Chapter 5 · Techniques of Integration

5.1 Trigonometric Integrals

Powers and products of sines, cosines, tangents and secants — tamed by three identities and one good substitution.

Integrals built from trigonometric functions turn up wherever something oscillates. The average power delivered by mains electricity is an integral of $\sin^2$; the energy in a vibrating string is an integral of products like $\sin(mx)\sin(nx)$; the height of a city on a navigation chart is an integral of $\sec x$. None of these yields to a single substitution straight away, and integration by parts on its own goes round in circles.

The good news is that a handful of identities — $\sin^2x+\cos^2x=1$, $1+\tan^2x=\sec^2x$, the half-angle formulas and the product-to-sum formulas — let us rewrite the integrand into a shape where an ordinary $u$-substitution (Calculus I, 4.3) finishes the job. In this section we learn to read an integrand and know, before writing anything, which rewrite to make.

Powers of Sine and Cosine

Consider $\displaystyle\int\sin^mx\cos^nx\,dx$ where $m$ and $n$ are non-negative integers. The key observation is that $\cos x\,dx$ is exactly $d(\sin x)$ and $\sin x\,dx$ is exactly $-d(\cos x)$. So if we can save one factor of $\cos x$ (or $\sin x$) and write everything else in terms of $\sin x$ (or $\cos x$), a substitution does the rest. An even power can always be converted using $\sin^2x+\cos^2x=1$; an odd power always leaves one factor over to save.

🧭 Problem-solving strategy — $\displaystyle\int\sin^mx\cos^nx\,dx$
  1. The power of cosine is odd ($n=2k+1$). Save one factor $\cos x$, write $\cos^{2k}x=\left(1-\sin^2x\right)^k$, and substitute $u=\sin x$, $du=\cos x\,dx$.
  2. The power of sine is odd ($m=2k+1$). Save one factor $\sin x$, write $\sin^{2k}x=\left(1-\cos^2x\right)^k$, and substitute $u=\cos x$, $du=-\sin x\,dx$.
  3. Both powers are even. Lower the powers with the half-angle identities $$\sin^2x=\frac{1-\cos2x}{2},\qquad\cos^2x=\frac{1+\cos2x}{2},\qquad\sin x\cos x=\frac12\sin2x,$$ repeating until every term is a single cosine of a multiple angle.
If both powers are odd, either of the first two options works. If one power is $0$ the same rules apply — for example $\cos^5x$ has odd cosine power.
✏️ Example 1 — An odd power of sine
Evaluate $\displaystyle\int\sin^3x\cos^2x\,dx$.
Plan The power of sine is odd ($m=3$). Save one $\sin x$ and convert the remaining $\sin^2x$ to cosines: $$\sin^3x\cos^2x=\left(1-\cos^2x\right)\cos^2x\,\sin x.$$
Substitute $u=\cos x$, $du=-\sin x\,dx$: $$\int\left(1-\cos^2x\right)\cos^2x\sin x\,dx=-\int\left(1-u^2\right)u^2\,du=-\int\left(u^2-u^4\right)du=-\frac{u^3}{3}+\frac{u^5}{5}+C.$$
Back to $x$ $$\int\sin^3x\cos^2x\,dx=-\frac{\cos^3x}{3}+\frac{\cos^5x}{5}+C.$$
Check $\dfrac{d}{dx}\left(-\tfrac13\cos^3x+\tfrac15\cos^5x\right)=\cos^2x\sin x-\cos^4x\sin x=\sin x\cos^2x\left(1-\cos^2x\right)=\sin^3x\cos^2x$. ✓
✏️ Example 2 — Both powers even
Evaluate $\displaystyle\int\sin^4x\,dx$, and hence find $\displaystyle\int_0^{\pi}\sin^4x\,dx$.
Plan Here $m=4$, $n=0$: both even, so saving a factor would leave $\sin^3x$ with no matching derivative. Use the half-angle identity instead.
Lower the power $$\sin^4x=\left(\frac{1-\cos2x}{2}\right)^2=\frac14\left(1-2\cos2x+\cos^22x\right).$$ The term $\cos^22x$ is still a square, so apply the identity again with angle $2x$: $\cos^22x=\dfrac{1+\cos4x}{2}$. Hence $$\sin^4x=\frac14-\frac12\cos2x+\frac18\left(1+\cos4x\right)=\frac38-\frac12\cos2x+\frac18\cos4x.$$
Integrate $$\int\sin^4x\,dx=\frac{3x}{8}-\frac{\sin2x}{4}+\frac{\sin4x}{32}+C.$$
Definite At $x=\pi$ and $x=0$ both sine terms vanish, so $\displaystyle\int_0^{\pi}\sin^4x\,dx=\frac{3\pi}{8}\approx1.1781$. Check: Interactive 1 with $m=4$, $n=0$ shows the same number when the slider reaches $\pi$.
Interactive 1 — Area under $\sin^m x\cos^n x$ with a Live Antiderivative Check

