Integration Using Reduction Formulas: Steps, Examples

What Is A Reduction Formula In Integration?

A reduction formula in integration is a rule that expresses an integral containing a whole-number parameter $n$, usually a power, in terms of the same kind of integral with a smaller parameter, such as $n - 1$ or $n - 2$. Integration using reduction formulas applies that rule repeatedly until the integral reaches a case you can integrate directly. It is one of the methods of integration, built on top of integration by parts.

Write the family of integrals as $I_n$. A typical reduction formula has the shape

$$I_n = B_n(x) + c_n,I_{n-k}$$

• $I_n$ is the integral with parameter $n$, for example $I_n = \int \sin^n x,dx$.

• $B_n(x)$ is the boundary term, an ordinary function that comes out of integration by parts.

• $c_n$ is a number that depends on $n$, such as $\frac{n-1}{n}$.

• $k$ is the step size: $1$ for $x^n e^x$ and $(\ln x)^n$, and $2$ for powers of trig functions.

The formula is valid only for the values of $n$ where it was derived (often $n \geq 2$), and the chain stops at a base case such as $I_0$ or $I_1$. A reduction formula is a recurrence relation for integrals. The answer to $I_8$ is found by climbing down to $I_0$ and then substituting back up.

How Do You Derive A Reduction Formula?

Most reduction formulas come from one recipe based on integration by parts, $\int u,dv = uv - \int v,du$:

1. Split off one easy factor. Write the power as a lower power times a factor you can integrate, such as $\sin^n x = \sin^{n-1} x \cdot \sin x$.

2. Integrate by parts. Let $u$ be the lower power and $dv$ the easy factor. Differentiating $u$ lowers the power by one.

3. Use an identity to bring back the original integral. For trig powers, $\cos^2 x = 1 - \sin^2 x$ or $\tan^2 x = \sec^2 x - 1$ turns the new integral into a mix of $I_n$ and $I_{n-2}$.

4. Solve for $I_n$. The original integral appears on both sides. Collect it on the left and divide.

Step 3 is where the method earns its name. The integral you started with comes back, and instead of going round in circles you treat it as an unknown and solve for it, exactly as you would solve $I = 5 - 2I$ for $I$. The identities used here are the Pythagorean trigonometric identities.

What Is The Reduction Formula For Powers Of Sine?

For every integer $n \geq 2$:

$$\int \sin^n x,dx = -\frac{1}{n}\sin^{n-1} x \cos x + \frac{n-1}{n}\int \sin^{n-2} x,dx$$

This answers a question that turns up on Q and A sites almost verbatim: "Give reduction formula for $\int \sin^n x,dx$." Here is the derivation, following the recipe.

Split $\sin^n x = \sin^{n-1} x \cdot \sin x$ and integrate by parts with

$$u = \sin^{n-1} x, \qquad du = (n-1)\sin^{n-2} x \cos x,dx$$

$$dv = \sin x,dx, \qquad v = -\cos x$$

Then

$$I_n = -\sin^{n-1} x \cos x + (n-1)\int \sin^{n-2} x \cos^2 x,dx$$

Replace $\cos^2 x$ with $1 - \sin^2 x$:

$$I_n = -\sin^{n-1} x \cos x + (n-1)\int \sin^{n-2} x,dx - (n-1)\int \sin^{n} x,dx$$

$$I_n = -\sin^{n-1} x \cos x + (n-1)I_{n-2} - (n-1)I_n$$

Move $(n-1)I_n$ to the left, so the left side becomes $n I_n$, and divide by $n$:

$$I_n = -\frac{1}{n}\sin^{n-1} x \cos x + \frac{n-1}{n}I_{n-2}$$

The base cases are $I_0 = \int 1,dx = x + C$ and $I_1 = \int \sin x,dx = -\cos x + C$. Even powers end at $I_0$ and odd powers end at $I_1$. Actually, you can stop the chain earlier if you already know a lower integral, such as $I_2 = -\frac{1}{2}\sin x\cos x + \frac{x}{2} + C$.

Geometrically, the boundary term $-\frac{1}{n}\sin^{n-1} x \cos x$ is zero at $x = 0$ and at $x = \frac{\pi}{2}$. That single fact turns the formula into Wallis' formula further down. For odd powers there is a faster substitution route, $u = \cos x$, covered with the other integration of trigonometric functions techniques. The reduction formula is the method that works for every power, odd or even.

