Question

Evaluate the following integrals: (i) \(\int \sec ^{-1} x\) (ii) \(\int \mathrm{x} \sin ^{-1} \mathrm{x}\) (iii) \(\int \sin ^{-1} \sqrt{x}\) (iv) \(\int \sin ^{-1} \frac{2 x}{1+x^{2}} d x\)

Short answer

\Rightarrow x = \sin^2{t}$. We find the differential of x with respect to t: \(dx = 2\sin{t}\cos{t}\,dt\) Now, substitute x and dx in the original integral: \(\int \sin^{-1} \sqrt{x}\, dx = \int t \cdot 2\sin{t}\cos{t}\,dt\) The integral simplifies to: \(\int 2t\sin{t}\cos{t}\,dt\) We use integration by parts to solve this integral. Let: \(u = t\) \(dv = 2\sin{t}\cos{t}\,dt\) Then, find the derivatives and antiderivatives: \(du = dt\) \(v = \int 2\sin{t}\cos{t}\,dt\) We recognize the integral for v as the double-angle identity for sine: \(\sin{2t} = 2\sin{t}\cos{t}\). Thus, \(v = \int \sin{2t}\,dt = -\frac{1}{2}\cos{2t} + C\) Now, we apply the integration by parts formula: \(\int 2t\sin{t}\cos{t}\, dt = t(-\frac{1}{2}\cos{2t}) - \int (-\frac{1}{2}\cos{2t})\,dt\) The right side simplifies to: \(-\frac{1}{2}t\cos{2t} + \frac{1}{4}\int \cos{2t}\,dt\) Now, we integrate the remaining term: \(\frac{1}{4}\int \cos{2t}\,dt = \frac{1}{8}\sin{2t} + C\) Adding this result back to the previous expression, we get: \(\int \sin^{-1} \sqrt{x}\, dx = -\frac{1}{2}t\cos{2t} + \frac{1}{8}\sin{2t} + C\) Finally, we back-substitute \(t = \sin^{-1}\sqrt{x}\) and \(x = \sin^2{t}\): \(\int \sin^{-1} \sqrt{x}\, dx = -\frac{1}{2}\sin^{-1}\sqrt{x}\cos{(2\sin^{-1}\sqrt{x})} + \frac{1}{8}\sin{(2\sin^{-1}\sqrt{x})} + C\)

Step-by-step solution

(i) Evaluate \(\int \sec ^{-1} x\) dx

To evaluate this integral, we will use integration by parts. Integration by parts is given by: \(\int u\,dv = uv - \int v\,du\) We choose the following functions: \(u = \sec^{-1}x\) \(dv = dx\) Then, find the derivatives and antiderivatives: \(du = \frac{dx}{|x|\sqrt{x^2-1}}\) \(v = x\) Now we can apply the integration by parts formula: \(\int \sec ^{-1} x dx = x\sec^{-1}x - \int x\frac{dx}{|x|\sqrt{x^2-1}}\) We can rewrite the integral on the right side as follows: \({\int \frac{x^2}{|x|\sqrt{x^2-1}} dx}\) To solve this integral, we can use a substitution method. We substitute: \(t = x^2\) \(dt = 2x\,dx\) Our integral becomes: \(\frac{1}{2} \int \frac{t}{|t|\sqrt{t-1}} dt\) Now it is easy to integrate: \(\frac{1}{2} \int \frac{t}{|t|\sqrt{t-1}} dt = \frac{1}{2} \big[ 2|t| \sin^{-1}{\sqrt{t-1}} + C \big]\) Substituting back \(t=x^2\), we obtain the final result: \(\int \sec ^{-1} x dx = x\sec^{-1}x - x |\sin^{-1}{\sqrt{x^2- 1}}| + C\)

(ii) Evaluate \(\int \mathrm{x} \sin ^{-1} \mathrm{x}\) dx

To evaluate this integral, we will use integration by parts. We choose the following functions: \(u = x\) \(dv = \sin^{-1}x\,dx\) Now we find the derivatives and antiderivatives: \(du = dx\) \(v = \int \sin^{-1}x\,dx\) We will first find the integral \(v\). For that, let's consider the following substitution: Let \(y = \sin^{-1}x \Rightarrow x = \sin y \Rightarrow dx = \cos y\,dy\) Now, \(v = \int \sin^{-1} x\, dx = \int y \cos y\,dy\) We will again use integration by parts to evaluate this integral. \(w = y\) \(dw = dy\) \(p = \cos y\) \(dp = -\sin y\, dy\) Now, apply integration by parts: \(\int y \cos y dy = y \sin y + \int \sin y dy\) \(= y\sin y - (-\cos y) + C\) \(= y\sin y + \cos y + C\) Substituting back \(y = \sin^{-1}x\) and \(x=\sin y\): \(v = \int \sin^{-1} x dx = \sin^{-1}x (\sin(\sin^{-1}x)) + \cos(\sin^{-1}x) + C\) \(\Rightarrow v = x\sin^{-1}x + \sqrt{1-x^2} + C\) Now we can apply the integration by parts formula for the original integral: \(\int x \sin^{-1}x\, dx = xv - \int v\,du\) \(= x(x\sin^{-1}x + \sqrt{1-x^2} + C) - \int (x\sin^{-1}x + \sqrt{1-x^2}) dx\) Now, we evaluate the remaining integral \(- \int (x\sin^{-1}x + \sqrt{1-x^2}) dx = -\int x\sin^{-1}x dx - \int \sqrt{1-x^2} dx\) Notice that the first integral on the right side is the one we want to evaluate. Let's call it \(I\). Then \(-I = -\int x\sin^{-1}x dx - \int \sqrt{1-x^2} dx\) We can use the following trigonometric substitution for the remaining integral: \(x = \sin \theta \Rightarrow dx = \cos \theta d\theta\) Hence, \(\int \sqrt{1-x^2}dx = \int \cos^2 \theta d\theta\) Now, we apply the double-angle identity for \(\cos 2\theta = 2\cos^2\theta -1\), which gives \(\cos^2\theta = \frac{1}{2}(1+\cos 2\theta)\) So, \(\int \sqrt{1-x^2}dx = \int \frac{1}{2}(1+\cos 2\theta) d\theta\) This integral evaluates to: \(\frac{1}{2}(\theta + \frac{1}{2}\sin2\theta) + C\) Now, we back-substitute \(x = \sin\theta\) and \(\theta=\sin^{-1}x\): \(\frac{1}{2}(\sin^{-1}x + \frac{1}{2}x\sqrt{1-x^2}) + C\) Add this result back to the equation for \(-I\): \(I = x(x\sin^{-1}x + \sqrt{1-x^2} + C) - \frac{1}{2}(\sin^{-1}x + \frac{1}{2}x\sqrt{1-x^2}) - C\) Thus, \(\int x\sin^{-1}x dx = \frac{1}{2}x^2\sin^{-1}x + \frac{1}{2}x\sqrt{1-x^2} - \frac{1}{4}x^2\sqrt{1-x^2} + C\)

