\,du\). The sign before the second integral is negative.
Common wrong choice: Choosing True usually means the minus sign has been forgotten.
17. For \[ \int x\tan^{-1}x\,dx, \] which choice is most suitable for integration by parts?
Tested fundamental: Choosing \(u\) for integration by parts.
Explanation: Inverse trigonometric functions are usually chosen as \(u\), since differentiating them gives rational algebraic functions.
Common wrong choice: A differentiates the algebraic factor instead and leaves the harder inverse trigonometric function to integrate.
18. For \[ \int e^{\sin^{-1}x}\,dx, \] which choice is most suitable for integration by parts?
Tested fundamental: Applying integration by parts directly to a non-standard integrand.
Explanation: Use integration by parts with \(u=e^{\sin^{-1}x}\) and \(\frac{dv}{dx}=1\). Then \(\frac{du}{dx}=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}\) and \(v=x\).
Hence \(\int e^{\sin^{-1}x}\,dx=xe^{\sin^{-1}x}-\int \frac{xe^{\sin^{-1}x}}{\sqrt{1-x^2}}\,dx.\)
For the remaining integral, apply integration by parts again with \(u=e^{\sin^{-1}x}\) and \(\frac{dv}{dx}=\frac{x}{\sqrt{1-x^2}}\). Then \(\frac{du}{dx}=\frac{e^{\sin^{-1}x}}{\sqrt{1-x^2}}\) and \(v=-\sqrt{1-x^2}\).
So \(\int \frac{xe^{\sin^{-1}x}}{\sqrt{1-x^2}}\,dx=-e^{\sin^{-1}x}\sqrt{1-x^2}+\int e^{\sin^{-1}x}\,dx.\)
Substituting this back, \(\int e^{\sin^{-1}x}\,dx=xe^{\sin^{-1}x}-\left[-e^{\sin^{-1}x}\sqrt{1-x^2}+\int e^{\sin^{-1}x}\,dx\right].\)
Therefore \(2\int e^{\sin^{-1}x}\,dx=e^{\sin^{-1}x}\left(x+\sqrt{1-x^2}\right)\).
Hence \(\int e^{\sin^{-1}x}\,dx=\frac12e^{\sin^{-1}x}\left(x+\sqrt{1-x^2}\right)+c.\)
Common wrong choice: B requires us to integrate \(e^{\sin^{-1}x}\) immediately, which is exactly the original problem.
19. For an integral like \[ \int e^{2x}\sin x\,dx, \] why may integration by parts need to be applied twice?
Tested fundamental: Recognising cyclic integration by parts.
Explanation: Products of exponential and trigonometric functions often cycle after integration by parts. After applying it twice, a multiple of the original integral appears and can be moved to the other side.
Common wrong choice: B is false because exponential functions can be integrated directly; the issue is the product with the trigonometric function.
Integration by Substitution
20. For a substitution \(x=f(t)\), it is necessary to replace \(dx\) by \(f'(t)\,dt\).
Tested fundamental: Replacing \(dx\) correctly during substitution.
Explanation: Since \(\frac{dx}{dt}=f'(t)\), we have \(dx=f'(t)\,dt\). All \(x\)-terms must also be rewritten in terms of \(t\).
Common wrong choice: Choosing False usually means the differential \(dx\) has not been converted properly.
21. Using the substitution \(u=\cos x\), write \(\sin2x\,dx\) in terms of \(u\) and \(du\).
Answer format: Give your answer in terms of \(u\) and \(du\), for example 3u du.
Answer: \(-2u\,du\)
Tested fundamental: Rewriting an integrand fully under substitution.
Explanation: Since \(\sin2x=2\sin x\cos x=2u\sin x\), and \(du=-\sin x\,dx\), we have \(\sin2x\,dx=-2u\,du\).
Common error: Forgetting the negative sign from \(du=-\sin x\,dx\).
22. Using the substitution \[ x=3\sin\theta,\qquad 0<\theta<\frac{\pi}{2}, \] find \[ \int \frac{x}{\sqrt{9-x^2}}\,dx \] in terms of \(x\).
Tested fundamental: Converting a trigonometric substitution answer back to the original variable.
Explanation: Since \(x=3\sin\theta\), we have \(dx=3\cos\theta\,d\theta\). Also, \(\sqrt{9-x^2}=\sqrt{9-9\sin^2\theta}=3\cos\theta\).
So \(\int \frac{x}{\sqrt{9-x^2}}\,dx=\int \frac{3\sin\theta}{3\cos\theta}\cdot3\cos\theta\,d\theta=\int 3\sin\theta\,d\theta=-3\cos\theta+c.\)
But the final answer must be in terms of \(x\). Since \(x=3\sin\theta\), we have \(\sin\theta=\frac{x}{3}\).
Using a right-angled triangle, \(\text{opposite}=x\), \(\text{hypotenuse}=3\), so \(\text{adjacent}=\sqrt{9-x^2}\).
Therefore \(\cos\theta=\frac{\sqrt{9-x^2}}{3}\), and \(-3\cos\theta=-\sqrt{9-x^2}\).
So the final answer is \(-\sqrt{9-x^2}+c\).
Common wrong choice: B is the answer before converting back to \(x\). C has the wrong sign and is also still in terms of \(\theta\).
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