What Is The Reduction Formula For Powers Of Cosine?

For every integer $n \geq 2$:

$$\int \cos^n x,dx = \frac{1}{n}\cos^{n-1} x \sin x + \frac{n-1}{n}\int \cos^{n-2} x,dx$$

The derivation mirrors the sine case. Split $\cos^n x = \cos^{n-1} x \cdot \cos x$, take $u = \cos^{n-1} x$ and $dv = \cos x,dx$, so $v = \sin x$ and $du = -(n-1)\cos^{n-2} x \sin x,dx$. Integration by parts gives $\cos^{n-1} x \sin x + (n-1)\int \cos^{n-2} x \sin^2 x,dx$, and $\sin^2 x = 1 - \cos^2 x$ brings back $I_n$.

Only one sign differs from the sine formula, the sign of the boundary term. The derivative of $\sin x$ is $\cos x$ with no minus sign, so $v$ carries no minus sign either.

What Is The Reduction Formula For Polynomial Times Exponential Integrals?

For every integer $n \geq 1$ and constant $a \neq 0$:

$$\int x^n e^{ax},dx = \frac{1}{a}x^n e^{ax} - \frac{n}{a}\int x^{n-1} e^{ax},dx$$

Here there is no identity step. Take $u = x^n$, so $du = n x^{n-1},dx$, and $dv = e^{ax},dx$, so $v = \frac{1}{a}e^{ax}$. One round of integration by parts drops the power by exactly one, and the step size is $1$. The base case is $\int e^{ax},dx = \frac{1}{a}e^{ax} + C$, from the rules for the integration of exponential functions.

With $a = 1$ the formula is $I_n = x^n e^x - n I_{n-1}$. Each step multiplies by a falling whole number, so $\int x^3 e^x,dx$ picks up the coefficients $1$, $3$, $3 \cdot 2 = 6$ and $3 \cdot 2 \cdot 1 = 6$ with alternating signs. Example 3 below shows it.

What Is The Reduction Formula For Powers Of Tangent?

For every integer $n \geq 2$:

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

This one needs no integration by parts. Split off $\tan^2 x$ and use $\tan^2 x = \sec^2 x - 1$:

$$\int \tan^n x,dx = \int \tan^{n-2} x,(\sec^2 x - 1),dx$$

$$= \int \tan^{n-2} x \sec^2 x,dx - \int \tan^{n-2} x,dx$$

The first integral is a straight substitution: with $w = \tan x$, $dw = \sec^2 x,dx$, it becomes $\int w^{n-2},dw = \frac{w^{n-1}}{n-1}$. The base cases are $\int 1,dx = x + C$ and $\int \tan x,dx = \ln\lvert\sec x\rvert + C$.

Notice two things that differ from the sine formula: there is no fraction in front of the remaining integral, and it is subtracted. That minus sign is where Example 2 goes wrong.

What Is The Reduction Formula For Powers Of Secant?

For every integer $n \geq 2$:

$$\int \sec^n x,dx = \frac{\sec^{n-2} x \tan x}{n-1} + \frac{n-2}{n-1}\int \sec^{n-2} x,dx$$

Split $\sec^n x = \sec^{n-2} x \cdot \sec^2 x$. Take $u = \sec^{n-2} x$, so $du = (n-2)\sec^{n-2} x \tan x,dx$, and $dv = \sec^2 x,dx$, so $v = \tan x$. By parts:

$$I_n = \sec^{n-2} x \tan x - (n-2)\int \sec^{n-2} x \tan^2 x,dx$$

Replace $\tan^2 x$ with $\sec^2 x - 1$:

$$I_n = \sec^{n-2} x \tan x - (n-2)I_n + (n-2)I_{n-2}$$

$$(n-1)I_n = \sec^{n-2} x \tan x + (n-2)I_{n-2}$$

Divide by $n - 1$. The base cases are $\int \sec^2 x,dx = \tan x + C$ and $\int \sec x,dx = \ln\lvert\sec x + \tan x\rvert + C$. The famous $\int \sec^3 x,dx$ in Example 4 is one step above that second base case.

What Is The Reduction Formula For Powers Of The Natural Logarithm?

