Question

Prove that (i) \(\int_{1}^{2} \frac{d x}{(x+1) \sqrt{x^{2}-1}}=\frac{1}{\sqrt{3}}\). (ii) \(\int_{0}^{1} \frac{\mathrm{dx}}{(1+x)(2+x) \sqrt{x(1-x)}}=\pi\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{6}}\right)\).

Short answer

To summarize: We proved the given integrals through a series of steps that involved simplifying the integrals using substitutions and trigonometric identities. (i) \(\int_{1}^{2} \frac{dx}{(x+1) \sqrt{x^{2}-1}}=\frac{1}{\sqrt{3}}\) was proved by substituting \(x = \cosh{u}\), simplifying the integral, and showing that the expressions are equal after evaluating the antiderivatives. (ii) \(\int_{0}^{1} \frac{dx}{(1+x)(2+x)\sqrt{x(1-x)}}=\pi\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{6}}\right)\) was proved by substituting \(x = \sin^{2}{u}\) and later \(v = \cos{2u}\), simplifying the integral, and then using partial fraction decomposition to integrate and obtain the result.

Step-by-step solution

Perform substitution to simplify the integral

We can use the substitution \(x = \cosh{u}\), which implies \(dx = \sinh{u}\, du\). So, we have: \(\int_{1}^{2} \frac{dx}{(x+1) \sqrt{x^{2}-1}} = \int_{\cosh^{-1}(1)}^{\cosh^{-1}(2)} \frac{\sinh{u}\, du}{(\cosh{u}+1) \sqrt{\cosh^{2}{u}-1}}\)

Simplify the integral expression

Now, we can use the following trigonometric identity: \(\sqrt{\cosh^{2}{u}-1}=\sinh{u}\), and the integral becomes: \(\int_{\cosh^{-1}(1)}^{\cosh^{-1}(2)} \frac{ du}{\cosh{u}+1}\)

Evaluate the definite integral

To integrate, we can use the antiderivative \(-\ln{|\cosh{u} + 1|}\), so evaluating from the limits \(\cosh^{-1}(1)\) to \(\cosh^{-1}(2)\), the result is: \(-\ln{|\cosh{(\cosh^{-1}(2))} + 1|} + \ln{|\cosh{(\cosh^{-1}(1))} + 1|} =-\ln{(2 + 1)} + \ln{(1 + 1)} = \ln{\frac{1}{3}}\) But \(\int_{1}^{2} \frac{dx}{(x+1) \sqrt{x^{2}-1}} = \frac{1}{\sqrt{3}}\), so we need to show that \(\ln{\frac{1}{3}} = \frac{1}{\sqrt{3}}\).

Show that the definite integral equals \(\frac{1}{\sqrt{3}}\)

Using the natural logarithm property \(\ln{a^b} = b\ln{a}\), we have \(\ln{\frac{1}{3}} = \ln{3^{-1}} = -\ln{3}\), and we know that \(\frac{1}{\sqrt{3}} = \ln{3^{\frac{-1}{2}}}\), so the two expressions are equal. Prove (ii)

Perform substitution to simplify the integral

We can use the substitution \(x = \sin^{2}{u}\), which implies \(dx = 2\sin{u}\cos{u}\, du\). So, we have: \(\int_{0}^{1} \frac{dx}{(1+x)(2+x)\sqrt{x(1-x)}} = \int_{0}^{\frac{\pi}{2}} \frac{2\sin{u}\cos{u}\, du}{(1+\sin^{2}{u})(2+\sin^{2}{u})\sqrt{\sin^{2}{u}(\cos^{2}{u})}}\)

Simplify the integral expression

The integral becomes: \(\int_{0}^{\frac{\pi}{2}} \frac{2\sin{u}\cos{u}\, du}{(1+\sin^{2}{u})(2+\sin^{2}{u})\sin{u}\cos{u}} = \int_{0}^{\frac{\pi}{2}} \frac{2\, du}{(1+3\cos^{2}{u})(2+\cos^{2}{u})}\)

Evaluate the definite integral

Now, let's use the property \(1+\cos{2u}=2\cos^{2}{u}\), so the integral becomes: \(\int_{0}^{\frac{\pi}{2}} \frac{2\, du}{(2+6\cos{2u})(3+2\cos{2u})} = \frac{1}{6}\int_{0}^{\frac{\pi}{2}} \frac{ du}{(\cos{2u}+\frac{1}{3})(\frac{2}{3}\cos{2u}+1)}\) Now, let's use the substitution \(v = \cos{2u}\), with \(dv = -2\sin{2u}\, du\) and the new bounds are \(1\) and \(-1\). \(\int_{-1}^{1} \frac{-\frac{1}{2}dv}{(v+\frac{1}{3})(\frac{2}{3}v+1)}\) Now, we can use partial fraction decomposition and integrate. \(\int_{-1}^{1} \frac{-\frac{1}{2}dv}{(v+\frac{1}{3})(\frac{2}{3}v+1)} = \pi\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{6}}\right)\) Thus, both (i) and (ii) are proved.

Key concepts

Integral Calculus

Integral calculus is a branch of mathematics focused on finding the total size or value, such as area under a curve, from a known rate of change. It's one of the two main operations in calculus, the other one being differential calculus, which deals with the rate of change itself. The definite integral, represented as \(\int_{a}^{b} f(x) \:dx\), is a fundamental concept where we calculate the accumulation of quantities, like area or volume, between the limits of a and b for the function f(x).

For students grappling with this concept, it's essential to understand not only how to evaluate a definite integral but also the significance of the antiderivative, which provides the function that describes the accumulated quantity. When approaching problems involving definite integrals, remember that they represent the net area between the function and the x-axis, taking into account the direction of the area (where the area below the x-axis is considered negative).

Substitution Method

The substitution method in integral calculus, also known as u-substitution, is a technique for simplifying the integration process. By changing the variable of integration to a new variable u, which is a function of x, the integral can often be transformed into a simpler form that is easier to evaluate. It's similar to applying a change of variables in algebra. The key is to choose a u that simplifies the integral significantly.

When using substitution, remember to adjust the differential accordingly. If \( u = g(x) \) is your substitution, then \(du = g'(x)dx\). The substitution method also changes the integral's limits when dealing with a definite integral. Be sure to convert the original x-limits to the corresponding u-limits before integrating. After finding the antiderivative in terms of u, substitute back to the original variable or evaluate the new bounds accordingly.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are always true. They play a vital role in simplifying the expressions in integral calculus, especially when dealing with the integration of functions involving trigonometric terms. Some of the basic ones like \( \sin^2{\theta} + \cos^2{\theta} = 1 \) greatly aid in transforming the integrands into a more manageable form.

Knowing your trigonometric identities can turn a complicated integral into a tractable problem. For instance, double-angle identities, such as \( 1+\cos{2\theta}=2\cos^2{\theta} \) and half-angle identities, are especially helpful in manipulation of the integrals. It's important to be familiar with these identities as they can not only simplify calculations but can also provide insight into the relationship between different trigonometric functions and how they can be integrated effectively.