CBSE Class 12 Mathematics Integrals Notes Set 03

Important Notes and Formulas

Indefinite Integral: Definition

If f(x) is the derivative of F(x), then F(x) is said to be an anti-derivative or integral or primitive of f(x). Therefore differentiation and integration are inverse processes of each other.
If \( \frac{d}{dx}[F(x)] = f(x) \) then \( \int f(x)\,dx = F(x) \)
Here the sign \( \int \) is used for integration, the letter x in dx denotes the variable of integration, f(x) is said to be the integrand and F(x) is the integral.

Integration Constant and Indefinite Integral

We know that the derivative of any constant is zero, therefore \( \frac{d}{dx}(C) = 0 \), where C is constant.
If \( \frac{d}{dx}[F(x)] = f(x) \) then \( \frac{d}{dx}[F(x) + C] = f(x) \)
Integrating both sides w.r.t. x, we get \( \int \left[\frac{d}{dx}\{F(x) + C\}\right] dx = \int f(x)\,dx \)
⇒ \( F(x) + C = \int f(x)\,dx \Rightarrow \int f(x)\,dx = F(x) + C \)
Here, C is said to be the constant of integration and the value of C is not definite, so F(x) + C is called an indefinite integral of f(x).
Note: It may happen that, by different methods of integration, we obtain different integrals of the same function but they differ from each other by a constant.

Standard Formulae of Integration

Basic Theorems on Integration

If f(x), g(x) are two functions of a variable x and k is a constant, then
(i) \( \int k\,f(x)\,dx = k\int f(x)\,dx \)
(ii) \( \int [f(x) \pm g(x)]\,dx = \int f(x)\,dx \pm \int g(x)\,dx \)
(iii) \( \frac{d}{dx}\left(\int f(x)\,dx\right) = f(x) \)
(iv) \( \int \left(\frac{d}{dx} f(x)\right) dx = f(x) + C \)
(v) If \( \int f(x)\,dx = \phi(x) + C \), then \( \int f(ax + b)\,dx = \frac{1}{a}\phi(ax + b) + C \)
\( \left(\because \frac{d}{dx}\left\{\frac{1}{a}\phi(ax + b)\right\} = \phi'(ax + b) = f(ax + b)\right) \)

Methods of Integration

(1) Integration by Substitution
(a) When the integrand is the product of two factors such that one is the derivative of the other, i.e.
\( I = \int f(x)\,f'(x)\,dx \)
In this case, we put f(x) = t to convert it into a standard integral.
(b) When the integrand is the product of two factors such that one is a derivative of a function and the other is defined in terms of that function.
If \( \int f(x)\,dx = g(x) \), then \( \int f[\phi(x)]\,\phi'(x)\,dx = g[\phi(x)] + C \)
Here, we put \( \phi(x) = t \) so that \( \phi'(x)\,dx = dt \).
(c) Some standard forms of integrals:
The following two forms are very useful to write the integral directly.
(i) \( \int [f(x)]^n f'(x)\,dx = \frac{[f(x)]^{n+1}}{n+1} + C,\ n \neq -1 \)    (ii) \( \int \frac{f'(x)}{f(x)}\,dx = \log|f(x)| + C \)

(d) Standard formulae:
(i) \( \int \tan x\,dx = \log|\sec x| + C = -\log|\cos x| + C \)
(ii) \( \int \cot x\,dx = \log|\sin x| + C = -\log|\text{cosec}\,x| + C \)
(iii) \( \int \sec x\,dx = \log|\sec x + \tan x| + C = -\log|\sec x - \tan x| + C = \log\tan\left(\frac{\pi}{4} + \frac{x}{2}\right) + C \)
(iv) \( \int \text{cosec}\,x\,dx = -\log|\text{cosec}\,x + \cot x| + C = \log|\text{cosec}\,x - \cot x| + C = \log\tan\left(\frac{x}{2}\right) + C \)

(e) Standard substitutions:
The following standard substitutions will be useful:
(i) Integrand form \( \sqrt{a^2 - x^2} \) or \( \frac{1}{\sqrt{a^2 - x^2}} \): substitute \( x = a\sin\theta \) or \( x = a\cos\theta \)
(ii) Integrand form \( \sqrt{x^2 + a^2} \) or \( \frac{1}{\sqrt{x^2 + a^2}} \): substitute \( x = a\tan\theta \) or \( x = a\cot\theta \)
(iii) Integrand form \( \sqrt{x^2 - a^2} \) or \( \frac{1}{\sqrt{x^2 - a^2}} \): substitute \( x = a\sec\theta \) or \( x = a\,\text{cosec}\,\theta \)

(f) Integrals of some particular functions:
(i) \( \int \frac{1}{x^2 - a^2}\,dx = \frac{1}{2a}\log\left|\frac{x - a}{x + a}\right| + C\ (x \gt a) \)
(ii) \( \int \frac{1}{a^2 - x^2}\,dx = \frac{1}{2a}\log\left|\frac{a + x}{a - x}\right| + C\ (x \lt a) \)
(iii) \( \int \frac{1}{x^2 + a^2}\,dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C = -\frac{1}{a}\cot^{-1}\left(\frac{x}{a}\right) + C \)
(iv) \( \int \frac{1}{x\sqrt{x^2 - a^2}}\,dx = \frac{1}{a}\sec^{-1}\frac{x}{a} + C \)
(v) \( \int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \sin^{-1}\left(\frac{x}{a}\right) + C = -\cos^{-1}\left(\frac{x}{a}\right) + C \)
(vi) \( \int \frac{1}{\sqrt{x^2 + a^2}}\,dx = \log\left|x + \sqrt{x^2 + a^2}\right| + C \)
(vii) \( \int \frac{1}{\sqrt{x^2 - a^2}}\,dx = \log\left|x + \sqrt{x^2 - a^2}\right| + C \)

(viii) To find the integral of the type \( \int \frac{1}{ax^2 + bx + c}\,dx \),
we write \( ax^2 + bx + c = a\left[x^2 + \frac{b}{a}x + \frac{c}{a}\right] = a\left[\left(x + \frac{b}{2a}\right)^2 + \left(\frac{c}{a} - \frac{b^2}{4a^2}\right)\right] \)
Now, put \( x + \frac{b}{2a} = t \), so that dx = dt, and writing \( \frac{c}{a} - \frac{b^2}{4a^2} = \pm k^2 \),
we find the integral reduced to the form \( \frac{1}{a}\int \frac{dt}{t^2 \pm k^2} \) depending upon the sign of \( \left(\frac{c}{a} - \frac{b^2}{4a^2}\right) \) and hence it can be evaluated.
(ix) To find the integral of the type \( \int \frac{px + q}{ax^2 + bx + c}\,dx \), where p, q, a, b, c are constants, we are to find real numbers A, B such that
\( px + q = A\frac{d}{dx}(ax^2 + bx + c) + B = A(2ax + b) + B \)
To determine A and B, we equate the coefficients of x and the constant terms from both sides. A and B are thus obtained and hence the integral is reduced to one of the known forms.