For every integer $n \geq 1$ and $x > 0$:

$$\int (\ln x)^n,dx = x(\ln x)^n - n\int (\ln x)^{n-1},dx$$

Take $u = (\ln x)^n$, so $du = \frac{n(\ln x)^{n-1}}{x},dx$, and $dv = dx$, so $v = x$. The $x$ from $v$ cancels the $\frac{1}{x}$ from $du$:

$$\int (\ln x)^n,dx = x(\ln x)^n - \int x \cdot \frac{n(\ln x)^{n-1}}{x},dx$$

$$= x(\ln x)^n - n\int (\ln x)^{n-1},dx$$

The base case is $\int 1,dx = x + C$, so even $\int \ln x,dx = x\ln x - x + C$ is just the $n = 1$ step. The condition $x > 0$ comes from the natural logarithm itself. A wider version handles $x^m(\ln x)^n$ and appears in the FAQ.

How Does Wallis' Formula Evaluate Powers Of Sine From 0 To Pi Over 2?

Wallis' formula evaluates the definite integral of $\sin^n x$ from $0$ to $\frac{\pi}{2}$ as a product of simple fractions:

$$W_n = \int_{0}^{\pi/2} \sin^n x,dx = \frac{n-1}{n},W_{n-2}, \qquad W_0 = \frac{\pi}{2}, \quad W_1 = 1$$

A Chegg-style exercise asks students to "use the reduction formula to show that" exactly this relation holds. Apply the sine reduction formula as a definite integral on $\left[0, \frac{\pi}{2}\right]$. The boundary term $-\frac{1}{n}\sin^{n-1} x \cos x$ is $0$ at $x = \frac{\pi}{2}$ (because $\cos \frac{\pi}{2} = 0$) and $0$ at $x = 0$ (because $\sin 0 = 0$ and $n - 1 \geq 1$). Only the multiplier survives.

Unrolling the recurrence gives two closed forms:

$$W_{2m} = \frac{1 \cdot 3 \cdot 5 \cdots (2m-1)}{2 \cdot 4 \cdot 6 \cdots (2m)} \cdot \frac{\pi}{2}$$

$$W_{2m+1} = \frac{2 \cdot 4 \cdot 6 \cdots (2m)}{3 \cdot 5 \cdot 7 \cdots (2m+1)}$$

Even powers carry a $\pi$; odd powers are rational. The same table works for $\cos^n x$, because the substitution $x \to \frac{\pi}{2} - x$ swaps sine and cosine without changing the interval, one of the properties of definite integrals.

The table also hides a famous result. On $\left[0, \frac{\pi}{2}\right]$, $0 \leq \sin x \leq 1$, so higher powers are smaller: $W_{2m+1} \leq W_{2m} \leq W_{2m-1}$. Squeezing the ratio $\frac{W_{2m}}{W_{2m+1}}$ between $1$ and $\frac{2m+1}{2m}$ forces it to $1$, which rearranges into the Wallis product:

$$\frac{\pi}{2} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdots$$

Which Reduction Formulas Should You Know?

The six derived above cover most course work. Reference lists add several more, all built with the same recipe. Every row below was checked by differentiating the right-hand side.

The last row is worth checking yourself. Some printed lists show $\sqrt{1+x^2}$ in that denominator. Differentiating shows it must be $1 + x^2$, because the derivative of $\arctan x$ is $\frac{1}{1+x^2}$. The habit of differentiating a formula before trusting it catches errors like this one.

Examples Of Integration Using Reduction Formulas

Example 1

Find $\int \sin^4 x,dx$.

This is a real Quora question: "What is the integration of $\sin^4 x,dx$ by reduction formula?" Use the sine formula with $n = 4$, then $n = 2$:

$$I_4 = -\frac{1}{4}\sin^3 x\cos x + \frac{3}{4}I_2$$

$$I_2 = -\frac{1}{2}\sin x\cos x + \frac{1}{2}I_0, \qquad I_0 = x$$

$$I_2 = -\frac{1}{2}\sin x\cos x + \frac{1}{2}x$$

Substitute back:

$$I_4 = -\frac{1}{4}\sin^3 x\cos x + \frac{3}{4}\left(-\frac{1}{2}\sin x\cos x + \frac{1}{2}x\right) + C$$

$$= -\frac{1}{4}\sin^3 x\cos x - \frac{3}{8}\sin x\cos x + \frac{3}{8}x + C$$

Final answer: $-\frac{1}{4}\sin^3 x\cos x - \frac{3}{8}\sin x\cos x + \frac{3}{8}x + C$

Example 2

Find $\int \tan^4 x,dx$.

