Hard Further Maths Further integration techniques Questions

Challenging, exam-style Further Maths Further integration techniques questions with worked solutions. Stretch yourself on the hardest arc-length, cartesian-form, integration, parametric-form problems.

arc-lengthcartesian-formintegrationparametric-formpolar-formsurface-of-revolution
Further Maths34 questionsStep-by-step solutions
Question 1
9 markschallenging
A curve has parametric equations x=6t2x=6 t^{2}, y=4t3y=4 t^{3}, for 0≤t≤10\le t\le 1. The arc is rotated through 2π2\pi radians about the xx-axis. Which of the following integrals gives the area of the curved surface generated?
Show worked solution

Worked solution

  1. Recall the surface-area formula for rotation about the xx-axis

    S=2π∫y dsS=2\pi\int y\,ds

    The radius of each band is the yy-coordinate of the curve.

  2. Write the element of arc length in terms of tt

    ds=(dxdt)2+(dydt)2 dt=12tt2+1 dtds=\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}\,dt=12 t \sqrt{t^{2} + 1}\,dt

    Here dxdt=12t\frac{dx}{dt}=12 t and dydt=12t2\frac{dy}{dt}=12 t^{2}.

  3. Substitute into the formula

    S=2π∫0148t4t2+1 dtS=2\pi \int_{0}^{1}48 t^{4} \sqrt{t^{2} + 1}\,dt

    Both the radius and the arc element are written in terms of the parameter.

  4. Reject the option with radius xx

    rotation about the x-axis gives radius y\text{rotation about the }x\text{-axis gives radius }y

    The radius xx belongs to rotation about the yy-axis.

  5. Reject the option that uses the cartesian arc element

    1+(dydt)2≠dsdt\sqrt{1+\left(\frac{dy}{dt}\right)^{2}}\neq\frac{ds}{dt}

    In parametric form both derivatives must be squared and added.

  6. Recall the cartesian arc-length formula

    s=∫ab1+(dydx)2 dxs=\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx

    This is the formula quoted in the formula book for a curve given as y=f(x)y=f(x).

  7. Recall where the arc-length formula comes from

    (δs)2≈(δx)2+(δy)2\left(\delta s\right)^{2}\approx\left(\delta x\right)^{2}+\left(\delta y\right)^{2}

    Pythagoras on a small element of the curve gives dsdx=1+(dydx)2\frac{ds}{dx}=\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}.

  8. Recall the parametric arc-length formula

    s=∫t1t2(dxdt)2+(dydt)2 dts=\int_{t_{1}}^{t_{2}}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}\,dt

    Dividing the element (δs)2≈(δx)2+(δy)2\left(\delta s\right)^{2}\approx\left(\delta x\right)^{2}+\left(\delta y\right)^{2} by (δt)2\left(\delta t\right)^{2} gives this form.

  9. Recall the polar arc-length formula

    s=∫αβr2+(drdθ)2 dθs=\int_{\alpha}^{\beta}\sqrt{r^{2}+\left(\frac{dr}{d\theta}\right)^{2}}\,d\theta

    It follows from the parametric formula with x=rcos⁡θx=r\cos\theta and y=rsin⁡θy=r\sin\theta.

  10. Recall the surface-area formula for rotation about the xx-axis

    S=2π∫y dsS=2\pi\int y\,ds

    Each element of arc sweeps out a thin band of radius yy and width dsds.

  11. Recall the surface-area formula for rotation about the yy-axis

    S=2π∫x dsS=2\pi\int x\,ds

    The radius of the band is now the distance xx from the yy-axis.

  12. Recall the integration-by-parts formula

    ∫abudvdx dx=[uv]ab−∫abvdudx dx\int_{a}^{b} u\frac{dv}{dx}\,dx=\left[uv\right]_{a}^{b}-\int_{a}^{b} v\frac{du}{dx}\,dx

    Integration by parts is what turns InI_{n} into an expression involving a lower index.

  13. Recall the Pythagorean identity for the tangent

    1+tan⁡2x=sec⁡2x1+\tan^{2}x=\sec^{2}x

    This identity is what lets a power of sec⁡x\sec x be split off during a reduction.

  14. Recall the Pythagorean identity

    sin⁡2x+cos⁡2x=1\sin^{2}x+\cos^{2}x=1

    It converts between powers of sin⁡x\sin x and powers of cos⁡x\cos x.

