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Evaluate the integral.

{eq}\displaystyle \int_1^5 \dfrac {dx} {x^2 \sqrt {4 + x^2}} {/eq}

Question:

Evaluate the integral.

{eq}\displaystyle \int_1^5 \dfrac {dx} {x^2 \sqrt {4 + x^2}} {/eq}

Definite Integrals:

Although there is no specific method to solve the definite or indefinite integrals, but when the integrand function is an expression of the following forms: {eq}\displaystyle \sqrt{a^2+u^2} {/eq}, {eq}\displaystyle \sqrt{u^2-a^2} {/eq}, {eq}\displaystyle \sqrt{a^2-u^2} {/eq}. In these cases, we can use the method of trigonometric substitution, in addition to the fundamental theorem of calculus.

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Answer and Explanation: 1

We can rewrite the integral:

{eq}\displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\int_1^5\frac{dx}{x^2\sqrt{2^2+x^2}}\\ {/eq}

Here, how the integrand is of the form {eq}\displaystyle \sqrt{a^2+u^2} {/eq}, we use the trigonometric substitution:

{eq}\displaystyle\,x=2\tan\theta\Rightarrow{\tan\theta=\frac{x}{2}} \\ \displaystyle\,dx=2\sec^2\theta\,d\theta \\ \displaystyle \sqrt{2^2+x^2}=\sqrt{2^2+ 2^2\tan^2\theta} \\ \displaystyle \sqrt{2^2+x^2}=\sqrt{2^2(1+ \tan^2\theta)} \\ \displaystyle \sqrt{2^2+x^2}=\sqrt{4}\cdot{\sqrt{1+\tan^2\theta}} \\ \displaystyle \sqrt{2^2+x^2}=2\cdot{\sqrt{1+\tan^2\theta}} \\ {/eq}

But we know by trigonometric identity that:

{eq}\displaystyle 1+\tan^2\theta=\sec^2\theta \\ {/eq}

Then:

{eq}\displaystyle \sqrt{2^2+x^2}=2\cdot{\sqrt{\sec^2\theta}} \\ \displaystyle \sqrt{2^2+x^2}=2\sec\theta \\ {/eq}

By replacing in the original integral:

{eq}\displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\int_1^5\frac{dx}{x^2\sqrt{2^2+x^2}}\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\int_1^5\frac{2\sec^2\theta}{4\tan^2\,2\sec\theta}d\theta\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\int_1^5\frac{2\sec^2\theta}{4\tan^2\theta\cdot{2\sec\theta}}d\theta\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\frac{1}{4}\int_1^5\frac{\sec\theta}{\tan^2\theta}d\theta\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\frac{1}{4}\int_1^5\frac{\frac{1}{\cos\theta}}{\frac{\sin^2\theta}{\cos^2\theta}}d\theta\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\frac{1}{4}\int_1^5\frac{\cos^2\theta}{\cos\theta\sin\theta}d\theta\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\frac{1}{4}\int_1^5\frac{\cos\theta}{\sin^2\theta}d\theta\\ {/eq}

Doing a change of variable:

{eq}\displaystyle\,w= \sin\theta\\ \displaystyle\,dw=\cos\theta\,d\theta\\ {/eq}

Substituting in the integral:

{eq}\displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\frac{1}{4}\int_1^5\frac{dw}{w^2}\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\frac{1}{4}\int_1^5 w^{-2}dw\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=-\frac{1}{4}\left(\frac{1}{w}\right)\bigg|_{1}^{5}\\ {/eq}

But:

{eq}\displaystyle\,w=\sin\theta\\ {/eq}

And

{eq}\displaystyle\sin\theta=\frac{x}{\sqrt{4+x^2}}\\ {/eq}

We take the original integral and substitute:

{eq}\displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=-\frac{1}{4}\left(\frac{1}{\frac{x}{\sqrt{4+x^2}}}\right)\bigg|_{1}^{5}\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=-\frac{1}{4}\left(\frac{\sqrt{4+x^2}}{x}\right)\bigg|_{1}^{5}\\ {/eq}

Applying the fundamental theorem of calculus:

{eq}\displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=-\frac{1}{4}\left(\frac{\sqrt{4+(5)2}}{5}- \frac{\sqrt{4+(1)^2}}{(1)}\right)\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=-\frac{1}{4}\left(\frac{\sqrt{29}}{5}-\sqrt{5}\right)\\ \displaystyle \int_1^5\frac{dx}{x^2\sqrt{4+x^2}}=\boxed{\frac{1}{4}\sqrt{5}-\frac{\sqrt{29}}{20}} {/eq}


Learn more about this topic:

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Evaluating Definite Integrals Using the Fundamental Theorem

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Chapter 16 / Lesson 2
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In calculus, the fundamental theorem is an essential tool that helps explain the relationship between integration and differentiation. Learn about evaluating definite integrals using the fundamental theorem, and work examples to gain understanding.


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