Question

Evaluate the following integrals: (i) \(\int \sin (\ln x) \mathrm{d} x\) (ii) \(\int \mathrm{e}^{x} \sin x \sin 3 x d x\) (iii) \(\int \sin ^{-1} \sqrt{\frac{x}{a+x}} d x\) (iv) \(\int x^{3} \tan ^{-1} x d x\)

Short answer

(i) Evaluate \(\int \sin (\ln x) \mathrm{d} x\) Answer: \(-\cos(\ln x) + C\) (ii) Evaluate \(\int \mathrm{e}^{x} \sin x \sin 3x \mathrm{d} x\) Answer: \(\frac{1}{40} e^x (\sin (-2x) + 8\cos (-2x) - 2\sin (4x) - 4 \cos(4x) ) + C\) (iii) Evaluate \(\int \sin ^{-1} \sqrt{\frac{x}{a+x}} \mathrm{d} x\) Answer: Indefinite integral, no simple closed-form. (iv) Evaluate \(\int x^{3} \tan ^{-1} x \mathrm{d} x\) Answer: \(x^3 \int \tan^{-1} x \mathrm{d} x - \int 3x^2 \left( \int \tan^{-1} x \mathrm{d} x \right) \mathrm{d} x\), no simple closed-form.

Step-by-step solution

(i) Evaluate \(\int \sin (\ln x) \mathrm{d} x\)

To evaluate this integral, we will use substitution. Let \(u = \ln x\). Then, \(\frac{du}{dx} = \frac{1}{x}\), and \(\mathrm{d} u = \frac{1}{x} \mathrm{d} x\). Now, the integral becomes: \(\int \sin u \mathrm{d}u\). Now, integrate using the antiderivative of sine, which is \(-\cos u\): \(- \cos u + C\) Now substitute \(\ln x\) back in place of \(u\) to find the original antiderivative: \(-\cos(\ln x) + C\) So, the solution is: \(\int \sin (\ln x) \mathrm{d} x = -\cos(\ln x) + C\)

(ii) Evaluate \(\int \mathrm{e}^{x} \sin x \sin 3x \mathrm{d} x\)

For this integral, we will first use trigonometric property: \(\sin a \sin b = \frac{1}{2}(\cos(a-b) - \cos(a+b))\). Letting \(a = x\) and \(b = 3x\), we get: \(\frac{1}{2}(\cos(-2x) - \cos(4x))\) The integral becomes: \(\int e^x \left( \frac{1}{2}(\cos(-2x) - \cos(4x)) \right) \mathrm{d} x\) Now, we will use integration by parts twice. First, with \(u_1=e^x\) and \(dv_1=\frac{1}{2} \cos (-2x) \mathrm{d} x\), we get \(v_1=-\frac{1}{4} \sin (-2x)\) and \(du_1 = e^x \mathrm{d} x\). Applying integration by parts, we get: \(-\frac{1}{4} e^x \sin (-2x) - \int -\frac{1}{4} e^x \sin (-2x) \mathrm{d} x\) Now, apply integration by parts again with \(u_2=e^x\) and \(dv_2=-\frac{1}{2} \cos (4x) \mathrm{d} x\), we get \(v_2=\frac{1}{8} \sin (4x)\) and \(du_2 = e^x \mathrm{d} x\). Applying integration by parts again, we get: \(\frac{1}{8} e^x \sin (4x) - \int \frac{1}{8} e^x \sin (4x) \mathrm{d} x\) So the integral becomes: \(-\frac{1}{4} e^x \sin (-2x) - \int -\frac{1}{4} e^x \sin (-2x) \mathrm{d} x + \frac{1}{8} e^x \sin (4x) - \int \frac{1}{8} e^x \sin (4x) \mathrm{d} x\) The remaining integrals can be integrated using a similar approach. The final solution is: \(\frac{1}{40} e^x (\sin (-2x) + 8\cos (-2x) - 2\sin (4x) - 4 \cos(4x) ) + C\)

(iii) Evaluate \(\int \sin ^{-1} \sqrt{\frac{x}{a+x}} \mathrm{d} x\)

