An integral by definition is the continuous analog of a sum, which is used to calculate areas, volumes, and their generalizations. Integration or anti derivative, is the process of computing an integral for a given function. Alongside differentiation, it’s a fundamental operation of calculus. The symbol for the integration is:
It is a fundamental concept in calculus that plays an indispensable role in pure and applied mathematics. Integration has many applications including: finding the area bounded by the graph of a function, solving differential equations in mathematical modeling, calculating probabilities and statistics, evaluating areas under curves and so on. Before you continue, you must check out “A guide to integration” to ensure that you are already familiar with the basics and key integration techniques as this article will work on improving your integration skills through six different difficult examples.
Example one
Integrate:
\(\int x^7 \sqrt{1 + x^4} \, dx\)
First thing you should do is rewrite the root as a power of half:
\(\int x^7 (1 + x^4)^{\frac{1}{2}} \, dx\)
This function is a product of a power of x and a composite function involving a square root. This is a perfect candidate for integration using u-substitution.
Step 1: Let’s choose a substitution
Let’s try:
\(u = 1 + x^4\)
Then:
\(\frac{du}{dx} = 4x^3\)
\(du = 4x^3 \, dx \)
But we currently have \( x^7 \, dx \) in the integral, so we’ll need to express that in terms of \( x^3 \, dx \), which means:
\(x^7 \, dx = x^4 \cdot x^3 \, dx\)
\(x^7 \, dx = x^4 \cdot \frac{du}{4}\)
Also, recall from the substitution:
\(x^4 = u – 1 \text{ from } u = 1 + x^4\)
Now put back everything into the integral
Step 2: Substituting into the integral
Remember that we started with this:
\(\int x^7 (1 + x^4)^{\frac{1}{2}} \, dx\)
Now it becomes this:
\(\int (x^4)(1 + x^4)^{\frac{1}{2}} \cdot x^3 \, dx\)
Using substitution:
\(\int (u – 1) \cdot u^{\frac{1}{2}} \cdot \frac{du}{4}\)
The constant can be factored out:
\(\frac{1}{4} \int (u – 1) \cdot u^{\frac{1}{2}} \, du\)
Step 3: Simplify
As you can see we can expand the expression now:
\((u – 1)(u^{\frac{1}{2}}) = u^{1.5} – u^{0.5}\)
So:
\(\frac{1}{4} \int (u^{1.5} – u^{0.5}) \, du\)
Now integrate each term:
\(\int u^{1.5} \, du = \frac{u^{2.5}}{2.5} = \frac{u^{2.5}}{\frac{5}{2}} = \frac{2}{5}u^{2.5}\)
\(\int u^{0.5} \, du = \frac{u^{1.5}}{1.5} = \frac{u^{1.5}}{\frac{3}{2}} = \frac{2}{3}u^{1.5}\)
So the whole thing becomes:
\(\frac{1}{4}\left[ \frac{2}{5}u^{2.5} – \frac{2}{3}u^{1.5} \right] + C\)
Further simplify – You can multiply everything here:
\(\frac{1}{4}\left(\frac{2}{5}\right)u^{2.5} – \frac{1}{4}\left(\frac{2}{3}\right)u^{1.5}\)
\(\frac{1}{10}u^{2.5} – \frac{1}{6}u^{1.5} + C\)
Step 4: Replace “u” back with \( 1 + x^4 \)
Final Answer:
\(\frac{1}{10}(1 + x^4)^{2.5} – \frac{1}{6}(1 + x^4)^{1.5} + C\)
Example two
Integrate:
\(\int e^x \cdot \cos(x) \, dx\)
This is the most basic example for needing integration by parts. Though this is a bit of a special one and you will see why as we go on.
Step 1: Identify your separate terms
This is multiplication between \( e^x \) and \( cos(x) \) so we can’t do substitution here, and integration by parts is the only way forward.
Step 2: Integration by parts formula
Integration by parts is one of the easier methods of integration among all because of it’s clear straightforward formula:
\(\int u \, dv = u \cdot v – \int v \, du\)
If you have done enough integration by parts, you might have already seen that we will have to apply it twice.
Step 3: First integration by parts
Let:
\(u = \cos(x) \text{, then } du = -\sin(x)\)
\(dv = e^x \, dx \text{, then } v = e^x\)
Now put back your values into the formula:
\(\int e^x \cos(x) \, dx = e^x \cdot \cos(x) – \int e^x (-\sin(x)) \, dx\)
\(= e^x \cdot \cos(x) + \int e^x \sin(x) \, dx\)
Now we have:
\(\int e^x \cos(x) \, dx = e^x \cos(x) + \int e^x \sin(x) \, dx – \text{ This will be our equation (1)}\)
Now we need to integrate \( int e^x \sin(x) \, dx \) again by parts. It’s a full circle and we are back to the original form.
Step 4: Second integration by parts
We now compute \( int e^x \sin(x) \, dx \) using integration by parts again.
