# Integration by trigonometric substitution Calculator

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### Difficult Problems

1

Solved example of integration by trigonometric substitution

$\int\sqrt{x^2+4}dx$
2

We can solve the integral $\int\sqrt{x^2+4}dx$ by applying integration method of trigonometric substitution using the substitution

$x=2\tan\left(\theta \right)$

Differentiate both sides of the equation $x=2\tan\left(\theta \right)$

$dx=\frac{d}{d\theta}\left(2\tan\left(\theta \right)\right)$

Find the derivative

$\frac{d}{d\theta}\left(2\tan\left(\theta \right)\right)$

The derivative of a function multiplied by a constant ($2$) is equal to the constant times the derivative of the function

$2\frac{d}{d\theta}\left(\tan\left(\theta \right)\right)$

The derivative of the tangent of a function is equal to secant squared of that function times the derivative of that function, in other words, if ${f(x) = tan(x)}$, then ${f'(x) = sec^2(x)\cdot D_x(x)}$

$2\sec\left(\theta \right)^2\frac{d}{d\theta}\left(\theta \right)$

The derivative of the linear function is equal to $1$

$2\sec\left(\theta \right)^2$
3

Now, in order to rewrite $d\theta$ in terms of $dx$, we need to find the derivative of $x$. We need to calculate $dx$, we can do that by deriving the equation above

$dx=2\sec\left(\theta \right)^2d\theta$
4

Substituting in the original integral, we get

$\int2\sqrt{4\tan\left(\theta \right)^2+4}\sec\left(\theta \right)^2d\theta$
5

Factor by the greatest common divisor $4$

$\int2\sqrt{4\left(\tan\left(\theta \right)^2+1\right)}\sec\left(\theta \right)^2d\theta$

$\int2\sqrt{2^2\tan\left(\theta \right)^2+4}\sec\left(\theta \right)^2d\theta$

Calculate the power $2^2$

$\int2\sqrt{4\tan\left(\theta \right)^2+4}\sec\left(\theta \right)^2d\theta$

$\int2\sqrt{4}\sqrt{\tan\left(\theta \right)^2+1}\sec\left(\theta \right)^2d\theta$

Calculate the power $\sqrt{4}$

$\int2\cdot 2\sqrt{\tan\left(\theta \right)^2+1}\sec\left(\theta \right)^2d\theta$

Multiply $2$ times $2$

$\int4\sqrt{\tan\left(\theta \right)^2+1}\sec\left(\theta \right)^2d\theta$
6

The power of a product is equal to the product of it's factors raised to the same power

$\int4\sqrt{\tan\left(\theta \right)^2+1}\sec\left(\theta \right)^2d\theta$
7

Applying the trigonometric identity: $\tan(x)^2+1=\sec(x)^2$

$\int4\sec\left(\theta \right)\sec\left(\theta \right)^2d\theta$
8

The integral of a constant by a function is equal to the constant multiplied by the integral of the function

$4\int\sec\left(\theta \right)\sec\left(\theta \right)^2d\theta$
9

When multiplying exponents with same base you can add the exponents: $\sec\left(\theta \right)\sec\left(\theta \right)^2$

$4\int\sec\left(\theta \right)^{3}d\theta$

$4\int\sec\left(\theta \right)^2\sec\left(\theta \right)^{\left(3-2\right)}d\theta$

Subtract the values $3$ and $-2$

$4\int\sec\left(\theta \right)^2\sec\left(\theta \right)^{1}d\theta$

Any expression to the power of $1$ is equal to that same expression

$4\int\sec\left(\theta \right)^2\sec\left(\theta \right)d\theta$
10

Rewrite $\sec\left(\theta \right)^{3}$ as the product of two secants

$4\int\sec\left(\theta \right)^2\sec\left(\theta \right)d\theta$
11

We can solve the integral $\int\sec\left(\theta \right)^2\sec\left(\theta \right)d\theta$ by applying integration by parts method to calculate the integral of the product of two functions, using the following formula

$\displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du$

Taking the derivative of secant function: $\frac{d}{dx}\left(\sec(x)\right)=\sec(x)\cdot\tan(x)\cdot D_x(x)$