A natural first attempt reuses the shape of the sine formula and adds the lower integral:

$$I_4 = \frac{\tan^3 x}{3} + I_2, \qquad I_2 = \tan x + I_0 = \tan x + x$$

$$\text{Attempt: } \frac{\tan^3 x}{3} + \tan x + x$$

Check it by differentiating. The derivative is $\tan^2 x\sec^2 x + \sec^2 x + 1$, and simplifying gives $\tan^4 x + 2\sec^2 x$. That is not $\tan^4 x$, so the attempt is wrong. The extra $2\sec^2 x$ comes from adding the lower integral where the formula subtracts it.

The correct formula is $I_n = \frac{\tan^{n-1} x}{n-1} - I_{n-2}$:

$$I_4 = \frac{\tan^3 x}{3} - I_2$$

$$I_2 = \tan x - I_0 = \tan x - x$$

$$I_4 = \frac{\tan^3 x}{3} - (\tan x - x) + C$$

$$= \frac{\tan^3 x}{3} - \tan x + x + C$$

Check: differentiating gives $\tan^2 x\sec^2 x - \sec^2 x + 1 = \tan^2 x(\tan^2 x + 1) - \tan^2 x = \tan^4 x$.

Final answer: $\frac{\tan^3 x}{3} - \tan x + x + C$

Example 3

Find $\int x^3 e^x,dx$.

Use $I_n = x^n e^x - n I_{n-1}$ three times, with $I_0 = e^x$:

$$I_3 = x^3 e^x - 3I_2$$

$$I_2 = x^2 e^x - 2I_1$$

$$I_1 = x e^x - I_0 = x e^x - e^x$$

$$I_2 = x^2 e^x - 2x e^x + 2e^x$$

$$I_3 = x^3 e^x - 3x^2 e^x + 6x e^x - 6e^x + C$$

Final answer: $e^x\left(x^3 - 3x^2 + 6x - 6\right) + C$

Example 4

Find $\int \sec^3 x,dx$.

Use the secant formula with $n = 3$:

$$I_3 = \frac{\sec x\tan x}{2} + \frac{1}{2}\int \sec x,dx$$

$$= \frac{1}{2}\sec x\tan x + \frac{1}{2}\ln\lvert\sec x + \tan x\rvert + C$$

This integral appears in arc-length problems, such as the length of a parabola, so it is worth knowing by heart.

Final answer: $\frac{1}{2}\sec x\tan x + \frac{1}{2}\ln\lvert\sec x + \tan x\rvert + C$

Example 5

Find $\int (\ln x)^3,dx$ for $x > 0$.

Use $I_n = x(\ln x)^n - n I_{n-1}$ with $I_0 = x$:

$$I_3 = x(\ln x)^3 - 3I_2$$

$$I_2 = x(\ln x)^2 - 2I_1$$

$$I_1 = x\ln x - I_0 = x\ln x - x$$

$$I_2 = x(\ln x)^2 - 2x\ln x + 2x$$

$$I_3 = x(\ln x)^3 - 3x(\ln x)^2 + 6x\ln x - 6x + C$$

The coefficients $1, 3, 6, 6$ are the same ones as in Example 3. Both formulas multiply by a falling whole number at each step.

Final answer: $x(\ln x)^3 - 3x(\ln x)^2 + 6x\ln x - 6x + C$

Example 6

Evaluate $\int_{0}^{\pi/2} \sin^6 x,dx$ and $\int_{0}^{\pi/2} \cos^7 x,dx$.

Even power, so the chain ends at $W_0 = \frac{\pi}{2}$:

$$W_6 = \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2}$$

$$= \frac{15}{48} \cdot \frac{\pi}{2} = \frac{5\pi}{32}$$

Odd power, so the chain ends at $W_1 = 1$, and cosine uses the same values as sine on this interval:

$$\int_{0}^{\pi/2} \cos^7 x,dx = \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3} \cdot 1$$

$$= \frac{48}{105} = \frac{16}{35}$$

Final answer: $\frac{5\pi}{32} \approx 0.4909$ and $\frac{16}{35} \approx 0.4571$

Why Do Reduction Formulas Exist?