  15. Select the correct integral

    2π∫0148t4t2+1 dt2\pi \int_{0}^{1}48 t^{4} \sqrt{t^{2} + 1}\,dt

    This is the parametric surface-area integral for rotation about the xx-axis.

Answer
2π∫0148t4t2+1 dt2\pi \int_{0}^{1}48 t^{4} \sqrt{t^{2} + 1}\,dt
Question 2
9 markschallenging
The arc of the curve y=2xy=2 \sqrt{x} between x=1x=1 and x=4x=4 is rotated through 2π2\pi radians about the xx-axis. Which of the following is the exact area of the curved surface generated?
Show worked solution

Worked solution

  1. Differentiate the equation of the curve

    dydx=1x\frac{dy}{dx}=\frac{1}{\sqrt{x}}

    The gradient is needed for the element of arc length.

  2. Simplify the element of arc length

    dsdx=x+1x\frac{ds}{dx}=\frac{\sqrt{x + 1}}{\sqrt{x}}

    The expression under the root simplifies to x+1x\frac{x + 1}{x}.

  3. Form the surface-area integral

    S=2π∫142x+1 dxS=2\pi\int_{1}^{4}2 \sqrt{x + 1}\,dx

    The radius of each band is yy.

  4. Evaluate the integral

    S=2π[4(x+1)323]14=8π(−22+55)3S=2\pi\left[\frac{4 \left(x + 1\right)^{\frac{3}{2}}}{3}\right]_{1}^{4}=\frac{8 \pi \left(- 2 \sqrt{2} + 5 \sqrt{5}\right)}{3}

    Substituting the limits and multiplying by 2π2\pi gives the exact area.

  5. Check the value numerically

    S≈69.968821S\approx 69.968821

    The decimal value rules out the options of the wrong size.

  6. Reject the option without the factor 2π2\pi

    the 2π comes from the circumference of each band\text{the }2\pi\text{ comes from the circumference of each band}

    Dropping it makes the answer far too small.

  7. Recall the cartesian arc-length formula

    s=∫ab1+(dydx)2 dxs=\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx

    This is the formula quoted in the formula book for a curve given as y=f(x)y=f(x).

  8. Recall where the arc-length formula comes from

    (δs)2≈(δx)2+(δy)2\left(\delta s\right)^{2}\approx\left(\delta x\right)^{2}+\left(\delta y\right)^{2}

    Pythagoras on a small element of the curve gives dsdx=1+(dydx)2\frac{ds}{dx}=\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}.

  9. Recall the parametric arc-length formula

    s=∫t1t2(dxdt)2+(dydt)2 dts=\int_{t_{1}}^{t_{2}}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}\,dt

    Dividing the element (δs)2≈(δx)2+(δy)2\left(\delta s\right)^{2}\approx\left(\delta x\right)^{2}+\left(\delta y\right)^{2} by (δt)2\left(\delta t\right)^{2} gives this form.

  10. Recall the polar arc-length formula

    s=∫αβr2+(drdθ)2 dθs=\int_{\alpha}^{\beta}\sqrt{r^{2}+\left(\frac{dr}{d\theta}\right)^{2}}\,d\theta

    It follows from the parametric formula with x=rcos⁡θx=r\cos\theta and y=rsin⁡θy=r\sin\theta.

  11. Recall the surface-area formula for rotation about the xx-axis

    S=2π∫y dsS=2\pi\int y\,ds

    Each element of arc sweeps out a thin band of radius yy and width dsds.

  12. Recall the surface-area formula for rotation about the yy-axis

    S=2π∫x dsS=2\pi\int x\,ds

    The radius of the band is now the distance xx from the yy-axis.

  13. Recall the integration-by-parts formula

    ∫abudvdx dx=[uv]ab−∫abvdudx dx\int_{a}^{b} u\frac{dv}{dx}\,dx=\left[uv\right]_{a}^{b}-\int_{a}^{b} v\frac{du}{dx}\,dx

    Integration by parts is what turns InI_{n} into an expression involving a lower index.

  14. Recall the Pythagorean identity for the tangent

    1+tan⁡2x=sec⁡2x1+\tan^{2}x=\sec^{2}x

    This identity is what lets a power of sec⁡x\sec x be split off during a reduction.

  15. Select the correct surface area

    S=8π(−22+55)3S=\frac{8 \pi \left(- 2 \sqrt{2} + 5 \sqrt{5}\right)}{3}

    This is the exact area of the curved surface.

