Integration Techniques

Some integrals are easy to evaluate, like the first 2 examples below.

What is the indefinite integral of $x^2$?

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We can write this as $\int x^2\ dx.$ Using the (integration) power rule, which states that
$$\int x^n\ dx = \frac{x^{n+1}}{n+1} + C,$$ where $C$ is the constant of integration (which will be used in all of other formulae below), $x^n =x^2 \implies n=2.$ Therefore,
$$\int x^2\ dx = \frac{x^{2+1}}{2+1} + C = \frac{x^{3}}{3} + C.\ _\square$$

What is the general anti-derivative of $666x$?

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We can write this as $\int 666x \ dx.$ Using the power rule, which states that
$$\int x^n\ dx = \frac{x^{n+1}}{n+1} + C,$$ where $C$ is the constant of integration, and the constant function rule, which states that
$$\int a\ f(x)\ dx = a \int f(x)\ dx,$$ $x^n =x=x^1 \implies n = 1$ and $a = 666.$ Therefore,
$$\int 666x\ dx = 666 \cdot \frac{x^{1+1}}{1+1} + C = \frac{666}{2} \cdot x^2 + C = 333x^2 + C.\ _\square$$

Here are fundamental theorems for simple evaluation of integrals.

The constant rule states that for an arbitrary constant $a$
$$\int a\ dx = ax + C.$$

What is the indefinite integral of $1$?

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Since $a=1$ in this case,
$$\int a\ dx = \int 1\ dx = \int dx = 1x + C = x + C.\ _\square$$

The constant function rule, more formally known as the constant multiple rule, states that for an arbitrary constant $a$ and an arbitrary function $f(x)$ $$\int a\ f(x)\ dx = a \int f(x)\ dx.$$

What is the indefinite integral of $2x$?

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Since $a=2$ and $f(x)=x$ in this case, we have $$\int a\ f(x)\ dx = \int 2x\ dx = 2 \int x\ dx = 2 \frac{x^{1+1}}{1+1} + C = 2 \frac{x^2}2 + C = \frac22x^2 + C = 1x^2 + C = x^2 + C.\ _\square$$ Note that we also used the power rule, which we'll be discussing in a later note.

The sum and difference rules state that for arbitrary functions $f(x)$ and $g(x)$
$$\int \big(f(x) \pm g(x)\big)\ dx = \int f(x)\ dx \pm \int g(x)\ dx.$$

What is the indefinite integral of $2x + 3x^2$?

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Since $f(x)=2x$ and $g(x)=3x^2$ in this case, we have $$\int \big(f(x) + g(x)\big)\ dx = \int \big(2x + 3x^2\big)\ dx = \int 2x\ dx + \int 3x^2\ dx = \frac2{1+1}x^{1+1} + \frac3{2+1}x^{2+1} + C = x^2 + x^3 + C.\ _\square$$

The power rule states that, for an arbitrary constant $n$ and an arbitrary variable $x$ with $n,x \neq 0,$
$$\int x^n\ dx = \frac{x^{n+1}}{n+1} + C.$$

What is the indefinite integral of $x$?

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Since $x^n=x=x^1 \implies n=1$ in this case, we have
$$\int x^n\ dx = \int x\ dx = \frac{x^{1+1}}{1+1} + C = \frac{x^2}{2} + C.\ _\square$$

Given an arbitrary constant $n$ and an arbitrary variable $x$ with $n,x \neq 0,$
$$\int \frac1{nx} dx = \frac{\ln |nx|}{n} + C.$$

What is the indefinite integral of $\frac1{2x}$?

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Since $nx=2x \implies n=2$ in this case, we have $$\int \frac1{nx}\ dx = \int \frac1{2x}\ dx = \frac{\ln |2x|}{2} + C.\ _\square$$

Given an arbitrary constant $n$ and an arbitrary variable $x$ with $n \neq 0,$
$$\int \sin\ nx\ dx = -\frac{1}{n}\ \cos\ nx + C.$$

What is the indefinite integral of $\sin\ 3x$?

