[CALCULUS] Seen questions

Worked solutions on trigonometric integrals, average values, reduction formulae, substitutions and partial fractions.

These past exam questions combine product-to-sum identities, tangent-secant integrals, reduction formulae, substitutions and partial fractions. The first chooses a method for each branch of a piecewise integral; the second combines a reduction formula with a substitution and partial fractions. In each problem, identify the structure that makes the method apply before starting the algebra and track interval and branch restrictions when changing variables.

[METHOD]

Product-to-sum

Convert the product to a sum, then integrate term by term:

\sin A\cos B=\tfrac12\bigl(\sin(A+B)+\sin(A-B)\bigr)

\cos A\cos B=\tfrac12\bigl(\cos(A+B)+\cos(A-B)\bigr)

\sin A\sin B=-\tfrac12\bigl(\cos(A+B)-\cos(A-B)\bigr)

[1]:Odd power of sine or cosine
[2]:Both powers even
[3]:Average value of a piecewise function
[4]: $\int \sin 3x \sin 2x \, dx$

[METHOD]

Powers of tangent and secant

To evaluate $\int \tan^m x \sec^n x \, dx$ :

If $n$ is a positive even integer, factor out $\sec^2 x$ and substitute $u=\tan x$ . Convert remaining powers of $\sec$ by $1+\tan^2 x=\sec^2 x$ .
If $n=0$ , use $\tan^2 x=\sec^2 x-1$ to lower the power of $\tan$ by two, with $\int dx$ and $\int\tan x\,dx$ as base cases.
If $n$ $n$ is odd:
- If $m$ is odd, factor out $\tan x\sec x$ and substitute $u=\sec x$ . Convert remaining powers of $\tan$ by $\tan^2 x=\sec^2 x-1$ .
- If $m$ is even, write $\tan^m x=(\sec^2 x-1)^{m/2}$ and expand into powers of $\sec x$ . Reduce odd secant powers by parts until only $\int\sec x\,dx$ remains.

[1]:Tangent and secant
[2]: $\int \sec^3 x \, dx$
[3]:Average value of a piecewise function
[4]: $\int \sec^4 x \tan^2 x \, dx$
[5]: $\int \sec x \tan^3 x \, dx$
1 more

[EXAMPLE]

Average value of a piecewise function

The average value of a function $f$ on $[a,b]$ is

\bar f=\frac{1}{b-a}\int_a^b f(x)\,dx.

Compute the average $\bar f$ on $\left[-\frac{\pi}{4},\frac{\pi}{4}\right]$ for

f(x)= \begin{cases} \cos 3x\sin 2x,&-\frac{\pi}{4}\le x\le0,\\[2mm] \tan^3x\sec^2x,&0<x\le\frac{\pi}{4}. \end{cases}

Solution

The interval is split at $x=0$ , so the integral of $f$ is the sum of one integral from each branch. Parts (a) and (b) find the two antiderivatives, part (c) applies them on the appropriate intervals and part (d) asks whether $f$ attains its average.

(a) The product-to-sum integral

The product-to-sum identity

\sin A\cos B=\frac12\bigl(\sin(A+B)+\sin(A-B)\bigr)

applies after writing $\cos3x\sin2x$ as $\sin2x\cos3x$ . Thus

\begin{aligned} I &=\int \cos3x\sin2x\,dx\\ &=\frac12\int\bigl(\sin(2x+3x)+\sin(2x-3x)\bigr)\,dx\\ &=\frac12\int\bigl(\sin5x-\sin x\bigr)\,dx\\ &=-\frac{\cos5x}{10}+\frac{\cos x}{2}+C. \end{aligned}

The difference of arguments is $-x$ and $\sin(-x)=-\sin x$ ; this sign is easy to lose.

(b) The tangent-secant integral

Use the substitution

u=\tan x, \qquad du=\sec^2x\,dx.

Then

J=\int\tan^3x\sec^2x\,dx =\int u^3\,du =\frac{u^4}{4}+C =\frac{\tan^4x}{4}+C.

For the definite integral on the second branch, the bounds transform as

x=0\Longrightarrow u=\tan0=0, \qquad x=\frac{\pi}{4}\Longrightarrow u=\tan\frac{\pi}{4}=1.