Answer
8π(−22+55)3\frac{8 \pi \left(- 2 \sqrt{2} + 5 \sqrt{5}\right)}{3}
Question 3
9 markschallenging
The integral InI_{n} is defined by In=∫0π4tan⁡n(x) dxI_{n}=\int_{0}^{\frac{\pi}{4}}\tan^{n}{\left(x \right)}\,dx for n≥2n\ge 2. Which of the following is the correct reduction formula for InI_{n}?
Show worked solution

Worked solution

  1. Set up the integration by parts

    tan⁡nx=tan⁡n−2x(sec⁡2x−1)\tan^{n}x=\tan^{n-2}x\left(\sec^{2}x-1\right)

    The split is chosen so that differentiating lowers the index.

  2. Identify the boundary term

    [tan⁡n−1(x)n−1]0π4=1n−1\left[\frac{\tan^{n - 1}{\left(x \right)}}{n - 1}\right]_{0}^{\frac{\pi}{4}}=\frac{1}{n - 1}

    This constant is the term that sits outside the recursion.

  3. Identify the coefficient of In−2I_{n-2}

    −1-1

    It comes from differentiating the index-carrying factor and rearranging.

  4. Test the candidate formula at n=3n=3

    I3=−−1+ln⁡(2)2,I1=ln⁡(2)2I_{3}=- \frac{-1 + \ln{\left(2 \right)}}{2},\qquad I_{1}=\frac{\ln{\left(2 \right)}}{2}

    Substituting known values eliminates the formulae that do not hold.

  5. Reject the option with the wrong sign

    a sign error changes the value of I4\text{a sign error changes the value of }I_{4}

    Only one of the five options reproduces the correct value of every InI_{n}.

  6. Recall the cartesian arc-length formula

    s=∫ab1+(dydx)2 dxs=\int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\,dx

    This is the formula quoted in the formula book for a curve given as y=f(x)y=f(x).

  7. Recall where the arc-length formula comes from

    (δs)2≈(δx)2+(δy)2\left(\delta s\right)^{2}\approx\left(\delta x\right)^{2}+\left(\delta y\right)^{2}

    Pythagoras on a small element of the curve gives dsdx=1+(dydx)2\frac{ds}{dx}=\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}.

  8. Recall the parametric arc-length formula

    s=∫t1t2(dxdt)2+(dydt)2 dts=\int_{t_{1}}^{t_{2}}\sqrt{\left(\frac{dx}{dt}\right)^{2}+\left(\frac{dy}{dt}\right)^{2}}\,dt

    Dividing the element (δs)2≈(δx)2+(δy)2\left(\delta s\right)^{2}\approx\left(\delta x\right)^{2}+\left(\delta y\right)^{2} by (δt)2\left(\delta t\right)^{2} gives this form.

  9. Recall the polar arc-length formula

    s=∫αβr2+(drdθ)2 dθs=\int_{\alpha}^{\beta}\sqrt{r^{2}+\left(\frac{dr}{d\theta}\right)^{2}}\,d\theta

    It follows from the parametric formula with x=rcos⁡θx=r\cos\theta and y=rsin⁡θy=r\sin\theta.

  10. Recall the surface-area formula for rotation about the xx-axis

    S=2π∫y dsS=2\pi\int y\,ds

    Each element of arc sweeps out a thin band of radius yy and width dsds.

  11. Recall the surface-area formula for rotation about the yy-axis

    S=2π∫x dsS=2\pi\int x\,ds

    The radius of the band is now the distance xx from the yy-axis.

  12. Recall the integration-by-parts formula

    ∫abudvdx dx=[uv]ab−∫abvdudx dx\int_{a}^{b} u\frac{dv}{dx}\,dx=\left[uv\right]_{a}^{b}-\int_{a}^{b} v\frac{du}{dx}\,dx

    Integration by parts is what turns InI_{n} into an expression involving a lower index.

  13. Recall the Pythagorean identity for the tangent

    1+tan⁡2x=sec⁡2x1+\tan^{2}x=\sec^{2}x

    This identity is what lets a power of sec⁡x\sec x be split off during a reduction.

  14. Recall the Pythagorean identity

    sin⁡2x+cos⁡2x=1\sin^{2}x+\cos^{2}x=1

    It converts between powers of sin⁡x\sin x and powers of cos⁡x\cos x.

  15. Select the correct reduction formula

    In=1n−1−In−2I_{n}=\frac{1}{n - 1}-I_{n-2}

    This is the formula obtained from the working above.