(2) Integration by Partial Fractions
Let the integrand be of the form \( \frac{f(x)}{g(x)} \), where both f(x) and g(x) are polynomials. If the degree of f(x) is equal to or greater than the degree of g(x), then first divide f(x) by g(x) till the degree of the remainder becomes less than the degree of g(x). Let Q(x) be the quotient and R(x) the remainder, then
\( \frac{f(x)}{g(x)} = Q(x) + \frac{R(x)}{g(x)} \)
Now in \( \frac{R(x)}{g(x)} \), factorize g(x) and then write partial fractions in the following manner:
(a) For every non-repeated linear factor in the denominator, write
\( \frac{1}{(x - a)(x - b)} = \frac{A}{(x - a)} + \frac{B}{(x - b)} \)
(b) For repeated linear factors in the denominator, write
\( \frac{1}{(x - a)^3(x - b)} = \frac{A}{(x - a)} + \frac{B}{(x - a)^2} + \frac{C}{(x - a)^3} + \frac{D}{(x - b)} \)
(c) For every non-repeated quadratic factor in the denominator, write
\( \frac{1}{(ax^2 + bx + c)(x - d)} = \frac{Ax + B}{(ax^2 + bx + c)} + \frac{C}{(x - d)} \)

(3) Integration by Parts
(a) If u and v are differentiable functions of x, then
\( \int u \cdot v\,dx = u\int v\,dx - \int \left[\left(\frac{d}{dx}(u)\right)\left(\int v\,dx\right)\right] dx \)
i.e. Integral of the product of two functions = (first function) × (integral of second function) − Integral of [(derivative of first function) × (integral of second function)]
(i) How to choose the Ist and IInd function: If two functions are of different types, take that function as Ist which comes first in the word ILATE, where I stands for inverse circular function, L stands for logarithmic function, A stands for algebraic functions, T stands for trigonometric and E stands for exponential functions.
(ii) For the integration of logarithmic or inverse trigonometric functions alone, take unity (1) as the second function.

(b) Integral of the type \( \int e^x[f(x) + f'(x)]\,dx \)
We have \( I = \int e^x[f(x) + f'(x)]\,dx = \int e^x f(x)\,dx + \int e^x f'(x)\,dx \)
Taking f(x) and \( e^x \) as the first function and second function respectively in the first integrand and integrating it by parts, we have
\( I = e^x f(x) - \int f'(x)e^x\,dx + \int e^x f'(x)\,dx + C = e^x f(x) + C \)
Thus, \( \int e^x[f(x) + f'(x)]\,dx = e^x f(x) + C \)

(c) Integrals of some more types:
(i) \( \int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C \)
(ii) \( \int \sqrt{x^2 + a^2}\,dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2}\log\left|x + \sqrt{x^2 + a^2}\right| + C \)
(iii) \( \int \sqrt{x^2 - a^2}\,dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\log\left|x + \sqrt{x^2 - a^2}\right| + C \)

Definite Integral: Definition

A definite integral is denoted by \( \int_{a}^{b} f(x)\,dx \), where a is called the lower limit of the integral and b is called the upper limit of the integral. If f has an anti-derivative F in the interval [a, b], then the value of the definite integral is the difference between the values of F at the end points, i.e., F(b) − F(a).

Fundamental Theorem of Integral Calculus

Let f be a continuous function defined on the closed interval [a, b] and F be an anti-derivative of f, then
\( \int_{a}^{b} f(x)\,dx = [F(x)]_{a}^{b} = F(b) - F(a) \)

Properties of Definite Integrals

(P-1) \( \int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(t)\,dt = \int_{a}^{b} f(u)\,du \)
(P-2) \( \int_{a}^{b} f(x)\,dx = -\int_{b}^{a} f(x)\,dx \)
(P-3) \( \int_{a}^{b} f(x)\,dx = \int_{a}^{c} f(x)\,dx + \int_{c}^{b} f(x)\,dx \)

(P-4) \( \int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a + b - x)\,dx \)
Proof: Let x = a + b − t, then dx = −dt
When x = a, t = b and when x = b, t = a
∴ \( \int_{a}^{b} f(x)\,dx = -\int_{b}^{a} f(a + b - t)\,dt \)
\( = \int_{a}^{b} f(a + b - t)\,dt \) (by P-2)
\( = \int_{a}^{b} f(a + b - x)\,dx \) (by P-1)
Particular case: \( \int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx \)

(P-5) \( \int_{-a}^{a} f(x)\,dx = \begin{cases} 2\int_{0}^{a} f(x)\,dx, & \text{if } f(-x) = f(x) \text{ i.e. f(x) is an even function} \\ 0, & \text{if } f(-x) = -f(x) \text{ i.e. f(x) is an odd function} \end{cases} \)

(P-6) \( \int_{0}^{2a} f(x)\,dx = \int_{0}^{a} f(x)\,dx + \int_{0}^{a} f(2a - x)\,dx = \begin{cases} 2\int_{0}^{a} f(x)\,dx, & \text{if } f(2a - x) = f(x) \\ 0, & \text{if } f(2a - x) = -f(x) \end{cases} \)
Proof: \( \int_{0}^{2a} f(x)\,dx = \int_{0}^{a} f(x)\,dx + \int_{a}^{2a} f(x)\,dx \) (by P-3)
In the second integral of the right hand side, put x = 2a − t, then dx = −dt
When x = a, t = a and when x = 2a, t = 0
∴ \( \int_{0}^{2a} f(x)\,dx = \int_{0}^{a} f(x)\,dx - \int_{a}^{0} f(2a - t)\,dt \)
\( = \int_{0}^{a} f(x)\,dx + \int_{0}^{a} f(2a - t)\,dt \) (by P-2)
\( = \int_{0}^{a} f(x)\,dx + \int_{0}^{a} f(2a - x)\,dx \) (by P-1)
\( = \begin{cases} 2\int_{0}^{a} f(x)\,dx, & \text{if } f(2a - x) = f(x) \\ 0, & \text{if } f(2a - x) = -f(x) \end{cases} \)