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Since $nx=3x \implies n=3$ in this case, we have $$\int \sin\ nx\ dx = \int \sin\ 3x\ dx = -\frac13\ \cos\ 3x + C.\ _\square$$

Given an arbitrary constant $n$ and an arbitrary variable $x$ with $n \neq 0,$
$$\int \cos\ nx\ dx = \frac{1}{n}\ \sin\ nx + C.$$

What is the indefinite integral of $\cos\ 4x$?

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Since $nx=4x \implies n=4$ in this case, we have $$\int \cos\ nx\ dx = \int \cos\ 4x\ dx = \frac14\ \sin\ 4x + C.\ _\square$$

Given an arbitrary constant $n$ and an arbitrary variable $x$ with $n, |\sec\ nx| \neq 0,$
$$\int \tan\ nx\ dx = \frac{1}{n}\ln |\sec\ nx| + C.$$

What is the indefinite integral of $\tan\ 5x$?

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Since $nx=5x \implies n=5$ in this case, we have
$$\int \tan\ nx\ dx = \int \tan\ 5x\ dx = \frac15\ln |\sec\ 5x| + C.\ _\square$$

Given an arbitrary constant $n$ and an arbitrary variable $x$ with $n, |\csc\ nx - \cot\ nx| \neq 0,$
$$\int \csc\ nx\ dx = \frac{1}{n}\ln |\csc\ nx - \cot\ nx| + C.$$

What is the indefinite integral of $\csc\ 6x$?

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Since $nx=6x \implies n=6$ in this case, we have $$\int \csc\ nx\ dx = \int \csc\ 6x\ dx = \frac16\ln |\csc\ 6x - \cot\ 6x| + C.\ _\square$$

Given an arbitrary constant $n$ and an arbitrary variable $x$ with $n \neq 0,$
$$\int \csc^2\ nx\ dx = -\frac{1}{n}\cot\ nx + C.$$

What is the indefinite integral of $\csc^2\ 7x$?

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Since $nx=7x \implies n=7$ in this case, we have
$$\int \csc^2\ nx\ dx = \int \csc^2\ 7x\ dx = \frac17\cot\ 7x + C.\ _\square$$

Given an arbitrary constant $n$ and an arbitrary variable $x$ with $n \neq 0,$
$$\int \csc\ (nx)\cot\ nx\ dx = -\frac{1}{n}\csc\ nx + C.$$

What is the indefinite integral of $\csc\ (8x)\ \cot\ 8x$?

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Since $nx=8x \implies n=8$ in this case, we have $$\int \csc\ (nx)\cot\ nx\ dx = \int \csc\ (8x)\ \cot\ 8x\ dx = -\frac18\csc\ 8x + C.\ _\square$$

Given an arbitrary constant $n$ and an arbitrary variable $x$ with $n, |\sec\ nx + \tan\ nx| \neq 0,$
$$\int \sec\ nx\ dx = \frac{1}{n}\ln |\sec\ nx + \tan\ nx| + C.$$

What is the indefinite integral of $\sec\ 9x$?

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Since $nx=9x \implies n=9$ in this case, we have
$$\int \sec\ nx\ dx = \int \sec\ 9x\ dx = \frac19\ln |\sec\ 9x + \tan\ 9x| + C.\ _\square$$

Given an arbitrary constant $n$ and an arbitrary variable $x$ with $n \neq 0,$
$$\int \sec^2\ nx\ dx = \frac{1}{n}\tan\ nx + C.$$

What is the indefinite integral of $\sec^2\ 10x$?

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Since $nx=10x \implies n=10$ in this case, we have
$$\int \sec^2\ nx\ dx = \int \sec^2\ 10x\ dx = \frac{1}{10}\tan\ 10x + C.\ _\square$$

Given an arbitrary constant $n$ and an arbitrary variable $x$ with $n \neq 0,$
$$\int \sec\ (nx)\ \tan\ nx\ dx = \frac{1}{n}\ \sec\ nx + C.$$

What is the indefinite integral of $\sec\ 11x$?

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Since $nx=11x \implies n=11$ in this case, we have
$$\int \sec\ (nx)\ \tan\ nx\ dx = \int \sec\ (11x)\ \tan\ (11x)\ dx = \frac{1}{11}\sec\ 11x + C.\ _\square$$

Given an arbitrary constant $n$ and an arbitrary variable $x$ with $n, |\sin\ nx| \neq 0,$
$$\int \cot\ nx\ dx = \frac{1}{n}\ln |\sin\ nx| + C.$$

What is the indefinite integral of $cot\ 12x$?