Choose the powers. The lab decides which rule of the strategy applies, builds the antiderivative $F$ exactly (shown below the graph), and sweeps the right end point $b$ from $0$ to $\pi$. The blue area counts positively and the red area negatively; the numerical area must always equal $F(b)-F(0)$.

$b$—
area by numerical sum—
$F(b)-F(0)$—
method—

Powers of Tangent and Secant

The same idea works for $\displaystyle\int\tan^mx\sec^nx\,dx$, now using the pair of facts $\dfrac{d}{dx}\tan x=\sec^2x$ and $\dfrac{d}{dx}\sec x=\sec x\tan x$, together with the identity $\tan^2x=\sec^2x-1$. First we need the two basic integrals, which are not on the Calculus I table.

📐 Theorem 5.1 — Integrals of $\tan x$ and $\sec x$
On any interval on which $\cos x\ne0$, $$\int\tan x\,dx=\ln\left|\sec x\right|+C,\qquad\int\sec x\,dx=\ln\left|\sec x+\tan x\right|+C.$$ Similarly, on any interval on which $\sin x\ne0$, $\displaystyle\int\cot x\,dx=\ln|\sin x|+C$ and $\displaystyle\int\csc x\,dx=-\ln|\csc x+\cot x|+C$.

Why. $\displaystyle\int\frac{\sin x}{\cos x}dx$ with $w=\cos x$ gives $-\ln|\cos x|=\ln|\sec x|$. For $\sec x$, multiply top and bottom by $\sec x+\tan x$: $$\int\sec x\,dx=\int\frac{\sec^2x+\sec x\tan x}{\sec x+\tan x}\,dx,$$ and the numerator is exactly the derivative of the denominator, so the integral is $\ln|\sec x+\tan x|+C$.