Answer
In=1n−1−In−2I_{n}=\frac{1}{n - 1}-I_{n-2}
Question 4
9 markschallenging
The integral InI_{n} is defined by In=∫0π4tan⁡n(x) dxI_{n}=\int_{0}^{\frac{\pi}{4}}\tan^{n}{\left(x \right)}\,dx for n≥2n\ge 2. By first establishing a reduction formula for InI_{n}, find the exact value of I5I_{5}.
Show worked solution

Worked solution

  1. Write down the integral to be reduced

    In=∫0π4tan⁡n(x) dxI_{n}=\int_{0}^{\frac{\pi}{4}}\tan^{n}{\left(x \right)}\,dx

    The index nn is the quantity that the reduction formula will lower.

  2. Split off two factors from the integrand

    tan⁡nx=tan⁡n−2xtan⁡2x\tan^{n}x=\tan^{n-2}x\tan^{2}x

    Two factors are removed so that the Pythagorean identity can be used.

  3. Use the Pythagorean identity

    tan⁡2x=sec⁡2x−1\tan^{2}x=\sec^{2}x-1

    This produces one integral that can be done by inspection and one that is In−2I_{n-2}.

  4. Write the integral as two integrals

    In=∫0π4tan⁡n−2xsec⁡2x dx−∫0π4tan⁡n−2x dxI_{n}=\int_{0}^{\frac{\pi}{4}}\tan^{n-2}x\sec^{2}x\,dx-\int_{0}^{\frac{\pi}{4}}\tan^{n-2}x\,dx

    The first integrand is ddx(tan⁡x)\frac{d}{dx}\left(\tan x\right) times a power of tan⁡x\tan x.

  5. Integrate the first part by inspection

    ∫tan⁡n−2xsec⁡2x dx=tan⁡n−1xn−1+c\int\tan^{n-2}x\sec^{2}x\,dx=\frac{\tan^{n-1}x}{n-1}+c

    The substitution w=tan⁡xw=\tan x makes this a power integral.

  6. Evaluate the resulting boundary term

    [tan⁡n−1xn−1]0π4=1n−1\left[\frac{\tan^{n-1}x}{n-1}\right]_{0}^{\frac{\pi}{4}}=\frac{1}{n - 1}

    At x=π4x=\frac{\pi}{4}, tan⁡x=1\tan x=1, and at x=0x=0, tan⁡x=0\tan x=0.

  7. Recognise the two integrals

    ∫0π4tan⁡n−2(x) dx=In−2,∫0π4tan⁡n(x) dx=In\int_{0}^{\frac{\pi}{4}}\tan^{n - 2}{\left(x \right)}\,dx=I_{n-2},\qquad\int_{0}^{\frac{\pi}{4}}\tan^{n}{\left(x \right)}\,dx=I_{n}

    Both are the defining integral with a different index.

  8. State the reduction formula

    In=1n−1−In−2I_{n}=\frac{1}{n - 1}-I_{n-2}

    This is valid for n≥2n\ge 2 and lowers the index by 22 each time.

  9. Apply the reduction formula with n=5n=5

    I5=14−I3I_{5}=\frac{1}{4}-I_{3}

    Each application lowers the index by 22.

  10. Apply the reduction formula with n=3n=3

    I3=12−I1I_{3}=\frac{1}{2}-I_{1}

    Each application lowers the index by 22.

  11. Evaluate the base integral I1I_{1}

    I1=∫0π4tan⁡(x) dx=ln⁡(2)2I_{1}=\int_{0}^{\frac{\pi}{4}}\tan{\left(x \right)}\,dx=\frac{\ln{\left(2 \right)}}{2}

    The recursion stops here, because I1I_{1} can be integrated directly.

  12. Substitute upwards to find I3I_{3}

    I3=−−1+ln⁡(2)2I_{3}=- \frac{-1 + \ln{\left(2 \right)}}{2}

    The value of I1I_{1} is now known, so I3I_{3} follows.

  13. Check the value numerically

    I5≈0.09657359I_{5}\approx 0.09657359

    A decimal check confirms the exact value is of the right size.

  14. Note the range of validity of the formula

    n≥2n\ge 2

    Below this the index n−2n-2 would be negative and the integral would change character.

  15. Substitute upwards once more to find I5I_{5}

    I5=−1+2ln⁡(2)4I_{5}=\frac{-1 + 2 \ln{\left(2 \right)}}{4}

    This is the exact value of the required integral.