$\sec\left(\theta \right)\tan\left(\theta \right)$
12

First, identify $u$ and calculate $du$

$\begin{matrix}\displaystyle{u=\sec\left(\theta \right)}\\ \displaystyle{du=\sec\left(\theta \right)\tan\left(\theta \right)d\theta}\end{matrix}$
13

Now, identify $dv$ and calculate $v$

$\begin{matrix}\displaystyle{dv=\sec\left(\theta \right)^2d\theta}\\ \displaystyle{\int dv=\int \sec\left(\theta \right)^2d\theta}\end{matrix}$
14

Solve the integral

$v=\int\sec\left(\theta \right)^2d\theta$
15

The integral of $\sec(x)^2$ is $\tan(x)$

$\tan\left(\theta \right)$

When multiplying two powers that have the same base ($\tan\left(\theta \right)$), you can add the exponents

$4\left(\tan\left(\theta \right)\sec\left(\theta \right)-\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta\right)$
16

Now replace the values of $u$, $du$ and $v$ in the last formula

$4\left(\tan\left(\theta \right)\sec\left(\theta \right)-\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta\right)$

$4\tan\left(\theta \right)\sec\left(\theta \right)+4\left(-1\right)\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta$

Multiply $4$ times $-1$

$4\tan\left(\theta \right)\sec\left(\theta \right)-4\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta$
17

Solve the product $4\left(\tan\left(\theta \right)\sec\left(\theta \right)-\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta\right)$

$4\tan\left(\theta \right)\sec\left(\theta \right)-4\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta$

Apply the formula: $\int\sec\left(x\right)\tan\left(x\right)^2dx$$=\int\sec\left(x\right)^3dx-\int\sec\left(x\right)dx, where x=\theta -4\int\sec\left(\theta \right)^3d\theta+4\int\sec\left(\theta \right)d\theta The integral of the secant function is given by the following formula, \displaystyle\int\sec(x)dx=\ln\left|\sec(x)+\tan(x)\right| -4\int\sec\left(\theta \right)^3d\theta+4\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right) Rewrite \sec\left(\theta \right)^3 as the product of two secants -4\int\sec\left(\theta \right)^2\sec\left(\theta \right)d\theta+4\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right) We can solve the integral \int\sec\left(\theta \right)^2\sec\left(\theta \right)d\theta by applying integration by parts method to calculate the integral of the product of two functions, using the following formula \displaystyle\int u\cdot dv=u\cdot v-\int v \cdot du First, identify u and calculate du \begin{matrix}\displaystyle{u=\sec\left(\theta \right)}\\ \displaystyle{du=\sec\left(\theta \right)\tan\left(\theta \right)d\theta}\end{matrix} Now, identify dv and calculate v \begin{matrix}\displaystyle{dv=\sec\left(\theta \right)^2d\theta}\\ \displaystyle{\int dv=\int \sec\left(\theta \right)^2d\theta}\end{matrix} Solve the integral v=\int\sec\left(\theta \right)^2d\theta The integral of \sec(x)^2 is \tan(x) \tan\left(\theta \right) Now replace the values of u, du and v in the last formula -4\left(\tan\left(\theta \right)\sec\left(\theta \right)-\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta\right)+4\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right) Solve the product -4\left(\tan\left(\theta \right)\sec\left(\theta \right)-\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta\right) -4\tan\left(\theta \right)\sec\left(\theta \right)-4\left(-1\right)\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta+4\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right) Multiply -4 times -1 -4\tan\left(\theta \right)\sec\left(\theta \right)+4\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta+4\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right) Apply the formula: \int\sec\left(x\right)\tan\left(x\right)^2dx$$=\int\sec\left(x\right)^3dx-\int\sec\left(x\right)dx$, where $x=\theta$

$-4\tan\left(\theta \right)\sec\left(\theta \right)+4\int\sec\left(\theta \right)^3d\theta-4\int\sec\left(\theta \right)d\theta+4\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$

The integral of the secant function is given by the following formula, $\displaystyle\int\sec(x)dx=\ln\left|\sec(x)+\tan(x)\right|$