Therefore

\int_0^{\pi/4}\tan^3x\sec^2x\,dx =\int_0^1u^3\,du =\left[\frac{u^4}{4}\right]_0^1 =\frac14.

(c) Computing the average

For the first branch, use the antiderivative from (a):

\begin{aligned} \int_{-\pi/4}^{0}\cos3x\sin2x\,dx &=\left[-\frac{\cos5x}{10}+\frac{\cos x}{2}\right]_{-\pi/4}^{0}\\ &=\frac25-\frac{3\sqrt2}{10}. \end{aligned}

The second branch contributes $\frac14$ by (b). Each formula for $f$ applies only on its own subinterval, so these two integrals give the total:

\begin{aligned} \int_{-\pi/4}^{\pi/4}f(x)\,dx &=\int_{-\pi/4}^{0}\cos3x\sin2x\,dx +\int_{0}^{\pi/4}\tan^3x\sec^2x\,dx\\ &=\left(\frac25-\frac{3\sqrt2}{10}\right)+\frac14\\ &=\frac{13-6\sqrt2}{20}. \end{aligned}

The interval length is

\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)=\frac{\pi}{2}.

Consequently,

{ \bar f =\frac{1}{\pi/2}\cdot\frac{13-6\sqrt2}{20} =\frac{13-6\sqrt2}{10\pi} }.

(d) Existence of a point attaining the average

The first branch is continuous on $\left[-\frac{\pi}{4},0\right]$ . At the endpoints of this subinterval,

f\left(-\frac{\pi}{4}\right) =\cos\left(-\frac{3\pi}{4}\right)\sin\left(-\frac{\pi}{2}\right) =\frac{\sqrt2}{2}, \qquad f(0)=0.

Also,

0<\frac{13-6\sqrt2}{10\pi}<\frac{\sqrt2}{2}.

Thus $\bar f$ lies strictly between $f(0)$ and $f\left(-\frac{\pi}{4}\right)$ . The Intermediate Value Theorem applied to the first branch gives a point

c\in\left(-\frac{\pi}{4},0\right) \subset\left(-\frac{\pi}{4},\frac{\pi}{4}\right)

such that $f(c)=\bar f$ . Therefore the answer is Yes.

Which branch owns the single point $x=0$ does not affect the integral or the average. What matters is which formula is integrated on each side.

[exam]

[1]:Product-to-sum
[2]:Powers of tangent and secant

[DEFINITION]

Reduction formula

A reduction formula expresses an integral $I_n$ in terms of $I_{n-k}$ for a smaller index, typically by integration by parts.

[THEOREM]

Partial fractions decomposition

For a fixed denominator after cancellation and factorisation, every proper rational function $f$ has a unique partial-fraction expansion with terms of the form

\frac{A}{(x-a)^k},\qquad \frac{Bx+C}{(x^2+bx+c)^k},

where the quadratic $x^2+bx+c$ is irreducible over the real numbers (i.e. $b^2-4c<0$ ).

[EXAMPLE]

Reduction formulae and substitution of a hyperbolic function

Consider

I_n=\int_0^1(7+12x)^n e^{7x+6x^2}\,dx.

It can be written as a reduction formula in the form

I_n=a_n+b_nI_{n-2},\qquad n\ge2,

where $a_n$ and $b_n$ depend on $n$ . Find $a_n$ and $b_n$ in terms of $n$ .

Next, consider

J=\int\frac{\cosh(6x)} {\sinh^4(6x)-343\sinh(6x)}\,dx.

Suggest a substitution so that

J=\frac16\int\frac{dt}{t^4-343t},

and then write the rational integrand as

\frac{A(t)}{t}+\frac{B(t)}{t-7} +\frac{C(t)}{t^2+7t+49},

where the degree of each numerator is less than the degree of its denominator. Find $A(t)$ , $B(t)$ and $C(t)$ .

Solution

In the first integral, the factor $7+12x$ is the derivative of the exponent:

\frac{d}{dx}(7x+6x^2)=7+12x.

Differentiating a product with one lower power of $7+12x$ therefore produces both an $I_n$ term and an $I_{n-2}$ term.

(a) Finding the reduction formula

For $n\ge2$ , differentiate

(7+12x)^{n-1}e^{7x+6x^2}.