🧭 Problem-solving strategy — $\displaystyle\int\tan^mx\sec^nx\,dx$
  1. The power of secant is even ($n=2k\ge2$). Save one factor $\sec^2x$, write the remaining $\sec^{2k-2}x=\left(1+\tan^2x\right)^{k-1}$, and substitute $u=\tan x$, $du=\sec^2x\,dx$.
  2. The power of tangent is odd ($m=2k+1$) and $n\ge1$. Save one factor $\sec x\tan x$, write $\tan^{2k}x=\left(\sec^2x-1\right)^k$, and substitute $u=\sec x$, $du=\sec x\tan x\,dx$.
  3. Tangent power even, secant power odd. Replace $\tan^2x$ by $\sec^2x-1$ to leave only odd powers of $\sec x$, then use $\int\sec x\,dx$ (Theorem 5.1) and integration by parts or the reduction formula (Theorem 5.2) for $\sec^3x,\sec^5x,\dots$
  4. Only a power of tangent ($n=0$). Split off $\tan^2x=\sec^2x-1$: $\displaystyle\int\tan^mx\,dx=\int\tan^{m-2}x\sec^2x\,dx-\int\tan^{m-2}x\,dx$, and repeat.
✏️ Example 3 — An even power of secant
Evaluate $\displaystyle\int\tan^3x\sec^4x\,dx$.
Plan The power of secant is even, so save $\sec^2x$ and convert the other $\sec^2x$: $$\tan^3x\sec^4x=\tan^3x\left(1+\tan^2x\right)\sec^2x.$$ (The tangent power is also odd, so rule 2 would work too — try it and compare.)
Substitute $u=\tan x$, $du=\sec^2x\,dx$: $$\int\left(u^3+u^5\right)du=\frac{u^4}{4}+\frac{u^6}{6}+C=\frac{\tan^4x}{4}+\frac{\tan^6x}{6}+C.$$
Check $\dfrac{d}{dx}\left(\tfrac14\tan^4x+\tfrac16\tan^6x\right)=\tan^3x\sec^2x+\tan^5x\sec^2x=\tan^3x\sec^2x\left(1+\tan^2x\right)=\tan^3x\sec^4x$. ✓
✏️ Example 4 — The integral of $\sec^3x$
Evaluate $\displaystyle\int\sec^3x\,dx$ on an interval where $\cos x\ne0$.
Plan Tangent power $0$ (even), secant power odd — rule 3. Integrate by parts, giving the easy-to-integrate factor $\sec^2x$ the role of $dv$: $$\begin{array}{ll}u=\sec x & dv=\sec^2x\,dx\\ du=\sec x\tan x\,dx & v=\tan x\end{array}$$
Parts $$\int\sec^3x\,dx=\sec x\tan x-\int\sec x\tan^2x\,dx=\sec x\tan x-\int\sec x\left(\sec^2x-1\right)dx.$$ So $$\int\sec^3x\,dx=\sec x\tan x-\int\sec^3x\,dx+\int\sec x\,dx.$$
Solve The original integral has come back (as with $\int e^x\sin x\,dx$ in Calculus I). Move it to the left side and use Theorem 5.1: $$2\int\sec^3x\,dx=\sec x\tan x+\ln|\sec x+\tan x|\ \Longrightarrow\ \int\sec^3x\,dx=\frac12\sec x\tan x+\frac12\ln|\sec x+\tan x|+C.$$
Check Numerically, $\displaystyle\int_0^{\pi/4}\sec^3x\,dx=\tfrac12\sqrt2+\tfrac12\ln\left(1+\sqrt2\right)\approx1.1478$, which agrees with a midpoint sum using 1000 strips. This integral reappears in 5.2 and in arc-length problems.
⚠️ Common mistakes
  • Power rule on a trig function. $\displaystyle\int\sin^2x\,dx\ne\frac{\sin^3x}{3}+C$: the derivative of $\tfrac13\sin^3x$ is $\sin^2x\cos x$, not $\sin^2x$. The missing $\cos x$ is exactly the "saved factor" — without it, use half-angle identities.
  • Losing the minus sign in $du=-\sin x\,dx$ when $u=\cos x$.
  • Forgetting the double angle. $\displaystyle\int\cos2x\,dx=\tfrac12\sin2x+C$, not $\sin2x+C$.
  • Dropping absolute values. $\ln|\sec x+\tan x|$ needs the bars: $\sec x+\tan x$ is negative on parts of its domain.

Reduction Formulas

For large powers, applying the strategy by hand gets long. Integration by parts gives a formula that lowers the power by two at each step — a reduction formula. Applied repeatedly, it reduces any power to one of $\int dx$, $\int\sin x\,dx$, $\int\cos x\,dx$, $\int\sec x\,dx$ or $\int\tan x\,dx$.

📐 Theorem 5.2 — Reduction formulas
For every integer $n\ge2$, $$\int\sin^nx\,dx=-\frac1n\sin^{n-1}x\cos x+\frac{n-1}{n}\int\sin^{n-2}x\,dx,$$ $$\int\cos^nx\,dx=\frac1n\cos^{n-1}x\sin x+\frac{n-1}{n}\int\cos^{n-2}x\,dx,$$ and, on any interval on which $\cos x\ne0$, $$\int\sec^nx\,dx=\frac{1}{n-1}\sec^{n-2}x\tan x+\frac{n-2}{n-1}\int\sec^{n-2}x\,dx,\qquad\int\tan^nx\,dx=\frac{\tan^{n-1}x}{n-1}-\int\tan^{n-2}x\,dx.$$

Proof of the first. Take $u=\sin^{n-1}x$, $dv=\sin x\,dx$, so $du=(n-1)\sin^{n-2}x\cos x\,dx$ and $v=-\cos x$. Then $$\int\sin^nx\,dx=-\sin^{n-1}x\cos x+(n-1)\int\sin^{n-2}x\cos^2x\,dx.$$ Write $\cos^2x=1-\sin^2x$: the right side becomes $-\sin^{n-1}x\cos x+(n-1)\int\sin^{n-2}x\,dx-(n-1)\int\sin^nx\,dx$. Add $(n-1)\int\sin^nx\,dx$ to both sides and divide by $n$. The others are proved the same way (Example 4 is the case $n=3$ for secant).