Let:
\(u = \sin(x) \text{ then, } du = \cos(x)\)
\(dv = e^x \, dx \text{ then, } v = e^x\)
Apply the formula again:
\(\int e^x \sin(x) \, dx = e^x \cdot \sin(x) – \int e^x \cdot \cos(x) \, dx\)
Now you can substitute this back into equation (1):
\(\int e^x \cos(x) \, dx = e^x \cos(x) + \left[ e^x \cdot \sin(x) – \int e^x \cos(x) \, dx \right]\)
Open the bracket:
\(\int e^x \cos(x) \, dx = e^x \cos(x) + e^x \sin(x) – \int e^x \cos(x) \, dx\)
Now notice something cool – the same integral appears on both sides. Set those two integrals into one value like so:
Let’s call \( I = \int e^x \cos(x) \, dx \)
Then we can make the entire expression look simplified by substituting \( I \):
\(I = e^x \cos(x) + e^x \sin(x) – I\)
Using your well established algebraic skills you should predict that now we will add \( I \) to both sides and simplify:
\(2I = e^x (\cos(x) + \sin(x))\)
Divide by 2:
\(I = \frac{1}{2} e^x (\sin(x) + \cos(x)) + C\)
Final Answer: \(\int e^x \cos(x) \, dx = \frac{1}{2} e^x (\sin(x) + \cos(x)) + C\)
Because of the repeating pattern, this method is specifically called “tabular integration” or double integration.
Example three
Integrate:
\(\int \sin^2(x) \cdot \cos^3(x) \, dx\)
For this question we are working with powers of sine and cosine, and the strategy depends on whether the power of sine or cosine is odd or even sp..
Step 1: Analyse the powers
\( sin(x) \) has an even power
\( cos(x) \) has an odd power
If one of them (sine or cosine) has an odd power, we can split off one factor of that function and use Pythagorean identities.
Step 2: Split off one factor of the odd powered function like so
Since \( cos^3(x) = \cos^2(x) \cdot \cos(x) \), we rewrite:
\(\int \sin^2(x) \cdot \cos^2(x) \cdot \cos(x) \, dx\)
We can now focus on:
Converting \( cos^2(x) \) to an expression involving \( sin(x) \) and saving one \( cos(x) \, dx \) for substitution
Step 3: Use identity
From the Pythagorean identity – we know \( cos^2(x) = 1 – sin^2(x) \)
So substitute that into our integral:
\(int \sin^2(x) \cdot (1 – \sin^2(x)) \cdot \cos(x) \, dx\)
Step 4: Use integration by u-substitution
Let:
\(u = \sin(x)\)
Then: \( du = \cos(x) \, dx \)
Our integral becomes:
\(int u^2(1 – u^2) \, du\)
Now expand:
\(u^2(1 – u^2) = u^2 – u^4\)
So:
\(int u^2 – u^4 \, du\)
\(= \int u^2 \, du – \int u^4 \, du\)
\(= \frac{1}{3}u^3 – \frac{1}{5}u^5 + C\)
Now go back to x:
\(u = \sin(x)\)
So:
\(\frac{1}{3} \sin^3(x) – \frac{1}{5} \sin^5(x) + C\)
Final Answer:
\(\int \sin^2(x) \cdot \cos^3(x) \, dx = \frac{1}{3} \sin^3(x) – \frac{1}{5} \sin^5(x) + C\)
Example four
Integrate:
\( \int \frac{1}{\sqrt[3]{1 + 2x}^2} \, dx \)
Same thing as:
\( \int (1 + 2x)^{-\frac{2}{3}} \, dx \)
Step 1: Identify the type of integral you are working with
This is a power of a linear expression so u-substitution will be the most suitable method.
Step 2: Let’s substitute
Let:
\( u = 1 + 2x \)
Then: \( \frac{du}{dx} = 2 \)
\( dx = \frac{du}{2} \)
Now rewrite the integral:
\( int (1 + 2x)^{-\frac{2}{3}} \, dx \)
\( = \int u^{-\frac{2}{3}} \cdot \left(\frac{du}{2}\right) \)
\( = \frac{1}{2} \int u^{-\frac{2}{3}} \, du \)
Step 3: Integrate using power rule
Recall the power rule for integration:
\( int u^n \, du = \frac{u^{n+1}}{n + 1} \), as long as \( n \neq -1 \)
Here, \( n = -\frac{2}{3} \)
\( n + 1 = \frac{1}{3} \)
So:
\( \int u^{-\frac{2}{3}} \, du = \frac{u^{\frac{1}{3}}}{\frac{1}{3}} = 3u^{\frac{1}{3}} \)
Don’t forget the \( \frac{1}{2} \) we factored out earlier:
\( \frac{1}{2} \times 3u^{\frac{1}{3}} = \frac{3}{2} u^{\frac{1}{3}} \)
Step 4: Replace u back
\( u = 1 + 2x \)
So:
\( \frac{3}{2}(1 + 2x)^{\frac{1}{3}} + C \)
Final Answer:
\( \int \frac{1}{\sqrt[3]{1 + 2x}^2} \, dx = \frac{3}{2}(1 + 2x)^{\frac{1}{3}} + C \)
Example five
Integrate:
\(\int \sec^2(x) \cdot \tan^5(x) \, dx\)
Step 1: Recognise the identities involved
We know:
\(\frac{d}{dx} [\tan(x)] = \sec^2(x)\)
So this is perfect for substitution.