Solved Examples

Illustration 1. Evaluate the following integrals: [1 Mark each]
(i) \( \int \tan^{-1}\left(\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}\right) dx \)    (ii) \( \int \left(x + \frac{1}{x}\right)\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) dx \)
(iii) \( \int \frac{2x}{(2x + 1)^2}\,dx \)    (iv) \( \int \sin 5x \sin 3x\,dx \)    (v) \( \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos\alpha}\,dx \)
Answer:
(i) \( \int \tan^{-1}\left(\sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}\right) dx = \int \tan^{-1}\left(\sqrt{\frac{2\sin^2 x}{2\cos^2 x}}\right) dx \)
\( = \int \tan^{-1}(\tan x)\,dx \) (We are assuming that \( 0 \lt x \lt \frac{\pi}{2} \))
\( = \int x\,dx = \frac{1}{2}x^2 + C \) [1]
(ii) \( \int \left(x + \frac{1}{x}\right)\left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) dx = \int \left(x\sqrt{x} + \frac{x}{\sqrt{x}} + \frac{\sqrt{x}}{x} + \frac{1}{x\sqrt{x}}\right) dx \)
\( = \int x^{3/2}\,dx + \int x^{1/2}\,dx + \int x^{-1/2}\,dx + \int x^{-3/2}\,dx \)
\( = \frac{2}{5}x^{5/2} + \frac{2}{3}x^{3/2} + 2\sqrt{x} - \frac{2}{\sqrt{x}} + C \) [1]
(iii) \( \int \frac{2x}{(2x + 1)^2}\,dx = \int \frac{(2x + 1) - 1}{(2x + 1)^2}\,dx = \int \left(\frac{1}{2x + 1} - \frac{1}{(2x + 1)^2}\right) dx \)
\( = \frac{\log|2x + 1|}{2} - \frac{(2x + 1)^{-1}}{(-1) \times 2} + C = \frac{1}{2}\left(\log|2x + 1| + \frac{1}{2x + 1}\right) + C \) [1]
(iv) \( \int \sin 5x \sin 3x\,dx = \frac{1}{2}\int 2\sin 5x \sin 3x\,dx = \frac{1}{2}\int (\cos 2x - \cos 8x)\,dx \)
\( = \frac{1}{2}\left(\frac{\sin 2x}{2} - \frac{\sin 8x}{8}\right) + C = \frac{1}{16}(4\sin 2x - \sin 8x) + C \) [1]
(v) \( \int \frac{\cos 2x - \cos 2\alpha}{\cos x - \cos\alpha}\,dx = \int \frac{(2\cos^2 x - 1) - (2\cos^2\alpha - 1)}{\cos x - \cos\alpha}\,dx = 2\int (\cos x + \cos\alpha)\,dx \)
\( = 2\sin x + 2x\cos\alpha + C \) [1]

Illustration 2. If \( f'(x) = x^2 - \frac{1}{x^2} \) and \( f(1) = \frac{1}{3} \), find f(x). [4 Marks]
Answer:
Given \( f'(x) = x^2 - \frac{1}{x^2} \),
\( f(x) = \int \left(x^2 - \frac{1}{x^2}\right) dx \) [By definition] [1]
\( = \int (x^2 - x^{-2})\,dx \)
\( = \frac{x^3}{3} - \frac{x^{-1}}{-1} + C = \frac{1}{3}x^3 + \frac{1}{x} + C \) [1]
Also \( f(1) = \frac{1}{3} \Rightarrow \frac{1}{3} \times 1^3 + \frac{1}{1} + C = \frac{1}{3} \)
⇒ \( \frac{1}{3} + 1 + C = \frac{1}{3} \Rightarrow C = -1 \) [1]
∴ \( f(x) = \frac{1}{3}x^3 + \frac{1}{x} - 1 \) [1]

Illustration 3. Evaluate the following integrals: [1 Mark each]
(i) \( \int \sec^4 x \tan x\,dx \)    (ii) \( \int \frac{\sqrt{\tan x}}{\sin x \cos x}\,dx \)    (iii) \( \int \frac{(x + 1)(x + \log x)^2}{x}\,dx \)    (iv) \( \int \frac{e^{4x}}{(1 + e^{4x})^2}\,dx \)
Answer:
(i) \( \int \sec^4 x \tan x\,dx = \int (\sec x)^3(\sec x \tan x)\,dx = \int (\sec x)^3\left\{\frac{d}{dx}(\sec x)\right\} dx \)
\( = \frac{(\sec x)^4}{4} + C \) [1]
(ii) \( \int \frac{\sqrt{\tan x}}{\sin x \cos x}\,dx = \int \frac{(\tan x)^{\frac{1}{2}}}{\tan x}\sec^2 x\,dx = \int (\tan x)^{-\frac{1}{2}}\left\{\frac{d}{dx}(\tan x)\right\} dx \)
\( = \frac{(\tan x)^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C = 2\sqrt{\tan x} + C \) [1]
(iii) \( \int \frac{(x + 1)(x + \log x)^2}{x}\,dx = \int (x + \log x)^2\left(1 + \frac{1}{x}\right) dx \)
\( = \int (x + \log x)^2 \frac{d}{dx}\{x + \log x\}\,dx \)
\( = \frac{(x + \log x)^3}{3} + C \) [1]
(iv) \( \int \frac{e^{4x}}{(1 + e^{4x})^2}\,dx = \frac{1}{4}\int (1 + e^{4x})^{-2}(4e^{4x})\,dx = \frac{1}{4}\int (1 + e^{4x})^{-2}\left\{\frac{d}{dx}(1 + e^{4x})\right\} dx \)
\( = -\frac{1}{4}(1 + e^{4x})^{-1} + C = -\frac{1}{4(1 + e^{4x})} + C \) [1]

Illustration 4. Evaluate the following integrals: [1 Mark each]
(i) \( \int \tan^3 x\,dx \)    (ii) \( \int \frac{\cos x}{\cos(x - a)}\,dx \)    (iii) \( \int \frac{\cos 2x}{\cos x}\,dx \)
Answer:
(i) \( \int \tan^3 x\,dx = \int \tan x(\sec^2 x - 1)\,dx = \int \tan x \sec^2 x\,dx - \int \tan x\,dx \)
\( = \frac{(\tan x)^2}{2} - (-\log|\cos x|) + C = \frac{1}{2}\tan^2 x + \log|\cos x| + C \) [1]
(ii) \( \int \frac{\cos x}{\cos(x - a)}\,dx = \int \frac{\cos[(x - a) + a]}{\cos(x - a)}\,dx \)
\( = \int [\cos a - \tan(x - a)\sin a]\,dx = \cos a\int 1\,dx - \sin a\int \tan(x - a)\,dx \)
\( = x\cos a + \sin a\log|\cos(x - a)| + C \) [1]
(iii) \( \int \frac{\cos 2x}{\cos x}\,dx = \int \frac{2\cos^2 x - 1}{\cos x}\,dx = \int (2\cos x - \sec x)\,dx \)
\( = 2\sin x - \log|\sec x + \tan x| + C \) [1]