✏️ Example 5 — A definite integral by reduction
Evaluate $\displaystyle\int_0^{\pi/2}\sin^5x\,dx$ using the reduction formula, then confirm it with the odd-power method.
Boundary term Write $I_n=\displaystyle\int_0^{\pi/2}\sin^nx\,dx$. The term $-\tfrac1n\sin^{n-1}x\cos x$ is $0$ at both $x=0$ (since $\sin0=0$) and $x=\tfrac\pi2$ (since $\cos\tfrac\pi2=0$). So for $n\ge2$ $$I_n=\frac{n-1}{n}I_{n-2}.$$
Reduce $I_1=\displaystyle\int_0^{\pi/2}\sin x\,dx=\Big[-\cos x\Big]_0^{\pi/2}=1$, so $$I_5=\frac45I_3=\frac45\cdot\frac23I_1=\frac{8}{15}\approx0.5333.$$
Check Odd power of sine: $\sin^5x=\left(1-\cos^2x\right)^2\sin x$, and with $u=\cos x$ ($u$ runs from $1$ to $0$): $$\int_1^0-\left(1-u^2\right)^2du=\int_0^1\left(1-2u^2+u^4\right)du=1-\frac23+\frac15=\frac8{15}.\ ✓$$
Pattern The same idea gives $\displaystyle\int_0^{\pi/2}\sin^nx\,dx=\frac{n-1}{n}\cdot\frac{n-3}{n-2}\cdots$, ending in $1$ (odd $n$) or $\tfrac\pi2$ (even $n$) — the formulas of John Wallis (1656), which also produce a famous infinite product for $\pi$.

Products $\sin(mx)\cos(nx)$ and the Product-to-Sum Identities

When two different frequencies are multiplied — as in sound synthesis, radio mixing or Fourier analysis — neither factor is the derivative of the other. Instead we turn the product into a sum of single sines or cosines, each of which integrates in one line.

📐 Theorem 5.3 — Product-to-sum identities
For all real $A$ and $B$, $$\sin A\cos B=\tfrac12\big[\sin(A-B)+\sin(A+B)\big],$$ $$\sin A\sin B=\tfrac12\big[\cos(A-B)-\cos(A+B)\big],$$ $$\cos A\cos B=\tfrac12\big[\cos(A-B)+\cos(A+B)\big].$$

Each follows by adding or subtracting the addition formulas $\sin(A\pm B)=\sin A\cos B\pm\cos A\sin B$ and $\cos(A\pm B)=\cos A\cos B\mp\sin A\sin B$.

✏️ Example 6 — A product of two frequencies
(a) Evaluate $\displaystyle\int\sin5x\cos2x\,dx$. (b) Show that $\displaystyle\int_{-\pi}^{\pi}\sin3x\sin5x\,dx=0$.
(a) Rewrite With $A=5x$, $B=2x$: $\sin5x\cos2x=\tfrac12\left(\sin3x+\sin7x\right)$.
(a) Integrate $$\int\sin5x\cos2x\,dx=\frac12\left(-\frac{\cos3x}{3}-\frac{\cos7x}{7}\right)+C=-\frac{\cos3x}{6}-\frac{\cos7x}{14}+C.$$
(b) $\sin3x\sin5x=\tfrac12\left[\cos(-2x)-\cos8x\right]=\tfrac12\left(\cos2x-\cos8x\right)$, so $$\int_{-\pi}^{\pi}\sin3x\sin5x\,dx=\frac12\left[\frac{\sin2x}{2}-\frac{\sin8x}{8}\right]_{-\pi}^{\pi}=0,$$ because $\sin(k\pi)=0$ for every integer $k$.
Meaning For positive integers $m\ne n$ the same calculation gives $\int_{-\pi}^{\pi}\sin mx\sin nx\,dx=0$, while $\int_{-\pi}^{\pi}\sin^2nx\,dx=\pi$. This orthogonality is what allows an engineer to pull a single frequency out of a complicated signal — the starting point of Fourier series.
Interactive 2 — Product-to-Sum Visualiser

The thick curve is the product. The two dashed waves are the halves of the product-to-sum identity, one at the difference frequency $|m-n|$ and one at the sum frequency $m+n$. Watch the tracer: at every $x$ the two dashed heights add up to the product. The shaded signed area builds $\int_{-\pi}^{x}$; over the full period it is $0$ unless the difference wave is a constant.