Reduction formulas exist because repeating integration by parts by hand does not scale, and one general rule does. Integrating $x^{10} e^x$ directly means ten rounds of integration by parts, each with its own chance of a dropped sign. The reduction formula $I_n = x^n e^x - n I_{n-1}$ does all ten rounds at once, as a pattern.

• Exact values that built mathematics. John Wallis worked with these integrals in his Arithmetica infinitorum of 1656 and reached his infinite product for $\pi$, the same product derived above. The Wallis biography at MacTutor quotes it as $\frac{1}{2}\pi = \frac{2 \cdot 2 \cdot 4 \cdot 4 \cdots}{1 \cdot 3 \cdot 3 \cdot 5 \cdots}$.

• Tables and software. Large integral tables list recurrences instead of thousands of separate answers, and computer algebra systems use the same recursions internally.

• Proof by induction. Because a reduction formula links $I_n$ to $I_{n-1}$ or $I_{n-2}$, results about every $n$ (such as the Wallis closed forms) follow by induction.

Where Are Reduction Formulas Used?

Reduction formulas turn up wherever powers of a function have to be integrated exactly.

• Electrical engineering. The average of $\sin^2 x$ over a quarter cycle is $\frac{W_2}{\pi/2} = \frac{\pi/4}{\pi/2} = \frac{1}{2}$. That one-half is why the effective (RMS) value of an alternating current is its peak divided by $\sqrt{2}$.

• Probability and statistics. Integrals of the form $\int_{0}^{1} x^m (1-x)^n,dx$ appear throughout the Beta distribution. A reduction formula evaluates them exactly; for instance, $\int_{0}^{1} x^2(1-x)^3,dx = \frac{1}{60}$.

• The Gamma function. The improper integral $\int_{0}^{\infty} x^n e^{-x},dx$ obeys the reduction $I_n = n I_{n-1}$ (the boundary term vanishes at both ends), so $I_n$ equals $n$ factorial. The NIST Digital Library of Mathematical Functions lists this recurrence as a basic property of the Gamma function.

• Fourier series and physics. Products of powers of sine and cosine are integrated term by term when signals are broken into frequencies, and Wallis integrals reappear in the volume of a ball in higher dimensions.

Why Do Students Get Reduction Formulas Wrong?

Most errors with reduction formulas are bookkeeping errors, not calculus errors. The formula is right; the steps around it slip. These four account for most lost marks.

Mistake 1: Stepping Down By The Wrong Amount

Where it slips in: Trig powers, where the formula links $I_n$ to $I_{n-2}$.

Don't do this: Write $I_5$ in terms of $I_4$, or keep stepping an odd power down until you reach $I_0$.

The correct way: Trig formulas drop by $2$, so even powers end at $I_0$ and odd powers end at $I_1$. Write the whole chain of indices first, for example $5 \to 3 \to 1$, before substituting anything.

Mistake 2: Losing The Multipliers When Substituting Back

Where it slips in: The back-substitution in Examples 1, 3 and 5.

Don't do this: Replace $I_2$ inside $I_4 = -\frac{1}{4}\sin^3 x\cos x + \frac{3}{4}I_2$ but forget to multiply every term of $I_2$ by $\frac{3}{4}$.

The correct way: Put brackets around each substituted integral, then expand. Each lower term carries the product of all the multipliers above it, which is why $\frac{3}{8} = \frac{3}{4} \cdot \frac{1}{2}$ sits on the $x$ term in Example 1.

The rusher loses marks here. The derivation goes smoothly, and then the final line drops a fraction. Students who write the chain of multipliers in a margin before expanding stop making this mistake almost at once, because the products are visible before the algebra starts.

Mistake 3: Using The Formula Below Its Valid Range

Where it slips in: Applying the tangent or secant formula with $n = 1$.

Don't do this: Put $n = 1$ into $\frac{\tan^{n-1} x}{n-1}$, which divides by zero.

The correct way: Treat $n = 0$ and $n = 1$ as base cases and integrate them directly: $\int \tan x,dx = \ln\lvert\sec x\rvert + C$ and $\int \sec x,dx = \ln\lvert\sec x + \tan x\rvert + C$.