Illustration 5. Evaluate: \( \int \frac{1}{1 + \tan x}\,dx \) [4 Marks]
Answer:
\( \int \frac{dx}{1 + \tan x} = \int \frac{\cos x\,dx}{\cos x + \sin x} \)
\( = \frac{1}{2}\int \frac{(\cos x + \sin x + \cos x - \sin x)\,dx}{\cos x + \sin x} \) [1]
\( = \frac{1}{2}\int dx + \frac{1}{2}\int \frac{\cos x - \sin x}{\cos x + \sin x}\,dx \) [½]
\( = \frac{x}{2} + \frac{C_1}{2} + \frac{1}{2}\int \frac{\cos x - \sin x}{\cos x + \sin x}\,dx \) ...(1) [½]
Now, consider \( I = \int \frac{\cos x - \sin x}{\cos x + \sin x}\,dx \)
Put cos x + sin x = t so that (cos x − sin x) dx = dt
Therefore \( I = \int \frac{dt}{t} = \log|t| + C_2 = \log|\cos x + \sin x| + C_2 \) [1]
Putting it in (1), we get
\( \int \frac{dx}{1 + \tan x} = \frac{x}{2} + \frac{C_1}{2} + \frac{1}{2}\log|\cos x + \sin x| + \frac{C_2}{2} \)
\( = \frac{x}{2} + \frac{1}{2}\log|\cos x + \sin x| + \frac{C_1}{2} + \frac{C_2}{2} \)
\( = \frac{x}{2} + \frac{1}{2}\log|\cos x + \sin x| + C,\ \left(C = \frac{C_1}{2} + \frac{C_2}{2}\right) \) [1]

Illustration 6. Evaluate the following integrals: [4 Marks each]
(i) \( \int \frac{x}{1 - 9x^4}\,dx \)    (ii) \( \int \frac{x^3 + x}{x^4 - 9}\,dx \) [Exemplar]
(iii) \( \int \frac{3x + 1}{x^2 + 2x + 3}\,dx \)    (iv) \( \int \frac{x}{x^4 + x^2 + 1}\,dx \)
(v) \( \int \frac{dx}{\sqrt{(x - \alpha)(\beta - x)}},\ \beta \gt \alpha \) [Exemplar]    (vi) \( \int \frac{1}{\sqrt{(x + 1)(x + 5)}}\,dx \)
Answer:
(i) Put \( x^2 = t \Rightarrow 2x\,dx = dt \Rightarrow x\,dx = \frac{1}{2}dt \),
∴ \( \int \frac{x}{1 - 9x^4}\,dx = \frac{1}{2}\int \frac{dt}{1 - 9t^2} \) [1]
\( = \frac{1}{2} \cdot \frac{1}{9}\int \frac{dt}{\left(\frac{1}{3}\right)^2 - t^2} \) [1]
\( = \frac{1}{18} \cdot \frac{1}{2 \times \frac{1}{3}}\log\left|\frac{\frac{1}{3} + t}{\frac{1}{3} - t}\right| + C = \frac{1}{12}\log\left|\frac{1 + 3t}{1 - 3t}\right| + C \) [1]
\( = \frac{1}{12}\log\left|\frac{1 + 3x^2}{1 - 3x^2}\right| + C \) [1]
(ii) \( \int \frac{x^3 + x}{x^4 - 9}\,dx = \int \frac{x^3}{x^4 - 9}\,dx + \int \frac{x}{x^4 - 9}\,dx \)
\( = \frac{1}{4}\int \frac{4x^3}{x^4 - 9}\,dx + \frac{1}{2}\int \frac{2x}{x^4 - 9}\,dx \) [1]
\( = \frac{1}{4}\log|x^4 - 9| + \frac{1}{2}\int \frac{dt}{t^2 - 9} \) [where \( t = x^2 \)] [1]
\( = \frac{1}{4}\log|x^4 - 9| + \frac{1}{2} \cdot \frac{1}{6}\log\left|\frac{t - 3}{t + 3}\right| + C \) [1]
\( = \frac{1}{4}\log|x^4 - 9| + \frac{1}{12}\log\left|\frac{x^2 - 3}{x^2 + 3}\right| + C \) [1]
(iii) \( \int \frac{3x + 1}{x^2 + 2x + 3}\,dx = \int \frac{\frac{3}{2}(2x + 2) - 2}{x^2 + 2x + 3}\,dx = \frac{3}{2}\int \frac{2x + 2}{x^2 + 2x + 3}\,dx - 2\int \frac{1}{x^2 + 2x + 3}\,dx \) [1]
\( = \frac{3}{2}\log|x^2 + 2x + 3| - 2\int \frac{1}{(x + 1)^2 + 2}\,dx \)
\( = \frac{3}{2}\log(x^2 + 2x + 3) - 2\int \frac{1}{(x + 1)^2 + (\sqrt{2})^2}\,dx \) (∵ \( x^2 + 2x + 3 = (x + 1)^2 + 2 \gt 0 \), ∀ x ∈ R) [1]
\( = \frac{3}{2}\log(x^2 + 2x + 3) - 2 \times \frac{1}{\sqrt{2}}\tan^{-1}\left(\frac{x + 1}{\sqrt{2}}\right) + C \) [1]
\( = \frac{3}{2}\log(x^2 + 2x + 3) - \sqrt{2}\tan^{-1}\left(\frac{x + 1}{\sqrt{2}}\right) + C \) [1]
(iv) Put \( x^2 = t \Rightarrow 2x\,dx = dt \Rightarrow x\,dx = \frac{1}{2}dt \),
∴ \( \int \frac{x}{x^4 + x^2 + 1}\,dx = \frac{1}{2}\int \frac{dt}{t^2 + t + 1} \) [1]
\( = \frac{1}{2}\int \frac{dt}{t^2 + t + \frac{1}{4} + \frac{3}{4}} = \frac{1}{2}\int \frac{dt}{\left(t + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} \) [1]
\( = \frac{1}{2} \cdot \frac{1}{\frac{\sqrt{3}}{2}}\tan^{-1}\left(\frac{t + \frac{1}{2}}{\frac{\sqrt{3}}{2}}\right) + C = \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2t + 1}{\sqrt{3}}\right) + C \) [1]
\( = \frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{2x^2 + 1}{\sqrt{3}}\right) + C \) [1]
(v) Put \( x - \alpha = t^2 \). Then dx = 2t dt. Now
\( I = \int \frac{2t\,dt}{\sqrt{t^2(\beta - \alpha - t^2)}} = \int \frac{2\,dt}{\sqrt{\beta - \alpha - t^2}} \) [1½]
⇒ \( I = 2\int \frac{dt}{\sqrt{k^2 - t^2}} \), where \( k^2 = \beta - \alpha \) [1]
\( = 2\sin^{-1}\frac{t}{k} + C = 2\sin^{-1}\sqrt{\frac{x - \alpha}{\beta - \alpha}} + C \) [1½]
(vi) \( \int \frac{1}{\sqrt{(x + 1)(x + 5)}}\,dx = \int \frac{1}{\sqrt{x^2 + 6x + 5}}\,dx \) [1]
\( = \int \frac{1}{\sqrt{(x^2 + 6x + 9) - 4}}\,dx = \int \frac{1}{\sqrt{(x + 3)^2 - 2^2}}\,dx \) [1]
\( = \log\left|x + 3 + \sqrt{(x + 3)^2 - 2^2}\right| + C \) [1]
\( = \log\left|x + 3 + \sqrt{x^2 + 6x + 5}\right| + C \) [1]