Using the product rule and the chain rule gives

\begin{aligned} \frac{d}{dx}\left((7+12x)^{n-1}e^{7x+6x^2}\right) &=12(n-1)(7+12x)^{n-2}e^{7x+6x^2}\\ &\quad +(7+12x)^{n}e^{7x+6x^2}. \end{aligned}

Integrate from $0$ to $1$ . The second term on the right is $I_n$ , while the first term is $12(n-1)I_{n-2}$ , so

\begin{aligned} \left[(7+12x)^{n-1}e^{7x+6x^2}\right]_0^1 &=12(n-1)I_{n-2}+I_n,\\ I_n &=\left[(7+12x)^{n-1}e^{7x+6x^2}\right]_0^1 -12(n-1)I_{n-2}. \end{aligned}

At the endpoints,

7+12(1)=19,\qquad 7+12(0)=7,

and

7(1)+6(1)^2=13,\qquad 7(0)+6(0)^2=0.

Therefore

I_n =\left(19^{n-1}e^{13}-7^{n-1}\right)-12(n-1)I_{n-2}.

Comparing this with $I_n=a_n+b_nI_{n-2}$ gives

{a_n=19^{n-1}e^{13}-7^{n-1}}, \qquad {b_n=-12(n-1)}.

(b) Substitution and partial fractions

The numerator $\cosh(6x)$ is the derivative of $\sinh(6x)$ up to a factor of $6$ , so this is an ordinary substitution, not a surd-clearing hyperbolic substitution. Set

t=\sinh(6x), \qquad dt=6\cosh(6x)\,dx.

Thus $\cosh(6x)\,dx=\frac16\,dt$ and

J =\frac16\int\frac{dt}{t^4-343t}.

Factor the denominator:

t^4-343t =t(t^3-343) =t(t-7)(t^2+7t+49).

The quadratic has discriminant

7^2-4(49)=-147<0,

so it is an irreducible real quadratic. The required partial-fraction form is

\frac{1}{t(t-7)(t^2+7t+49)} =\frac{A}{t}+\frac{B}{t-7} +\frac{Ct+D}{t^2+7t+49}.

The question writes the last numerator as $C(t)$ ; here $C(t)=Ct+D$ .

Multiplying through by the denominator gives the polynomial identity

1=A(t-7)(t^2+7t+49) +Bt(t^2+7t+49) +(Ct+D)t(t-7).

Since

(t-7)(t^2+7t+49)=t^3-343,

substitution of $t=0$ gives

1=A(-7)(49)=-343A, \qquad A=-\frac1{343}.

Substitution of $t=7$ gives

1=B(7)(7^2+7\cdot7+49) =B(7)(147)=1029B, \qquad B=\frac1{1029}.

To find the quadratic numerator, expand the identity and compare the coefficients of $t$ and $t^2$ :

1=A(t^3-343)+B(t^3+7t^2+49t) +C t^3+(D-7C)t^2-7Dt.

The coefficient of $t$ gives

49B-7D=0 \quad\Longrightarrow\quad D=7B=\frac7{1029}.

The coefficient of $t^2$ then gives

7B+D-7C=0 \quad\Longrightarrow\quad C=\frac{7B+D}{7} =\frac2{1029}.

Hence the requested numerators are

{A(t)=-\frac1{343}}, \qquad {B(t)=\frac1{1029}}, \qquad {C(t)=\frac{2t+7}{1029}}.

As a check, the decomposition is

\frac{1}{t^4-343t} = -\frac{1}{343t} +\frac{1}{1029(t-7)} +\frac{2t+7}{1029(t^2+7t+49)}.

Consequently, on any interval that does not contain a point where $\sinh(6x)=0$ or $\sinh(6x)=7$ , an antiderivative is

J =-\frac1{2058}\ln|t| +\frac1{6174}\ln|t-7| +\frac1{6174}\ln(t^2+7t+49)+C, \qquad t=\sinh(6x).

The first part is a reduction formula: differentiating a product with one lower power gives a recurrence that drops the index by two. The second is a substitution followed by partial fractions for a proper rational function.

[exam]

[1]:Reduction formula
[2]:Partial fractions decomposition

Seen questions

Product-to-sum

Powers of tangent and secant

Reduction formula

Partial fractions decomposition

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