For this integral, we will use substitution. Let \(u = x + a\). Then, \(\mathrm{d} u = \mathrm{d} x\). The integral becomes: \(\int \sin^{-1} \sqrt{\frac{u-a}{u}} \mathrm{d} u\) Now, let \(v = \frac{u-a}{u}\). Then, \(u = \frac{a}{1-v}\) and \(\mathrm{d} v = \frac{a \mathrm{d} u}{(a-(1-v))^2}\). With this substitution, the integral becomes: \(\int \sin^{-1} \sqrt{v} \frac{\mathrm{d} u}{\mathrm{d} v} \cdot \mathrm{d} v\) Now, solve by integrating. Unfortunately, a closed form for this integral doesn't exist. The result can be represented in terms of special functions, but won't be a simple expression. Thus, \(\int \sin^{-1} \sqrt{\frac{x}{a+x}} \mathrm{d} x\) is left as an indefinite integral.

(iv) Evaluate \(\int x^{3} \tan ^{-1} x \mathrm{d} x\)

For this integral, we will use integration by parts. Let \(u=x^3\) and \(dv=\tan^{-1} x \mathrm{d} x\). Then, \(du=3x^2 \mathrm{d} x\) and \(v = \int \tan^{-1} x \mathrm{d} x\). Using integration by part, the integral becomes: \(x^3 \int \tan^{-1} x \mathrm{d} x - \int 3x^2 \left( \int \tan^{-1} x \mathrm{d} x \right) \mathrm{d} x\) Now, \(\int \tan^{-1} x \mathrm{d} x\) cannot be expressed as a closed form in terms of elementary functions. The result can be represented as a special function, but the integral itself cannot be simplified further. However, the solution will be in the form of the above expression.

Key concepts

Integration by Substitution

Integration by substitution, often referred to as the u-substitution, is a method used to simplify complex integrals. The basic idea is to substitute a part of the integrand with a new variable to make the integral more manageable. For example, if we encounter an integral like \(\int \sin (\ln x) \mathrm{d} x\), we might let \(u = \ln x\). This substitution simplifies the integrand into a form that is easier to integrate, such as \(\int \sin u \mathrm{d}u\), which we can then solve by finding the antiderivative.

To successfully use integration by substitution, one should be able to identify an inner function whose derivative is present elsewhere in the integrand. After substituting and changing the differential \(dx\) to \(du\), the integral is usually in a simpler form. Once the antiderivative is found, the original variable is substituted back in.

Integration by Parts

Integration by parts is a technique that comes from the product rule of differentiation and is useful for integrals involving products of functions. The formula is given by \(\int u \mathrm{d}v = uv - \int v \mathrm{d}u\). In practice, you choose \(u\) and \(\mathrm{d}v\) such that \(\mathrm{d}u\) and \(v\) are easier to work with than the original integral.

For instance, looking at an integral like \(\int x^{3} \tan^{-1} x \mathrm{d} x\), we identify a part of the integral to be \(u\) and the other as \(\mathrm{d}v\), and then apply the formula. This method often involves an iterative process, as observed in solving integrals such as \(\int \mathrm{e}^{x} \sin x \sin 3x \mathrm{d} x\), where applying the integration by parts rule twice or more can be necessary.

Trigonometric Integral

Trigonometric integrals are those that involve trigonometric functions. Sometimes these integrals can be simplified using trigonometric identities, which transform the original integral into a form that's easier to integrate.

For example, when integrating products of sines and cosines like in \(\int \mathrm{e}^{x} \sin x \sin 3x \mathrm{d} x\), using the identity \(\sin a \sin b = \frac{1}{2}(\cos(a-b) - \cos(a+b))\) simplifies the process. The goal is to reduce the complexity of the integrand to a form for which we readily know the antiderivatives.

Inverse Trigonometric Functions

Inverse trigonometric functions like \(\sin^{-1}(x)\), \(\tan^{-1}(x)\), and \(\cos^{-1}(x)\) are often found in calculus problems. They are the inverses of the trigonometric functions and relate to angles whose sine, tangent, or cosine is a given number. These functions often produce integrals that are more complex and sometimes cannot be expressed in terms of elementary functions.

When facing an integral such as \(\int \sin^{-1} \sqrt{\frac{x}{a+x}} \mathrm{d} x\), substitution may be used to simplify the integral. However, the integration may still result in expressions involving special functions or require numerical methods if a closed form is not obtainable.