Illustration 7. (a) Evaluate the following integrals: [4 Marks each]
(i) \( \int \frac{x^2\,dx}{x^4 + x^2 - 2} \) [Exemplar]    (ii) \( \int \frac{3x + 1}{(x - 2)^2(x + 2)}\,dx \)
(b) Evaluate the following integrals: [6 Marks each]
(i) \( \int \frac{1}{(e^x + 1)^3}\,dx \)    (ii) \( \int \frac{x^3}{(x - 1)(x - 2)}\,dx \)
Answer:
(a) (i) Let \( x^2 = t \). Then
\( \frac{x^2}{x^4 + x^2 - 2} = \frac{t}{t^2 + t - 2} = \frac{t}{(t + 2)(t - 1)} = \frac{A}{(t + 2)} + \frac{B}{(t - 1)} \)
So t = A(t − 1) + B(t + 2) [1]
Comparing coefficients, we get \( A = \frac{2}{3},\ B = \frac{1}{3} \) [1]
So \( \frac{x^2}{x^4 + x^2 - 2} = \frac{2}{3} \cdot \frac{1}{(x^2 + 2)} + \frac{1}{3} \cdot \frac{1}{(x^2 - 1)} \)
Therefore,
\( \int \frac{x^2}{x^4 + x^2 - 2}\,dx = \frac{2}{3}\int \frac{1}{x^2 + 2}\,dx + \frac{1}{3}\int \frac{dx}{x^2 - 1} \) [1]
\( = \frac{2}{3} \cdot \frac{1}{\sqrt{2}}\tan^{-1}\frac{x}{\sqrt{2}} + \frac{1}{6}\log\left|\frac{x - 1}{x + 1}\right| + C \) [1]
(ii) \( \frac{3x + 1}{(x - 2)^2(x + 2)} = \frac{A}{(x - 2)} + \frac{B}{(x - 2)^2} + \frac{C}{(x + 2)} \)
⇒ \( 3x + 1 = A(x - 2)(x + 2) + B(x + 2) + C(x - 2)^2 \) ...(i) [1]
When x = 2, \( B = \frac{7}{4} \)
and x = −2, \( C = -\frac{5}{16} \) [½]
Equating coefficients of \( x^2 \) on both sides of (i), we get
\( 0 = A + C \Rightarrow A = -C = \frac{5}{16} \) [½]
∴ \( \frac{3x + 1}{(x - 2)^2(x + 2)} = \frac{5}{16} \cdot \frac{1}{(x - 2)} + \frac{7}{4} \cdot \frac{1}{(x - 2)^2} - \frac{5}{16} \cdot \frac{1}{(x + 2)} \)
∴ \( \int \frac{3x + 1}{(x - 2)^2(x + 2)}\,dx = \int \left[\frac{5}{16} \cdot \frac{1}{x - 2} + \frac{7}{4} \cdot \frac{1}{(x - 2)^2} - \frac{5}{16} \cdot \frac{1}{x + 2}\right] dx \)
\( = \frac{5}{16}\int \frac{1}{(x - 2)}\,dx + \frac{7}{4}\int (x - 2)^{-2}\,dx - \frac{5}{16}\int \frac{1}{(x + 2)}\,dx \) [1]
\( = \frac{5}{16}\log|x - 2| + \frac{7}{4} \cdot \frac{(x - 2)^{-1}}{-1} - \frac{5}{16}\log|x + 2| + C \)
\( = \frac{5}{16}\log\left|\frac{x - 2}{x + 2}\right| - \frac{7}{4} \cdot \frac{1}{(x - 2)} + C \) [1]
(b) (i) Put \( e^x = t \Rightarrow e^x\,dx = dt \Rightarrow dx = \frac{1}{t}dt \)
∴ \( \int \frac{1}{(e^x + 1)^3}\,dx = \int \frac{1}{t(t + 1)^3}\,dt \) [1]
Let \( \frac{1}{t(t + 1)^3} = \frac{A}{t} + \frac{B}{(t + 1)} + \frac{C}{(t + 1)^2} + \frac{D}{(t + 1)^3} \)
⇒ \( 1 = A(t + 1)^3 + Bt(t + 1)^2 + Ct(t + 1) + Dt \) ...(i)
When t = 0 ⇒ A = 1 and t = −1 ⇒ D = −1
Equating coefficients of \( t^3 \) on the two sides of (i), we get 0 = A + B ⇒ B = −1
Equating coefficients of \( t^2 \), we get 0 = 3A + 2B + C ⇒ C = −3A − 2B = −3(1) − 2(−1) = −1
Hence, \( \frac{1}{t(t + 1)^3} = \frac{1}{t} - \frac{1}{(t + 1)} - \frac{1}{(t + 1)^2} - \frac{1}{(t + 1)^3} \) [1]
⇒ \( \int \frac{1}{(e^x + 1)^3}\,dx = \int \left(\frac{1}{t} - \frac{1}{(t + 1)} - \frac{1}{(t + 1)^2} - \frac{1}{(t + 1)^3}\right) dt \) [1]
\( = \log|t| - \log|t + 1| - \frac{(t + 1)^{-1}}{-1} - \frac{(t + 1)^{-2}}{-2} + C \) [1]
\( = \log e^x - \log(e^x + 1) + \frac{1}{e^x + 1} + \frac{1}{2} \cdot \frac{1}{(e^x + 1)^2} + C \) (∵ \( t = e^x \gt 0 \)) [1]
\( = \log\left(\frac{e^x}{e^x + 1}\right) + \frac{1}{e^x + 1} + \frac{1}{2(e^x + 1)^2} + C \) [1]
(ii) ∵ \( \frac{x^3}{(x - 1)(x - 2)} = \frac{x^3}{x^2 - 3x + 2} = x + 3 + \frac{7x - 6}{(x - 1)(x - 2)} \) [1]
∴ \( \int \frac{x^3}{(x - 1)(x - 2)}\,dx = \int \left[x + 3 + \frac{7x - 6}{(x - 1)(x - 2)}\right] dx \) [1]
\( = \int (x + 3)\,dx + \int \frac{7x - 6}{(x - 1)(x - 2)}\,dx \) [1]
\( = \frac{x^2}{2} + 3x + \int \frac{7x - 6}{(x - 1)(x - 2)}\,dx \) [1]
\( = \frac{x^2}{2} + 3x + \int \left(\frac{8}{(x - 2)} - \frac{1}{(x - 1)}\right) dx \) [by partial fractions] [1]
\( = \frac{x^2}{2} + 3x + 8\log|x - 2| - \log|x - 1| + C \) [1]