$-4\tan\left(\theta \right)\sec\left(\theta \right)+4\int\sec\left(\theta \right)^3d\theta-4\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)+4\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$

Cancel like terms $-4\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$ and $4\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$

$-4\tan\left(\theta \right)\sec\left(\theta \right)+4\int\sec\left(\theta \right)^3d\theta$

Simplify the integral $\int\sec\left(\theta \right)^3d\theta$ applying the reduction formula, $\displaystyle\int\sec(x)^{n}dx=\frac{\sin(x)\sec(x)^{n-1}}{n-1}+\frac{n-2}{n-1}\int\sec(x)^{n-2}dx$

$-4\tan\left(\theta \right)\sec\left(\theta \right)+4\left(\frac{\sin\left(\theta \right)\sec\left(\theta \right)^{2}}{2}+\frac{1}{2}\int\sec\left(\theta \right)d\theta\right)$

Solve the product $4\left(\frac{\sin\left(\theta \right)\sec\left(\theta \right)^{2}}{2}+\frac{1}{2}\int\sec\left(\theta \right)d\theta\right)$

$-4\tan\left(\theta \right)\sec\left(\theta \right)+2\sin\left(\theta \right)\sec\left(\theta \right)^{2}+2\int\sec\left(\theta \right)d\theta$

Simplifying

$-2\tan\left(\theta \right)\sec\left(\theta \right)+2\int\sec\left(\theta \right)d\theta$

The integral of the secant function is given by the following formula, $\displaystyle\int\sec(x)dx=\ln\left|\sec(x)+\tan(x)\right|$

$-2\tan\left(\theta \right)\sec\left(\theta \right)+2\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$

Express the variable $\theta$ in terms of the original variable $x$

$-2\left(\frac{x}{2}\right)\left(\frac{\sqrt{x^2+4}}{2}\right)+2\ln\left(\frac{\sqrt{x^2+4}}{2}+\frac{x}{2}\right)$

Multiplying the fraction by $-2$

$\frac{-2x}{2}\frac{\sqrt{x^2+4}}{2}+2\ln\left(\frac{\sqrt{x^2+4}}{2}+\frac{x}{2}\right)$

Take $\frac{-2}{2}$ out of the fraction

$-\frac{1}{2}x\sqrt{x^2+4}+2\ln\left(\frac{\sqrt{x^2+4}}{2}+\frac{x}{2}\right)$

Add fraction's numerators with common denominators: $\frac{\sqrt{x^2+4}}{2}$ and $\frac{x}{2}$

$-\frac{1}{2}x\sqrt{x^2+4}+2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)$
18

The integral $-4\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta$ results in: $-\frac{1}{2}x\sqrt{x^2+4}+2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)$

$-\frac{1}{2}x\sqrt{x^2+4}+2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)$
19

Gather the results of all integrals

$4\tan\left(\theta \right)\sec\left(\theta \right)+2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)-\frac{1}{2}x\sqrt{x^2+4}$
20

Express the variable $\theta$ in terms of the original variable $x$

$4\left(\frac{x}{2}\right)\left(\frac{\sqrt{x^2+4}}{2}\right)+2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)-\frac{1}{2}x\sqrt{x^2+4}$
21

Multiplying the fraction by $4$

$\frac{4x}{2}\frac{\sqrt{x^2+4}}{2}+2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)-\frac{1}{2}x\sqrt{x^2+4}$
22

Take $\frac{4}{2}$ out of the fraction

$x\sqrt{x^2+4}+2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)-\frac{1}{2}x\sqrt{x^2+4}$
23

Adding $-\frac{1}{2}\sqrt{x^2+4}x$ and $\sqrt{x^2+4}x$

$2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)+\frac{1}{2}x\sqrt{x^2+4}$
24

As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$

$2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)+\frac{1}{2}x\sqrt{x^2+4}+C_0$
25

Simplify the expression by applying logarithm properties

$\frac{1}{2}x\sqrt{x^2+4}+2\ln\left(\sqrt{x^2+4}+x\right)+C_1$

$\frac{1}{2}x\sqrt{x^2+4}+2\ln\left(\sqrt{x^2+4}+x\right)+C_1$