The memorizer is most exposed here. They can recite the formula but not where it came from, so they cannot tell when it stops applying. (This was the point that took me longest to see: the $n - 1$ in the denominator is not decoration; it comes from dividing both sides in the derivation, and dividing by zero is exactly what $n = 1$ would ask.)

Mistake 4: Trusting A Recurrence Blindly On A Calculator

Where it slips in: Computing definite values such as $I_n = \int_{0}^{1} x^n e^{x-1},dx$ with the exact recurrence $I_n = 1 - n I_{n-1}$, starting from $I_0 = 1 - e^{-1}$.

Don't do this: Run the recurrence forward in rounded decimals all the way to $n = 20$. In standard double precision, $I_{10}$ still comes out as $0.0839$, but $I_{20}$ comes out as about $-30.19$, which is impossible because the integrand is never negative.

The correct way: The true value is $I_{20} \approx 0.0455$. Each step multiplies the rounding error in $I_0$ by $n$, so the error grows like $n$ factorial. Use the recurrence exactly (with fractions and $e$), or run it backwards from a large $n$, where the error shrinks instead of growing.

The real-world version of Mistake 4 cost lives. On 25 February 1991, a Patriot missile battery at Dhahran, Saudi Arabia, failed to intercept an incoming missile, and 28 American soldiers were killed. The US General Accounting Office report, Patriot Missile Defense: Software Problem Led to System Failure at Dhahran, traced the failure to a tiny truncation in the system clock that grew with every hour of operation. After about 100 hours it amounted to roughly a third of a second, which was enough to point the radar at the wrong part of the sky.

A small error repeated many times is not small, whether it sits in a missile clock or in a recurrence run forward on a calculator.

Practice Problems On Integration Using Reduction Formulas

Work each one before reading the answer. Every answer was checked by differentiation or by direct evaluation.

1. Find $\int \cos^4 x,dx$. Answer to Question 1: $\frac{1}{4}\cos^3 x\sin x + \frac{3}{8}\cos x\sin x + \frac{3}{8}x + C$

2. Find $\int x^2 e^{3x},dx$. Answer to Question 2: $e^{3x}\left(\frac{x^2}{3} - \frac{2x}{9} + \frac{2}{27}\right) + C$

3. Find $\int \tan^5 x,dx$. Answer to Question 3: $\frac{\tan^4 x}{4} - \frac{\tan^2 x}{2} + \ln\lvert\sec x\rvert + C$

4. Evaluate $\int_{0}^{\pi/2} \sin^5 x,dx$. Answer to Question 4: $\frac{4}{5} \cdot \frac{2}{3} \cdot 1 = \frac{8}{15}$

5. Evaluate $\int_{0}^{\pi/2} \cos^8 x,dx$. Answer to Question 5: $\frac{7}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{35\pi}{256}$

6. Evaluate $\int_{1}^{e} (\ln x)^2,dx$. Answer to Question 6: $\big[x(\ln x)^2 - 2x\ln x + 2x\big]_{1}^{e} = e - 2$

What To Remember About Reduction Formulas

• Integration using reduction formulas expresses $I_n$ in terms of $I_{n-1}$ or $I_{n-2}$ and walks down to a base case.

• Almost every reduction formula comes from one recipe: split, integrate by parts, use an identity, solve for $I_n$.

• Trig powers step down by $2$, while $x^n e^x$ and $(\ln x)^n$ step down by $1$.

• On $\left[0, \frac{\pi}{2}\right]$ the boundary term vanishes, giving Wallis' formula $W_n = \frac{n-1}{n}W_{n-2}$.

• The costly mistakes are a wrong sign in the tangent formula, dropped multipliers when substituting back, and using a formula below its valid $n$.

A Practical Next Step

Work through Practice Problems 1 to 6. If Problem 3 trips you up, return to Example 2 and check the minus sign in the tangent formula; if Problem 5 does, reread the Wallis section and write out the chain of multipliers first. Want a teacher to check your next set? Book a free demo class with a Bhanzu trainer.

Read More

• Table of integrals for quick reference

• Mathematical induction, the proof method behind recurrences

• Improper integrals and infinite limits

• Hyperbolic functions and their identities

• Applications of integration to area, volume and work