Step 2: Use substitution
Let: \( u = \tan(x) \)
Then: \( du = \sec^2(x) \, dx \)
That means:
\(\sec^2(x) \, dx = du\)
\(\tan^5(x) = u^5\)
So the integral becomes:
\(\int u^5 \, du = \frac{1}{6} u^6 + C\)
Now go back to x:
\( u = \tan(x) \rightarrow u^6 = \tan^6(x) \)
Final Answer:
\(\int \sec^2(x) \cdot \tan^5(x) \, dx = \frac{1}{6} \tan^6(x) + C\)
Example six
Integrate:
\(\int \frac{\sqrt{2 – x^2}}{x^2} \, dx\)
That’s the same as:
\(\int (2 – x^2)^{\frac{1}{2}} \cdot x^{-2} \, dx\)
This may look like a u-substitution situation but if you have done integration examples and are familiar with the “trig-subisituiton” method you will immediately notice that it is shaped perfectly for a trig substitution technique because of the square root of a difference involving \( x^2 \).
Step 1: Recognise the form
We see \( \sqrt{2 – x^2} \). That’s very similar to the standard trig identity:
\( \sin^2(\theta) = \cos^2(\theta) \quad \text{and} \quad \sin^2(\theta) + \cos^2(\theta) = 1\)
So for \( \sqrt{a^2 – x^2} \), we typically use:
Let \( x = a \sin(\theta) \)
Here, \( a = \sqrt{2} \)
So:
Let \( x = \sqrt{2} \sin(\theta) \)
Then:
\( dx = \sqrt{2} \cos(\theta) \, d\theta \)
Step 2: Replace everything into the integral
But first start by converting and simplifying each part:
1. \( \sqrt{2 – x^2} \)
If \( x = \sqrt{2} \sin(\theta) \), then:
\(2 – x^2 = 2 – 2\sin^2(\theta) = 2(1 – \sin^2(\theta)) = 2\cos^2(\theta)\)
So:
\(\sqrt{2 – x^2} = \sqrt{2\cos^2(\theta)} = \sqrt{2} \cdot \cos(\theta)\)
2. \( x^2 = (\sqrt{2} \sin(\theta))^2 = 2\sin^2(\theta) \)
So:
\(\frac{1}{x^2} = \frac{1}{2\sin^2(\theta)}\)
3. \( dx = \sqrt{2} \cos(\theta) \, d\theta \)
Now put it all into the integral:
\(\int \left[ \sqrt{2} \cdot \cos(\theta) \right] \cdot \left[ \frac{1}{2\sin^2(\theta)} \right] \cdot \left[ \sqrt{2} \cos(\theta) \, d\theta \right]\)
Simplify step by step:
\(\sqrt{2} \times \sqrt{2} = 2\)
So you get:
\(\int \frac{2\cos^2(\theta)}{2\sin^2(\theta)} \, d\theta\)
\(= \int \frac{\cos^2(\theta)}{\sin^2(\theta)} \, d\theta\)
\(= \int \cot^2(\theta) \, d\theta\)
Step 3: Integrate \( \cot^2(\theta) \)
We know:
\(\cot^2(\theta) = \csc^2(\theta) – 1\)
So:
\(\int \cot^2(\theta) \, d\theta = \int (\csc^2(\theta) – 1) \, d\theta\)
\(= -\cot(\theta) – \theta + C\)
Step 4: Back substitute to \( x \)
We used:
\(x = \sqrt{2} \sin(\theta)\)
\(\sin(\theta) = \frac{x}{\sqrt{2}}\)
\(\theta = \arcsin\left(\frac{x}{\sqrt{2}}\right)\)
But now we must express \( \cot(\theta) \) in terms of \( x \) because that is how our original integral was.
What you will do is draw out a right triangle for \( \sin(\theta) = \frac{x}{\sqrt{2}} \) where:
Opposite = \( x \)
Hypotenuse = \( \sqrt{2} \)
Adjacent = \( \sqrt{2 – x^2} \)
So using the formula:
\(\cot(\theta) = \frac{\text{adjacent}}{\text{opposite}} = \frac{\sqrt{2 – x^2}}{x}\)
And: \( \theta = \arcsin\left(\frac{x}{\sqrt{2}}\right) \)
Final Answer:
\(\int \frac{\sqrt{2 – x^2}}{x^2} \, dx = -\frac{\sqrt{2 – x^2}}{x} – \arcsin\left(\frac{x}{\sqrt{2}}\right) + C\)
When it comes to integration functions all you have to do is analyze what kind of integral you are working with. Familiarizing yourself with complex questions will help you faster understand the whole concept and trick to integration in an overall sense while helping you improve your algebra and calculus skills so keep practising and hopefully these were helpful.