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Since $nx=12x \implies n=12$ in this case
$$\int \cot\ nx\ dx = \int \cot\ 12x\ dx = \frac{1}{12}\ln |\sin\ 12x| + C.\ _\square$$

The exponential constant rule states that, for arbitrary constants $a$ and $n$ and an arbitrary variable $x$ with $a,n,x \neq 0,$
$$\int a^{nx}\ dx = \frac{a^{nx}}{n\ ln\ a} + C.$$

What is the indefinite integral of $13^{4x}$?

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Since $a=13$ and $nx=4x \implies n=4$ in this case, we have
$$\int a^{nx}\ dx = \int 13^{4x} dx = \frac{13^{4x}}{4\ ln\ 13} + C.\ _\square$$

Given arbitrary constants $a$ and $n$ and an arbitrary variable $x$ with $a,n,x \neq 0,$ we can derive the integral of $e^{nx}$ $\big(a^{nx}=e^{nx} \implies a=e\big),$ where $e$ is the base of the natural log, as follows:
$$\int a^{nx}\ dx = \int e^{nx}\ dx = \frac{e^{nx}}{n\ ln\ e} + C = \frac{e^{nx}}n + C.$$ Note that $\ln e = 1.$

What is the indefinite integral of $e^{14x}$?

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Since $nx=14x \implies n=14$ in this case,
$$\int e^{nx}\ dx = \int e^{14x} dx = \frac{e^{14x}}{14} + C.\ _\square$$

The natural log rule states that, for an arbitrary constant $n$ and an arbitrary variable $x$ with $n,x \neq 0,$
$$\int \ln nx\ dx = x\ln |x| - x + C.$$

What is the indefinite integral of $\ln\ 15x$?

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Since $nx=15x \implies n=15$ in this case, we have
$$\int \ln nx\ dx = \int \ln 15x\ dx = x\ \ln |x| - x + C.\ _\square$$

However, other integrals, such as $\int xe^x dx$ and $\int \sin x \cos x\ dx$, may not be ordinarily evaluated.
For this reason, we need to use appropriate techniques to make the evaluation of such integrals possible.

U-substitution, a.k.a. the (integration) chain rule, can be done in one of the following 2 methods:

Method 1: Direct Substitution

1. Identify which part of the integrand is the original function $f(x)$ of the other $\big($its derivative $f'(x)\big),$ and let it be $u:$ $u = f(x).$
2. Identify which part is the derivative of $u$, and let it be $du: du = f'(x)\ dx.$
3. Replace the integrand in terms of $u.$
4. Evaluate the resulting integral from (3) in terms of $u$ and $du.$
5. Substitute the value of $u$ in the result from (4) to convert it into an expression in terms of $x.$

What is the indefinite integral of $\big(2x + 4x^2\big)(2 + 8x)?$

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We can write this as $\int \big(2x + 4x^2\big)(2 + 8x)\, dx.$

Let $u=f(x)= 2x + 4x^2,$ then $du= f'(x)\ dx = (2 + 8x)\ dx,$ which gives $$\int \big(2x + 4x^2\big)(2 + 8x)\ dx = \int u\ du = \frac{u^{1+1}}{1+1} + C = \frac{u^2}2 + C = \frac{(2x + 4x^2)^2}2 + C.\ _\square$$

Method 2: Indirect Substitution

1. Identify which part of the integrand is the "more complicated" function $f(x)$, and let it be $u:$ $u = f(x).$
2. Identify which part is the "less complicated" function $g(x)$, and transform it in terms of $u.$
3. Replace the integrand in terms of $u.$
4. Evaluate the resulting integral from (3) in terms of $u$ and $du.$
5. Substitute the value of $u$ in the result from (4) to convert it into an expression in terms of $x.$

What is the indefinite integral of $\big(3x^3\big)(x+1)^8$?

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We can write this as $\int \big(3x^3\big)(x+1)^8\, dx$.