Illustration 8. (a) Evaluate the following integrals: [1 Mark each]
(i) \( \int \log(x + 1)\,dx \)    (ii) \( \int (1 + x)\log x\,dx \)    (iii) \( \int x^{3/2}\log x\,dx \)
(b) Evaluate the following integrals: [4 Marks each]
(i) \( \int \frac{x\sin^{-1} x}{\sqrt{1 - x^2}}\,dx \)    (ii) \( \int x^2\tan^{-1} x\,dx \) [Exemplar]    (iii) \( \int e^x \sin x\,dx \)
Answer:
(a) (i) \( \int \log(x + 1)\,dx = \int \log(x + 1) \times 1\,dx \)
\( = \log(x + 1) \cdot x - \int \left(\frac{1}{x + 1}\right) x\,dx = x\log(x + 1) - \int \left(1 - \frac{1}{x + 1}\right) dx \)
\( = x\log(x + 1) - x + \log(x + 1) + C = (x + 1)\log(x + 1) - x + C \) [1]
(ii) \( \int (1 + x)\log x\,dx = \log x\left(x + \frac{x^2}{2}\right) - \int \frac{1}{x}\left(x + \frac{x^2}{2}\right) dx + C \)
\( = \frac{x}{2}(2 + x)\log x - \frac{1}{2}\int (2 + x)\,dx + C = \frac{x}{2}(2 + x)\log x - \frac{1}{2}\left(2x + \frac{x^2}{2}\right) + C \)
\( = \frac{x}{2}(2 + x)\log x - \frac{1}{4}(4x + x^2) + C \) [1]
(iii) \( \int x^{3/2}\log x\,dx = (\log x)\left(\frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}\right) - \int \frac{1}{x} \cdot \frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}\,dx + C \)
\( = \frac{2}{5}x^{5/2}(\log x) - \frac{2}{5}\int x^{3/2}\,dx + C \)
\( = \frac{2}{5}x^{5/2}(\log x) - \frac{2}{5}\left(\frac{x^{\frac{3}{2} + 1}}{\frac{3}{2} + 1}\right) + C = \frac{2}{5}x^{5/2}(\log x) - \frac{4}{25}x^{5/2} + C \) [1]
(b) (i) Put \( \sin^{-1} x = t \Rightarrow x = \sin t \Rightarrow dx = \cos t\,dt \)
∴ \( \int \frac{x\sin^{-1} x}{\sqrt{1 - x^2}}\,dx = \int \frac{t\sin t}{\sqrt{1 - \sin^2 t}}\cos t\,dt = \int \frac{t\sin t\cos t}{\cos t}\,dt = \int t\sin t\,dt \) [1]
\( = t(-\cos t) - \int 1(-\cos t)\,dt \) [1]
\( = -t\cos t + \sin t + C \) [1]
\( = -\sqrt{1 - x^2}\,\sin^{-1} x + x + C \) [1]
(ii) \( I = \int x^2\tan^{-1} x\,dx \)
\( = \tan^{-1} x\int x^2\,dx - \int \frac{1}{1 + x^2} \cdot \frac{x^3}{3}\,dx \) [1½]
\( = \frac{x^3}{3}\tan^{-1} x - \frac{1}{3}\int \left(x - \frac{x}{1 + x^2}\right) dx \) [1]
\( = \frac{x^3}{3}\tan^{-1} x - \frac{x^2}{6} + \frac{1}{6}\log|1 + x^2| + C \) [1½]
(iii) Let \( I = \int e^x \sin x\,dx \)
\( I = e^x(-\cos x) - \int e^x(-\cos x)\,dx \) [1]
\( I = -e^x\cos x + \int e^x\cos x\,dx \)
\( I = -e^x\cos x + e^x\sin x - \int e^x\sin x\,dx \) [1]
\( I = e^x(\sin x - \cos x) - I \) [1]
\( 2I = e^x(\sin x - \cos x) \)
\( I = \frac{e^x}{2}(\sin x - \cos x) + C \) [1]

Illustration 9. Evaluate the following integrals: [4 Marks each]
(i) \( \int \frac{e^x(x + 1)}{(x + 3)^3}\,dx \)    (ii) \( \int e^x\left(\frac{\sin 4x - 4}{1 - \cos 4x}\right) dx \)
Answer:
(i) \( \int \frac{e^x(x + 1)}{(x + 3)^3}\,dx = \int e^x\left\{\frac{(x + 3) - 2}{(x + 3)^3}\right\} dx \) [1]
\( = \int e^x\left\{\frac{1}{(x + 3)^2} - \frac{2}{(x + 3)^3}\right\} dx \) [1]
\( = \int e^x\{f(x) + f'(x)\}\,dx \) \( \left[\text{where } f(x) = \frac{1}{(x + 3)^2} \Rightarrow f'(x) = -\frac{2}{(x + 3)^3}\right] \) [1]
\( = e^x f(x) + C = \frac{e^x}{(x + 3)^2} + C \) [1]
(ii) \( \int e^x\left(\frac{\sin 4x - 4}{1 - \cos 4x}\right) dx = \int e^x\left(\frac{2\sin 2x\cos 2x - 4}{2\sin^2 2x}\right) dx \) [1]
\( = \int e^x\left(\frac{2\sin 2x\cos 2x}{2\sin^2 2x} - \frac{4}{2\sin^2 2x}\right) dx = \int e^x(\cot 2x - 2\,\text{cosec}^2 2x)\,dx \) [1]
\( = \int e^x(f(x) + f'(x))\,dx \) [where \( f(x) = \cot 2x \Rightarrow f'(x) = -2\,\text{cosec}^2 2x \)] [1]
\( = e^x f(x) + C = e^x\cot 2x + C \) [1]

Illustration 10. (a) Evaluate the following integrals: [1 Mark each]
(i) \( \int_{0}^{4} \sqrt{16 - x^2}\,dx \)    (ii) \( \int_{0}^{1} \frac{1}{e^x + e^{-x}}\,dx \)
(b) Evaluate the following integrals: [4 Marks each]
(i) \( \int_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{(1 - \cos x)^{5/2}}\,dx \) [Exemplar]    (ii) \( \int_{0}^{1/2} \frac{dx}{(1 + x^2)\sqrt{1 - x^2}} \) [Exemplar]
Answer:
(a) (i) \( \int_{0}^{4} \sqrt{16 - x^2}\,dx = \left[\frac{x}{2}\sqrt{(4)^2 - x^2} + \frac{(4)^2}{2}\sin^{-1}\frac{x}{4}\right]_{0}^{4} \)
\( = 8\sin^{-1}(1) = 8 \times \frac{\pi}{2} = 4\pi \) [1]
(ii) Put \( e^x = t \Rightarrow e^x\,dx = dt \)
when x = 0 ⇒ t = 1 and x = 1 ⇒ t = e
∴ \( \int_{0}^{1} \frac{1}{e^x + e^{-x}}\,dx = \int_{0}^{1} \frac{e^x}{e^{2x} + 1}\,dx = \int_{1}^{e} \frac{dt}{t^2 + 1} \)
\( = \left[\tan^{-1} t\right]_{1}^{e} = \tan^{-1} e - \tan^{-1} 1 = \tan^{-1} e - \frac{\pi}{4} \) [1]
(b) (i) \( I = \int_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{(1 - \cos x)^{5/2}}\,dx \)
\( I = \int_{\pi/3}^{\pi/2} \frac{\sqrt{1 + \cos x}}{(1 - \cos x)^{5/2}} \times \frac{\sqrt{1 - \cos x}}{\sqrt{1 - \cos x}}\,dx \)
\( I = \int_{\pi/3}^{\pi/2} \frac{\sin x\,dx}{(1 - \cos x)^3} \) [1½]
Put 1 − cos x = t, sin x dx = dt; at \( x = \frac{\pi}{3},\ t = \frac{1}{2} \) and at \( x = \frac{\pi}{2},\ t = 1 \) [1]
\( I = \int_{1/2}^{1} \frac{dt}{t^3} = -\frac{1}{2}\left[\frac{1}{t^2}\right]_{1/2}^{1} = -\frac{1}{2}[1 - 4] = \frac{3}{2} \) [1½]
(ii) \( I = \int_{0}^{1/2} \frac{1}{(1 + x^2)\sqrt{1 - x^2}}\,dx \)
Put x = sin t, dx = cos t dt; x = 0 ⇒ t = 0 and \( x = \frac{1}{2} \Rightarrow t = \frac{\pi}{6} \)
\( I = \int_{0}^{\pi/6} \frac{\cos t\,dt}{(1 + \sin^2 t)\cos t} \)
\( I = \int_{0}^{\pi/6} \frac{\sec^2 t\,dt}{1 + 2\tan^2 t} \) (dividing numerator and denominator by \( \cos^2 t \)) [1½]
Put tan t = u, \( \sec^2 t\,dt = du \); t = 0 ⇒ u = 0 and \( t = \frac{\pi}{6} \Rightarrow u = \frac{1}{\sqrt{3}} \)
\( I = \int_{0}^{1/\sqrt{3}} \frac{du}{1 + (\sqrt{2}u)^2} = \frac{1}{\sqrt{2}}\left[\tan^{-1}\sqrt{2}u\right]_{0}^{1/\sqrt{3}} \) [1½]
\( = \frac{1}{\sqrt{2}}\tan^{-1}\sqrt{\frac{2}{3}} \) [1]

