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Solved example of integration by trigonometric substitution
$\int\sqrt{x^2+4}dx$
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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 $x=2\tan\left(\theta \right)$
$dx=\frac{d}{d\theta}\left(2\tan\left(\theta \right)\right)$
$\frac{d}{d\theta}\left(2\tan\left(\theta \right)\right)$
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 $1$
$2\sec\left(\theta \right)^2$
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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
$dx=2\sec\left(\theta \right)^2d\theta$
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Substituting in the original integral, we get
$\int2\sqrt{4\tan\left(\theta \right)^2+4}\sec\left(\theta \right)^2d\theta$
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Factor the polynomial $4\tan\left(\theta \right)^2+4$ by it's GCF: $4$
$\int2\sqrt{4\left(\tan\left(\theta \right)^2+1\right)}\sec\left(\theta \right)^2d\theta$
The power of a product is equal to the product of it's factors raised to the same power
$\int2\sqrt{2^2\tan\left(\theta \right)^2+4}\sec\left(\theta \right)^2d\theta$
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 $\sqrt{4}$
$\int2\cdot 2\sqrt{\tan\left(\theta \right)^2+1}\sec\left(\theta \right)^2d\theta$
$\int4\sqrt{\tan\left(\theta \right)^2+1}\sec\left(\theta \right)^2d\theta$
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The power of a product is equal to the product of it's factors raised to the same power
$\int4\sqrt{\tan\left(\theta \right)^2+1}\sec\left(\theta \right)^2d\theta$
Applying the trigonometric identity: $\tan(x)^2+1=\sec(x)^2$
$\int4\sqrt{\sec\left(\theta \right)^2}\sec\left(\theta \right)^2d\theta$
Applying the power of a power property
$\int4\sec\left(\theta \right)\sec\left(\theta \right)^2d\theta$
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Applying the trigonometric identity: $\tan(x)^2+1=\sec(x)^2$
$\int4\sec\left(\theta \right)\sec\left(\theta \right)^2d\theta$
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The integral of a constant by a function is equal to the constant multiplied by the integral of the function
$4\int\sec\left(\theta \right)\sec\left(\theta \right)^2d\theta$
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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
$4\int\sec\left(\theta \right)^2\sec\left(\theta \right)d\theta$
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Rewrite $\sec\left(\theta \right)^{3}$ as the product of two secants
$4\int\sec\left(\theta \right)^2\sec\left(\theta \right)d\theta$
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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: $\frac{d}{dx}\left(\sec(x)\right)=\sec(x)\cdot\tan(x)\cdot D_x(x)$
$\sec\left(\theta \right)\tan\left(\theta \right)$
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First, identify $u$ and calculate $du$
$\begin{matrix}\displaystyle{u=\sec\left(\theta \right)}\\ \displaystyle{du=\sec\left(\theta \right)\tan\left(\theta \right)d\theta}\end{matrix}$
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Now, identify $dv$ and calculate $v$
$\begin{matrix}\displaystyle{dv=\sec\left(\theta \right)^2d\theta}\\ \displaystyle{\int dv=\int \sec\left(\theta \right)^2d\theta}\end{matrix}$
$v=\int\sec\left(\theta \right)^2d\theta$
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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
$4\left(\tan\left(\theta \right)\sec\left(\theta \right)-\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta\right)$
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Now replace the values of $u$, $du$ and $v$ in the last formula
$4\left(\tan\left(\theta \right)\sec\left(\theta \right)-\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta\right)$
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Multiply the single term $4$ by each term of the polynomial $\left(\tan\left(\theta \right)\sec\left(\theta \right)-\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta\right)$
$4\tan\left(\theta \right)\sec\left(\theta \right)-4\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta$
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, $\displaystyle\int\sec(x)dx=\ln\left|\sec(x)+\tan(x)\right|$
$-4\int\sec\left(\theta \right)^3d\theta+4\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$
Simplify the integral $\int\sec\left(\theta \right)^3d\theta$ applying the reduction formula, $\displaystyle\int\sec(x)^{n}dx=\frac{\sin(x)\sec(x)^{n-1}}{n-1}+\frac{n-2}{n-1}\int\sec(x)^{n-2}dx$
$-4\left(\frac{\sin\left(\theta \right)\sec\left(\theta \right)^{2}}{2}+\frac{1}{2}\int\sec\left(\theta \right)d\theta\right)+4\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$
Solve the product $-4\left(\frac{\sin\left(\theta \right)\sec\left(\theta \right)^{2}}{2}+\frac{1}{2}\int\sec\left(\theta \right)d\theta\right)$
$-4\left(\frac{\sin\left(\theta \right)\sec\left(\theta \right)^{2}}{2}\right)-2\int\sec\left(\theta \right)d\theta+4\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$
$-2\tan\left(\theta \right)\sec\left(\theta \right)-2\int\sec\left(\theta \right)d\theta+4\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$
The integral of the secant function is given by the following formula, $\displaystyle\int\sec(x)dx=\ln\left|\sec(x)+\tan(x)\right|$
$-2\tan\left(\theta \right)\sec\left(\theta \right)-2\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)+4\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$
Combining like terms $-2\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$ and $4\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$
$-2\tan\left(\theta \right)\sec\left(\theta \right)+2\ln\left(\sec\left(\theta \right)+\tan\left(\theta \right)\right)$
Express the variable $\theta$ in terms of the original variable $x$
$-2\frac{x}{2}\frac{\sqrt{x^2+4}}{2}+2\ln\left(\frac{\sqrt{x^2+4}}{2}+\frac{x}{2}\right)$
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 $\frac{x}{2}$
$-\frac{1}{2}x\sqrt{x^2+4}+2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)$
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The integral $-4\int\tan\left(\theta \right)^2\sec\left(\theta \right)d\theta$ results in: $-\frac{1}{2}x\sqrt{x^2+4}+2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)$
$-\frac{1}{2}x\sqrt{x^2+4}+2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)$
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Gather the results of all integrals
$4\tan\left(\theta \right)\sec\left(\theta \right)+2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)-\frac{1}{2}x\sqrt{x^2+4}$
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Express the variable $\theta$ in terms of the original variable $x$
$4\frac{x}{2}\frac{\sqrt{x^2+4}}{2}+2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)-\frac{1}{2}x\sqrt{x^2+4}$
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Multiplying the fraction by $4$
$\frac{4x}{2}\frac{\sqrt{x^2+4}}{2}+2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)-\frac{1}{2}x\sqrt{x^2+4}$
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Take $\frac{4}{2}$ out of the fraction
$x\sqrt{x^2+4}+2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)-\frac{1}{2}x\sqrt{x^2+4}$
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Combining like terms $x\sqrt{x^2+4}$ and $-\frac{1}{2}x\sqrt{x^2+4}$
$\frac{1}{2}x\sqrt{x^2+4}+2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)$
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As the integral that we are solving is an indefinite integral, when we finish integrating we must add the constant of integration $C$
$\frac{1}{2}x\sqrt{x^2+4}+2\ln\left(\frac{\sqrt{x^2+4}+x}{2}\right)+C_0$
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Simplify the expression by applying logarithm properties
$2\ln\left(\sqrt{x^2+4}+x\right)+C_1+\frac{1}{2}x\sqrt{x^2+4}$
$2\ln\left(\sqrt{x^2+4}+x\right)+C_1+\frac{1}{2}x\sqrt{x^2+4}$