Let $u=f(x)= x+1,$ then $du= 1\ dx \implies du=dx.$
Also let $g(x)=3x^3,$ then $u= x+1 \implies x= u-1 \implies 3x^3= 3(u-1)^3,$ which gives

$$\begin{aligned} \int 3(u-1)^3\ u^8\ dx &= 3 \int \big(u^3 - 3u^2 + 3u - 1\big)\ u^8\ du \\ &= 3 \int \big(u^{11} - 3u^{10} + 3u^9 - u^8\big)\ du \\ &= 3\ \left(\int u^{11}\ du - \int 3u^{10}\ du + \int 3u^9\ du - \int u^8\ du\right) \\ &= 3\ \left(\frac{u^{11+1}}{11+1} - 3\frac{u^{10+1}}{10+1} + 3\frac{u^{9+1}}{9+1} - \frac{u^{8+1}}{8+1} + C\right) \\ &= 3\ \left(\frac{u^{11+1}}{11+1} - \frac3{10+1}u^{10+1} + \frac3{9+1}u^{9+1} - \frac{u^{8+1}}{8+1} + C\right) \\ &= 3\ \left(\frac{u^{12}}{12} - \frac3{11}u^{11} + \frac3{10}u^{10} - \frac{u^9}9 + C\right) \\ &= \frac3{12}u^{12} - \frac{3^2}{11}u^{11} + \frac{3^2}{10}u^{10} - \frac39u^9 + C \\\\ &= \frac14u^{12} - \frac9{11}u^{11} + \frac9{10}u^{10} - \frac13u^9 + C \\\\ &= \frac{u^{12}}4 - \frac9{11}u^{11} + \frac9{10}u^{10} - \frac{u^9}3 + C \\\\ &= \frac{(x+1)^{12}}4 - \frac9{11}(x+1)^{11} + \frac9{10}(x+1)^{10} - \frac{(x+1)^9}3 + C.\ _\square \end{aligned}$$

Another technique that we can use to evaluate tricky-looking integrals is integration by parts. This is similar to the $u$-substitution methods. Given arbitrary functions $f(x)$ and $g(x)$ with $u=f(x)$ and $v' = g(x)\ dx$, and an arbitrary differential $dx,$
$$\int uv'\ dx = uv - \int u'v\ dx.$$ This formula was derived from the product rule in differentiation.

1. Identify which part of the integrand is $u$: this may be the "written" part $f(x)$. However, if there are 2 arbitrary functions $f(x)$ and $g(x)$ multiplied together, let the function that is easier to differentiate be $u.$
2. Identify which part of the integrand is $v'$: this may be the "non-written" part $1$. However, if there are 2 arbitrary functions $f(x)$ and $g(x)$ multiplied together, let the function that is easier to integrate be $v'.$
3. Integrate the value of $v'$ to get $v.$
4. Integrate $u'v$ for the subtrahend $\int u'v.$
5. Subtract the result in (4) from the value of $uv.$
6. Substitute the values of $u$, $v$ and $u'$ in the result from (5).

What is the indefinite integral of $x^2\ln x$ $($with $x=|n|,$ where $n \neq 0)?$

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We can write this as $\int x^2\ln x\ dx.$

Let $u= \ln x,$ then $u'=x^{-1}.$
Also let $v'=x^2,$ then $\frac{dv}{dx}=x^2, \int \frac{dv}{dx}\ dx = \int dv = v= \int x^2\ dx = \frac{x^{2+1}}{2+1} + C = \frac{x^3}3 + C \implies v=\frac{x^3}3,$ which implies

$$\begin{aligned} \int x^2\ln x\ dx &= \int uv'\ dx \\ &= uv - \int u'v\ dx \\ &= \ln x\ \frac{x^3}3 - \int \frac{1}{x} \cdot \frac{x^3}3\ dx \\ &= \frac{x^3}3 \ln\ x - \int \frac{x^{3-1}}3\ dx \\ &= \frac{x^3}3 \ln\ x - \int \frac{x^2}3\ dx \\ &= \frac{x^3}3 \ln\ x - \frac13 \int x^2\ dx \\ &= \frac{x^3}3 \ln\ x - \frac13 \cdot \frac{x^{2+1}}{2+1} + C \\ &= \frac{x^3}3 \ln\ x - \frac13 \cdot \frac{x^3}3 + C \\ &= \frac{x^3}3 \ln\ x - \frac{x^3}{9} + C.\ _\square \end{aligned}$$