32
$x$—
product at $x$—
sum of the two halves—
$\int_{-\pi}^{x}$ product—
$\int_{-\pi}^{\pi}$ product—

Exercises

I
Powers of sine and cosine. Evaluate each integral.
  • 1 $\displaystyle\int\sin^3x\,dx$
  • 2 $\displaystyle\int\cos^3x\,dx$
  • 3 $\displaystyle\int\sin^2x\cos^3x\,dx$
  • 4 $\displaystyle\int\sin^5x\cos^2x\,dx$
  • 5 $\displaystyle\int\cos^2x\,dx$
  • 6 $\displaystyle\int\sin^2(3x)\,dx$
  • 7 $\displaystyle\int\sin^2x\cos^2x\,dx$
  • 8 $\displaystyle\int\cos^4x\,dx$
  • 9 $\displaystyle\int_0^{\pi/2}\sin^2x\cos x\,dx$
  • 10 $\displaystyle\int_0^{\pi}\sin^3x\,dx$
  • 11 $\displaystyle\int\frac{\sin^3x}{\sqrt{\cos x}}\,dx$
Check: the derivative of the answer (right panel) must retrace the integrand
$F$ (the answer) with its tangent
$F'$ being traced — the integrand
$x$—
$F(x)$—
$F'(x)$—
$F$ is…—
1 Odd sine power: $\sin^3x=\left(1-\cos^2x\right)\sin x$; $u=\cos x$: $-\displaystyle\int\left(1-u^2\right)du$, so $$\int\sin^3x\,dx=-\cos x+\frac{\cos^3x}{3}+C.$$
2 Odd cosine power: $\cos^3x=\left(1-\sin^2x\right)\cos x$; $u=\sin x$: $$\int\cos^3x\,dx=\sin x-\frac{\sin^3x}{3}+C.$$
3 Save $\cos x$: $\sin^2x\left(1-\sin^2x\right)\cos x$; $u=\sin x$: $\displaystyle\int\left(u^2-u^4\right)du$, so $$\int\sin^2x\cos^3x\,dx=\frac{\sin^3x}{3}-\frac{\sin^5x}{5}+C.$$
4 Save $\sin x$: $\left(1-\cos^2x\right)^2\cos^2x\sin x$; $u=\cos x$: $-\displaystyle\int\left(u^2-2u^4+u^6\right)du$, so $$\int\sin^5x\cos^2x\,dx=-\frac{\cos^3x}{3}+\frac{2\cos^5x}{5}-\frac{\cos^7x}{7}+C.$$
5 Half-angle: $\cos^2x=\tfrac12(1+\cos2x)$, so $$\int\cos^2x\,dx=\frac x2+\frac{\sin2x}{4}+C.$$
6 $\sin^23x=\tfrac12(1-\cos6x)$, so $$\int\sin^2(3x)\,dx=\frac x2-\frac{\sin6x}{12}+C.$$
7 $\sin x\cos x=\tfrac12\sin2x$, so $\sin^2x\cos^2x=\tfrac14\sin^22x=\tfrac18(1-\cos4x)$, and $$\int\sin^2x\cos^2x\,dx=\frac x8-\frac{\sin4x}{32}+C.$$
8 $\cos^4x=\tfrac14(1+\cos2x)^2=\tfrac14\left(1+2\cos2x+\tfrac12(1+\cos4x)\right)=\tfrac38+\tfrac12\cos2x+\tfrac18\cos4x$, so $$\int\cos^4x\,dx=\frac{3x}{8}+\frac{\sin2x}{4}+\frac{\sin4x}{32}+C.$$
9 $u=\sin x$ runs from $0$ to $1$: $\displaystyle\int_0^1u^2\,du=\frac13$.
10 By part 1: $\Big[-\cos x+\tfrac13\cos^3x\Big]_0^{\pi}=\left(1-\tfrac13\right)-\left(-1+\tfrac13\right)=\dfrac43\approx1.3333$.
11 Odd sine power, on an interval where $\cos x\gt0$: $u=\cos x$ gives $-\displaystyle\int\left(1-u^2\right)u^{-1/2}du=-\int\left(u^{-1/2}-u^{3/2}\right)du$, so $$\int\frac{\sin^3x}{\sqrt{\cos x}}\,dx=-2\sqrt{\cos x}+\frac25\cos^{5/2}x+C.$$
II
Powers of tangent and secant. Evaluate each integral (assume $\cos x\ne0$ on the interval used).
  • 1 $\displaystyle\int\tan^2x\,dx$
  • 2 $\displaystyle\int\sec^4x\,dx$
  • 3 $\displaystyle\int\tan x\sec^3x\,dx$
  • 4 $\displaystyle\int\tan^3x\sec x\,dx$
  • 5 $\displaystyle\int\tan^2x\sec^4x\,dx$
  • 6 $\displaystyle\int\tan^5x\sec^3x\,dx$
  • 7 $\displaystyle\int\tan^3x\,dx$
  • 8 $\displaystyle\int_0^{\pi/4}\sec^4x\,dx$
  • 9 $\displaystyle\int_0^{\pi/3}\sec x\,dx$