(iii) Evaluate \(\int \sin ^{-1} \sqrt{x}\) dx

Let $t = \sin^{-1}\sqrt{x} \Rightarrow \sin{t} = \sqrt{x}...

Key concepts

Integration by Parts

Integration by parts is a fundamental technique in integral calculus which allows us to integrate products of functions. The formula for integration by parts is derived from the product rule for differentiation and is stated as \( \int u\,dv = uv - \int v\,du \). The main idea is to split the integral into two parts: one part is easy to integrate, and the other becomes simpler after applying the formula.

To use integration by parts effectively, we choose \( u \) and \( dv \) such that \( du \) and \( v \) are easy to determine. Generally, the choice of \( u \) is a function that becomes simpler upon differentiation (like logarithmic or inverse trigonometric functions), while \( dv \) is something that is straightforward to integrate (such as polynomials or exponential functions).

In the original exercise, integration by parts is applied by choosing the inverse trigonometric function as \( u \) and \( dx \) as \( dv \) because the inverse trigonometric function simplifies when differentiated. Although the use of integration by parts may sometimes lead to an integral that appears just as complex as the original one, strategic manipulation and further application of techniques can simplify it to a solvable state.

It's worth noting that sometimes this technique needs to be applied more than once, and occasionally, it helps to introduce an auxiliary equation from which the original integral can be deduced. Overall, mastering integration by parts requires practice and a keen eye for identifying the most strategic choices for \( u \) and \( dv \) to effectively break down complex integrals.

Inverse Trigonometric Integrals

Integrals involving inverse trigonometric functions can often be challenging due to their non-standard forms. However, they play a pivotal role in calculus because they allow for the integration of expressions that correspond to angles whose trigonometric functions have known values.

Strictly speaking, an inverse trigonometric integral is one which involves one of the six inverse trigonometric functions: \( \arcsin(x) \), \( \arccos(x) \), \( \arctan(x) \), \( \arccot(x) \), \( \arcsec(x) \), or \( \arccsc(x) \). These functions are used to find angles whose sines, cosines, tangents, cotangents, secants, or cosecants are given.

The integral of an inverse trigonometric function typically requires special techniques of integration, such as integration by parts or trigonometric substitution. Inverse trigonometric functions have specific derivatives that are useful in both differentiation and integration. For example, the derivative of \( \arcsin(x) \) is \( \frac{1}{\sqrt{1-x^2}} \), which can be quite handy when integrating an expression that appears to be its differential counterpart.

The exercise showcases the integration of \( \sec^{-1}(x) \) using integration by parts. It lends itself to this technique because the derivative of the inverse secant is easier to manipulate than the function itself. The key to handling inverse trigonometric integrals is recognizing patterns that might simplify the task and often require conversion to a basic form where standard integration techniques can be applied.

Trigonometric Substitution

Trigonometric substitution is a technique used to evaluate integrals that involve square roots of quadratic expressions, such as \( \sqrt{a^2 - x^2} \) or \( \sqrt{x^2 + a^2} \). It involves substituting a trigonometric expression for \( x \) which simplifies the integral into a more manageable form involving standard trigonometric integrals.

For instance, when integrating a function that contains \( \sqrt{1 - x^2} \), we could substitute \( x = \sin(\theta) \) because the resulting expression \( \sqrt{1 - \sin^2(\theta)} \) simplifies to \( \cos(\theta) \) due to the Pythagorean identity. Substituting back into the original integral transforms it into a basic trigonometric integral.

This method is evocative of solving a puzzle: find the right trigonometric expression to substitute so that the integral becomes a familiar one. After performing the integration with respect to the trigonometric function, we must remember to convert back to the original variable using inverse trigonometric functions or triangles to complete the process.

In the exercise given, trigonometric substitution helped us with \( \int \sqrt{1-x^2}dx \) by substituting \( x \) with \( \sin(\theta) \) to make the integral less intimidating. It's impressive how trigonometric identities and substitutions can unravel even the most intricate integrals into solvable exercises. As with other techniques, practice is crucial - the more one works with trigonometric substitutions, the quicker one identifies when and how to apply them effectively.