Illustration 11. (a) Evaluate the following integrals: [4 Marks each]
(i) \( \int_{2}^{3} x\sqrt{5 - x}\,dx \)    (ii) \( \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cot x}}{1 + \sqrt{\cot x}}\,dx \)
(b) [6 Marks each] (i) Show that \( \int_{0}^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x}\,dx = \frac{1}{\sqrt{2}}\log(\sqrt{2} + 1) \) [Exemplar]
(ii) Evaluate: \( \int_{0}^{\pi/2} \frac{x}{1 + \sin x + \cos x}\,dx \)    (iii) Evaluate: \( \int_{0}^{1} \frac{\log(1 + x)}{1 + x^2}\,dx \)
Answer:
(a) (i) \( \int_{2}^{3} x\sqrt{5 - x}\,dx = \int_{2}^{3} (2 + 3 - x)\sqrt{5 - (2 + 3 - x)}\,dx \) \( \left(\because \int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a + b - x)\,dx\right) \)
\( = \int_{2}^{3} (5 - x)\sqrt{x}\,dx = \int_{2}^{3} (5x^{1/2} - x^{3/2})\,dx \) [1]
\( = 5\left[\frac{x^{3/2}}{3/2}\right]_{2}^{3} - \left[\frac{x^{5/2}}{5/2}\right]_{2}^{3} \) [1]
\( = \frac{10}{3}(3^{3/2} - 2^{3/2}) - \frac{2}{5}(3^{5/2} - 2^{5/2}) \)
\( = \frac{10}{3}(3\sqrt{3} - 2\sqrt{2}) - \frac{2}{5}(9\sqrt{3} - 4\sqrt{2}) \) [1]
\( = \left(10 - \frac{18}{5}\right)\sqrt{3} + \left(\frac{8}{5} - \frac{20}{3}\right)\sqrt{2} = \frac{32\sqrt{3}}{5} - \frac{76\sqrt{2}}{15} \) [1]
(ii) Let \( I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cot x}}{1 + \sqrt{\cot x}}\,dx = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx \) ...(1)
\( I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos\left(\frac{\pi}{2} - x\right)}}{\sqrt{\sin\left(\frac{\pi}{2} - x\right)} + \sqrt{\cos\left(\frac{\pi}{2} - x\right)}}\,dx \) \( \left(\because \int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a + b - x)\,dx\right) \) [1]
\( I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}}\,dx \) ...(2) [1]
On adding equation (1) and equation (2), we get
\( 2I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}}\,dx = \int_{\pi/6}^{\pi/3} 1\,dx \) [1]
\( 2I = [x]_{\pi/6}^{\pi/3} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \)
∴ \( I = \frac{\pi}{12} \) [1]
(b) (i) We have \( I = \int_{0}^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x}\,dx = \int_{0}^{\pi/2} \frac{\sin^2\left(\frac{\pi}{2} - x\right)}{\sin\left(\frac{\pi}{2} - x\right) + \cos\left(\frac{\pi}{2} - x\right)}\,dx \)
⇒ \( I = \int_{0}^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x}\,dx \) [1]
Adding, \( 2I = \int_{0}^{\pi/2} \frac{\sin^2 x + \cos^2 x}{\sin x + \cos x}\,dx = \int_{0}^{\pi/2} \frac{dx}{\sin x + \cos x} \), and since \( \sin x + \cos x = \sqrt{2}\cos\left(x - \frac{\pi}{4}\right) \),
thus we get \( 2I = \frac{1}{\sqrt{2}}\int_{0}^{\pi/2} \frac{dx}{\cos\left(x - \frac{\pi}{4}\right)} \) [1]
\( = \frac{1}{\sqrt{2}}\int_{0}^{\pi/2} \sec\left(x - \frac{\pi}{4}\right) dx = \frac{1}{\sqrt{2}}\left[\log\left(\sec\left(x - \frac{\pi}{4}\right) + \tan\left(x - \frac{\pi}{4}\right)\right)\right]_{0}^{\pi/2} \) [2]
\( = \frac{1}{\sqrt{2}}\left[\log\left(\sec\frac{\pi}{4} + \tan\frac{\pi}{4}\right) - \log\left\{\sec\left(-\frac{\pi}{4}\right) + \tan\left(-\frac{\pi}{4}\right)\right\}\right] \)
\( = \frac{1}{\sqrt{2}}\left[\log(\sqrt{2} + 1) - \log(\sqrt{2} - 1)\right] = \frac{1}{\sqrt{2}}\log\left|\frac{\sqrt{2} + 1}{\sqrt{2} - 1}\right| \)
\( = \frac{1}{\sqrt{2}}\log\left(\frac{(\sqrt{2} + 1)^2}{1}\right) = \frac{2}{\sqrt{2}}\log(\sqrt{2} + 1) \) [2]
Hence \( I = \frac{1}{\sqrt{2}}\log(\sqrt{2} + 1) \).
(ii) Let \( I = \int_{0}^{\pi/2} \frac{x}{1 + \sin x + \cos x}\,dx \) ...(1)
\( I = \int_{0}^{\pi/2} \frac{\left(\frac{\pi}{2} - x\right)}{1 + \sin\left(\frac{\pi}{2} - x\right) + \cos\left(\frac{\pi}{2} - x\right)}\,dx \) \( \left(\because \int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx\right) \)
\( I = \int_{0}^{\pi/2} \frac{\left(\frac{\pi}{2} - x\right)}{1 + \cos x + \sin x}\,dx \) ...(2) [1]
Adding (1) and (2), we get
\( 2I = \int_{0}^{\pi/2} \frac{\frac{\pi}{2}}{1 + \sin x + \cos x}\,dx \Rightarrow I = \frac{\pi}{4}\int_{0}^{\pi/2} \frac{dx}{1 + \sin x + \cos x} \) [1]
\( = \frac{\pi}{4}\int_{0}^{\pi/2} \frac{\sec^2\frac{x}{2}\,dx}{1 + \tan^2\frac{x}{2} + 2\tan\frac{x}{2} + 1 - \tan^2\frac{x}{2}} = \frac{\pi}{4}\int_{0}^{\pi/2} \frac{\sec^2\frac{x}{2}\,dx}{2\left(1 + \tan\frac{x}{2}\right)} \) [1]
Substitute \( \tan\frac{x}{2} = t \Rightarrow \frac{1}{2}\sec^2\frac{x}{2}\,dx = dt \Rightarrow \sec^2\frac{x}{2}\,dx = 2\,dt \)
When x = 0, t = 0 and \( x = \frac{\pi}{2} \), t = 1.
∴ \( I = \frac{\pi}{4}\int_{0}^{1} \frac{2\,dt}{2(1 + t)} = \frac{\pi}{4}\int_{0}^{1} \frac{1}{1 + t}\,dt \) [1]
\( = \frac{\pi}{4}\left[\log|1 + t|\right]_{0}^{1} \) [1]
\( = \frac{\pi}{4}(\log 2 - \log 1) = \frac{\pi}{4}\log 2 \) [1]
(iii) Let \( I = \int_{0}^{1} \frac{\log(1 + x)}{1 + x^2}\,dx \)
Substitute \( x = \tan\theta \Rightarrow dx = \sec^2\theta\,d\theta \).
When x = 0, θ = 0 and when x = 1, \( \theta = \frac{\pi}{4} \)
∴ \( I = \int_{0}^{\pi/4} \frac{\log(1 + \tan\theta)}{1 + \tan^2\theta}\sec^2\theta\,d\theta = \int_{0}^{\pi/4} \log(1 + \tan\theta)\,d\theta \) ...(1) [1]
\( I = \int_{0}^{\pi/4} \log\left(1 + \tan\left(\frac{\pi}{4} - \theta\right)\right) d\theta \) \( \left(\because \int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx\right) \) [1]
\( = \int_{0}^{\pi/4} \log\left(1 + \frac{1 - \tan\theta}{1 + \tan\theta}\right) d\theta \)
\( = \int_{0}^{\pi/4} \log\left(\frac{2}{1 + \tan\theta}\right) d\theta = \int_{0}^{\pi/4} \{\log 2 - \log(1 + \tan\theta)\}\,d\theta \) ...(2) [1]
Adding (1) and (2), we get
\( 2I = \int_{0}^{\pi/4} \log 2\,d\theta = \log 2\int_{0}^{\pi/4} 1\,d\theta \) [1]
\( = (\log 2)[\theta]_{0}^{\pi/4} \) [1]
\( 2I = (\log 2)\left(\frac{\pi}{4} - 0\right) \Rightarrow 2I = \frac{\pi}{4}\log 2 \)
\( I = \frac{\pi}{8}\log 2 \) [1]