  • 10 $\displaystyle\int\tan^2x\sec x\,dx$
Check: the derivative of the answer (right panel) must retrace the integrand
$F$ (the answer) with its tangent
$F'$ being traced — the integrand
$x$—
$F(x)$—
$F'(x)$—
$F$ is…—
1 $\tan^2x=\sec^2x-1$, so $$\int\tan^2x\,dx=\tan x-x+C.$$
2 Even secant power: $\sec^4x=\left(1+\tan^2x\right)\sec^2x$; $u=\tan x$: $$\int\sec^4x\,dx=\tan x+\frac{\tan^3x}{3}+C.$$
3 Odd tangent power: save $\sec x\tan x$, leaving $\sec^2x$; $u=\sec x$: $\displaystyle\int u^2\,du$, so $$\int\tan x\sec^3x\,dx=\frac{\sec^3x}{3}+C.$$
4 Save $\sec x\tan x$: $\tan^2x=\sec^2x-1$; $u=\sec x$: $\displaystyle\int\left(u^2-1\right)du$, so $$\int\tan^3x\sec x\,dx=\frac{\sec^3x}{3}-\sec x+C.$$
5 Save $\sec^2x$: $\tan^2x\left(1+\tan^2x\right)\sec^2x$; $u=\tan x$: $$\int\tan^2x\sec^4x\,dx=\frac{\tan^3x}{3}+\frac{\tan^5x}{5}+C.$$
6 Save $\sec x\tan x$: $\tan^4x\sec^2x=\left(\sec^2x-1\right)^2\sec^2x$; $u=\sec x$: $\displaystyle\int\left(u^6-2u^4+u^2\right)du$, so $$\int\tan^5x\sec^3x\,dx=\frac{\sec^7x}{7}-\frac{2\sec^5x}{5}+\frac{\sec^3x}{3}+C.$$
7 $\tan^3x=\tan x\left(\sec^2x-1\right)$: $\displaystyle\int\tan x\sec^2x\,dx-\int\tan x\,dx$, so $$\int\tan^3x\,dx=\frac{\tan^2x}{2}-\ln|\sec x|+C.$$
8 By part 2: $\Big[\tan x+\tfrac13\tan^3x\Big]_0^{\pi/4}=1+\tfrac13=\dfrac43$.
9 Theorem 5.1: $\Big[\ln|\sec x+\tan x|\Big]_0^{\pi/3}=\ln\left(2+\sqrt3\right)-\ln1=\ln\left(2+\sqrt3\right)\approx1.3170$.
10 Tangent power even, secant power odd: $\tan^2x\sec x=\sec^3x-\sec x$. With Example 4 and Theorem 5.1, $$\int\tan^2x\sec x\,dx=\frac12\sec x\tan x+\frac12\ln|\sec x+\tan x|-\ln|\sec x+\tan x|=\frac12\sec x\tan x-\frac12\ln|\sec x+\tan x|+C.$$
III
Product-to-sum identities and reduction formulas.
  • 1 $\displaystyle\int\sin5x\cos3x\,dx$
  • 2 $\displaystyle\int\sin3x\sin x\,dx$
  • 3 $\displaystyle\int\cos4x\cos2x\,dx$
  • 4 $\displaystyle\int_0^{\pi}\sin2x\cos3x\,dx$
  • 5 $\displaystyle\int_{-\pi}^{\pi}\sin4x\sin6x\,dx$
  • 6 $\displaystyle\int_{-\pi}^{\pi}\sin^2(4x)\,dx$
  • 7 Use Theorem 5.2 to evaluate $\displaystyle\int_0^{\pi/2}\cos^6x\,dx$.
  • 8 Show that $\displaystyle\int_0^{\pi/2}\sin^nx\,dx=\frac{n-1}{n}\int_0^{\pi/2}\sin^{n-2}x\,dx$ for $n\ge2$, and use it to evaluate $\displaystyle\int_0^{\pi/2}\sin^7x\,dx$.
1 $\sin5x\cos3x=\tfrac12(\sin2x+\sin8x)$, so $$\int\sin5x\cos3x\,dx=-\frac{\cos2x}{4}-\frac{\cos8x}{16}+C.$$
2 $\sin3x\sin x=\tfrac12(\cos2x-\cos4x)$, so $$\int\sin3x\sin x\,dx=\frac{\sin2x}{4}-\frac{\sin4x}{8}+C.$$
3 $\cos4x\cos2x=\tfrac12(\cos2x+\cos6x)$, so $$\int\cos4x\cos2x\,dx=\frac{\sin2x}{4}+\frac{\sin6x}{12}+C.$$
4 $\sin2x\cos3x=\tfrac12\left[\sin(-x)+\sin5x\right]=\tfrac12(\sin5x-\sin x)$. Then $\displaystyle\int_0^{\pi}\sin5x\,dx=\tfrac25$ and $\displaystyle\int_0^{\pi}\sin x\,dx=2$, so the integral is $\tfrac12\left(\tfrac25-2\right)=-\dfrac45$.
5 $\sin4x\sin6x=\tfrac12(\cos2x-\cos10x)$; both sine antiderivatives vanish at $\pm\pi$, so the integral is $0$ (orthogonality, $m\ne n$).