Answer
−1+2ln⁡(2)4\frac{-1 + 2 \ln{\left(2 \right)}}{4}
Question 5
9 markschallenging
The integral InI_{n} is defined by In=∫01(1−x2)n dxI_{n}=\int_{0}^{1}\left(1 - x^{2}\right)^{n}\,dx for n≥1n\ge 1. By first establishing a reduction formula for InI_{n}, find the exact value of I4I_{4}.
Show worked solution

Worked solution

  1. Choose the two parts for integration by parts

    u=(1−x2)n,dvdx=1u=\left(1 - x^{2}\right)^{n},\qquad \frac{dv}{dx}=1

    Differentiating uu lowers the index, which is exactly what a reduction formula needs.

  2. Differentiate uu and integrate dvdx\frac{dv}{dx}

    dudx=−2nx(1−x2)n−1,v=x\frac{du}{dx}=- 2 n x \left(1 - x^{2}\right)^{n - 1},\qquad v=x

    These are the two pieces the parts formula needs.

  3. Apply the integration-by-parts formula

    In=[x(1−x2)n]01−∫01−2nx2(1−x2)n−1 dxI_{n}=\left[x \left(1 - x^{2}\right)^{n}\right]_{0}^{1}-\int_{0}^{1}- 2 n x^{2} \left(1 - x^{2}\right)^{n - 1}\,dx

    The boundary term is evaluated at once; the new integral is the one that must be recognised.

  4. Evaluate the boundary term

    [x(1−x2)n]01=0\left[x \left(1 - x^{2}\right)^{n}\right]_{0}^{1}=0

    Substitute the two limits into uvuv and subtract.

  5. Rewrite the remaining integrand in terms of the index

    −2nx2(1−x2)n−1=−2n(1−x2)n−1+2n(1−x2)n- 2 n x^{2} \left(1 - x^{2}\right)^{n - 1}=-2 n\left(1 - x^{2}\right)^{n - 1}+2 n\left(1 - x^{2}\right)^{n}

    The identity turns the leftover integrand into a combination of the integrands of In−1I_{n-1} and InI_{n}.

  6. Substitute back into the expression for InI_{n}

    In=2nIn−1−2nInI_{n}=2 nI_{n-1}-2 nI_{n}

    Every piece is now expressed with the integrals InI_{n} and In−1I_{n-1}; note that InI_{n} itself has reappeared.

  7. Collect the terms in InI_{n}

    (2n+1)In=2nIn−1\left(2 n + 1\right)I_{n}=2 nI_{n-1}

    The InI_{n} on the right is moved across and the coefficient is divided out.

  8. State the reduction formula

    In=2n2n+1In−1I_{n}=\frac{2 n}{2 n + 1}I_{n-1}

    This is valid for n≥1n\ge 1 and lowers the index by 11 each time.

  9. Apply the reduction formula with n=4n=4

    I4=89I3I_{4}=\frac{8}{9}I_{3}

    Each application lowers the index by 11.

  10. Apply the reduction formula with n=3n=3

    I3=67I2I_{3}=\frac{6}{7}I_{2}

    Each application lowers the index by 11.

  11. Apply the reduction formula with n=2n=2

    I2=45I1I_{2}=\frac{4}{5}I_{1}

    Each application lowers the index by 11.

  12. Apply the reduction formula with n=1n=1

    I1=23I0I_{1}=\frac{2}{3}I_{0}

    Each application lowers the index by 11.

  13. Evaluate the base integral I0I_{0}

    I0=∫011 dx=1I_{0}=\int_{0}^{1}1\,dx=1

    The recursion stops here, because I0I_{0} can be integrated directly.

  14. Substitute upwards to find I1I_{1}

    I1=23I_{1}=\frac{2}{3}

    The value of I0I_{0} is now known, so I1I_{1} follows.

  15. Substitute upwards to find I2I_{2}

    I2=815I_{2}=\frac{8}{15}

    The value of I1I_{1} is now known, so I2I_{2} follows.

  16. Substitute upwards to find I3I_{3}

    I3=1635I_{3}=\frac{16}{35}

    The value of I2I_{2} is now known, so I3I_{3} follows.

  17. Substitute upwards once more to find I4I_{4}

    I4=128315I_{4}=\frac{128}{315}

    This is the exact value of the required integral.

Answer
128315\frac{128}{315}

Unlock 29 more Further integration techniques questions

Create a free account to work through every Further Maths Further integration techniques question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Further integration techniques practice

Practice QuestionsWorked Solutions

Related Further Pure topics

Conic sections 1Conic sections 2Further complex numbersGroupsInequalitiesFurther matrix algebra