6 $\sin^24x=\tfrac12(1-\cos8x)$: $\Big[\tfrac x2-\tfrac{\sin8x}{16}\Big]_{-\pi}^{\pi}=\pi$.
7 The boundary term $\tfrac1n\cos^{n-1}x\sin x$ vanishes at $0$ and $\tfrac\pi2$, so $J_n=\tfrac{n-1}{n}J_{n-2}$ with $J_0=\tfrac\pi2$: $$J_6=\frac56\cdot\frac34\cdot\frac12\cdot\frac\pi2=\frac{5\pi}{32}\approx0.4909.$$
8 Apply Theorem 5.2 between $0$ and $\tfrac\pi2$; the boundary term $-\tfrac1n\sin^{n-1}x\cos x$ is $0$ at both ends, which gives the relation. Then $$\int_0^{\pi/2}\sin^7x\,dx=\frac67\cdot\frac45\cdot\frac23\cdot1=\frac{16}{35}\approx0.4571.$$
IV
Applications and understanding.
  • 1 Applied (electrical engineering). The mains voltage in the UK and much of Europe and Africa is $V(t)=325\sin(100\pi t)$ volts, with period $T=\tfrac1{50}$ s. The root-mean-square voltage is $V_{\text{rms}}=\sqrt{\dfrac1T\displaystyle\int_0^TV(t)^2\,dt}$. Find it.
  • 2 Find the area of the region between $y=\sin x$ and $y=\sin^3x$ for $0\le x\le\pi$.
  • 3 Applied (cartography). On a Mercator map drawn from a globe of radius $R=10$ cm, the parallel of latitude $\varphi$ is drawn at height $y=R\displaystyle\int_0^{\varphi}\sec t\,dt$ above the equator. How high are the parallels at $45^\circ$ and $60^\circ$?
  • 4 Show that $\displaystyle\int_0^{2\pi}\sin mx\cos nx\,dx=0$ for all positive integers $m$ and $n$ (consider $m=n$ separately).
  • 5 Conceptual. A student writes $\displaystyle\int\cos^2x\,dx=\frac{\cos^3x}{3}+C$. Explain, by differentiating, why this is wrong, and give the correct answer.
1 $V^2=325^2\sin^2(100\pi t)=\tfrac{325^2}{2}\left(1-\cos200\pi t\right)$. Over one period the cosine integrates to $0$, so $\dfrac1T\displaystyle\int_0^TV^2\,dt=\dfrac{325^2}{2}$ and $V_{\text{rms}}=\dfrac{325}{\sqrt2}\approx229.8$ V — the familiar "230 V" supply.
2 On $[0,\pi]$, $0\le\sin x\le1$, so $\sin x\ge\sin^3x$. Area $=\displaystyle\int_0^{\pi}\left(\sin x-\sin^3x\right)dx=\int_0^{\pi}\sin x\cos^2x\,dx=\Big[-\tfrac13\cos^3x\Big]_0^{\pi}=\dfrac23$ square units.
3 By Theorem 5.1, $y=10\ln(\sec\varphi+\tan\varphi)$ cm. At $45^\circ$: $10\ln\left(\sqrt2+1\right)\approx8.81$ cm. At $60^\circ$: $10\ln\left(2+\sqrt3\right)\approx13.17$ cm. (Equal $15^\circ$ steps in latitude are drawn further and further apart.)
4 If $m\ne n$: $\sin mx\cos nx=\tfrac12\left[\sin(m-n)x+\sin(m+n)x\right]$, and $\int_0^{2\pi}\sin kx\,dx=\left[-\tfrac{\cos kx}{k}\right]_0^{2\pi}=0$ for every non-zero integer $k$. If $m=n$: $\sin nx\cos nx=\tfrac12\sin2nx$, which also integrates to $0$ over $[0,2\pi]$.
5 $\dfrac{d}{dx}\dfrac{\cos^3x}{3}=-\cos^2x\sin x\ne\cos^2x$ — the power rule needs the inner derivative, which is missing. Correct: $\displaystyle\int\cos^2x\,dx=\frac x2+\frac{\sin2x}{4}+C$ (Ex. I.5).
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. Why does Greenland look about as large as Africa on many wall maps — and exactly how much does the Mercator projection stretch high latitudes?

Why it matters. Mercator's 1569 projection is still used for marine navigation and by most web map services, because a straight line on it is a line of constant compass bearing. Its integral of $\sec\varphi$ shapes how billions of people picture the size of countries.

What to do.

  1. On a Mercator map built from a globe of radius $R$, the parallel at latitude $\varphi$ is drawn at height $y(\varphi)=R\int_0^{\varphi}\sec t\,dt$. Use Theorem 5.1 to write $y$ in closed form.
  2. With $R=1$, tabulate $y$ for $\varphi=15^\circ,30^\circ,45^\circ,60^\circ,75^\circ$ and compare with the equal-spacing values $R\varphi$ (in radians).
  3. The local scale factor at latitude $\varphi$ is $\sec\varphi$ in both directions, so areas are magnified by $\sec^2\varphi$. Compute this for the equator, for Nairobi ($1^\circ$S), Muscat ($23.6^\circ$N), London ($51.5^\circ$N) and central Greenland ($72^\circ$N).
  4. Real areas: Greenland about $2.17$ million km², Africa about $30.4$ million km². Explain, with your numbers, why they look similar on the map.

Tools: Calculator or spreadsheet; any online Mercator world map.

Hand in: A half-page note with your table, the magnification factors and a two-sentence explanation of the Greenland–Africa illusion.

Going further. Look up the Gall–Peters equal-area projection: there the height is proportional to $\sin\varphi$. What is gained and what is lost compared with Mercator?

Instructor note: $y=R\ln(\sec\varphi+\tan\varphi)$; for $R=1$, $y(60^\circ)\approx1.317$ against $\varphi\approx1.047$, and $\sec^2 72^\circ\approx10.47$, so Greenland is drawn about ten times too large relative to the equator while Africa (straddling the equator) is barely stretched.

🧪 Self-check quiz

0/5 correct

Q1. The best first move for $\displaystyle\int\sin^2x\cos^5x\,dx$ is to

Q2. $\displaystyle\int\sin^2x\,dx=$

Q3. $\displaystyle\int\sec x\,dx=$

Q4. $\displaystyle\int_0^{\pi/2}\sin^3x\,dx=$

Q5. $\sin4x\cos x=$

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