Do Yourself

Do Yourself 1
Evaluate the following integrals:
(i) (a) \( \int \frac{1}{1 + \sin x}\,dx \)    (b) \( \int \cos^{-1}(\sin x)\,dx;\ -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \)
(ii) (a) \( \int 3^{-2x} e^{-2x}\,dx \)    (b) \( \int \frac{(a^x + b^x)^2}{a^x b^x}\,dx \)
(iii) \( \int \frac{x^3}{x + 1}\,dx \)

Do Yourself 2
Evaluate the following integrals:
(i) \( \int \frac{1}{x\log x \log(\log x)}\,dx \)    (ii) \( \int \frac{\cos 2x}{(\cos x + \sin x)^2}\,dx \)
(iii) \( \int \frac{(\tan^{-1} x)^2}{1 + x^2}\,dx \)    (iv) \( \int \frac{x^{e-1} + e^{x-1}}{x^e + e^x}\,dx \)

Do Yourself 3
Evaluate the following integrals:
(i) \( \int \frac{1}{\sin(x + a)\cos(x + b)}\,dx \)    (ii) \( \int \frac{\sin 2x + 2\cos^2 x}{\sin^2 x}\,dx \)
(iii) \( \int \frac{\sin 2x}{\sin 4x}\,dx \)    (iv) \( \int \tan^3 2x \sec 2x\,dx \)

Do Yourself 4
Evaluate the following integrals:
(i) \( \int \frac{1}{e^x - e^{-x}}\,dx \)    (ii) \( \int \frac{\sin x}{16 - 9\cos^2 x}\,dx \)
(iii) \( \int \frac{x}{x^2 + 4x + 5}\,dx \)    (iv) \( \int \frac{1}{\sqrt{2x^2 + 1}}\,dx \)

Do Yourself 5
Evaluate the following integrals:
(i) \( \int \frac{2x}{(x^2 + 1)(x^2 + 2)}\,dx \)    (ii) \( \int \frac{\cos x}{(1 + \sin x)(2 + \sin x)}\,dx \)
(iii) \( \int \frac{1 - x^2}{x(1 - 2x)}\,dx \)

Do Yourself 6
Evaluate the following integrals:
(i) \( \int \sin^{-1}\sqrt{x}\,dx \)    (ii) \( \int \frac{\log|x|}{(x + 1)^2}\,dx \)

Do Yourself 7
Evaluate the following integrals:
(i) \( \int e^{2x}\left(\frac{2x - 1}{4x^2}\right) dx \)    (ii) \( \int e^x\left(\frac{1 - x}{1 + x^2}\right)^2 dx \)

Do Yourself 8
Evaluate the following integral:
\( \int \frac{x^2}{\sqrt{x^2 + 6}}\,dx \)

Do Yourself 9
Evaluate the following integrals:
(i) \( \int_{0}^{\pi/2} \frac{\sin 2x}{\sin^4 x + \cos^4 x}\,dx \)    (ii) \( \int_{0}^{1/2} \frac{\sin^{-1} x}{(1 - x^2)^{3/2}}\,dx \)    (iii) \( \int_{0}^{\pi/4} 2\tan^3 x\,dx \)

Do Yourself 10
Evaluate the following integrals:
(i) \( \int_{1}^{3} |x^2 - 2x|\,dx \)    (ii) \( \int_{0}^{\pi} \frac{x\tan x}{\sec x\,\text{cosec}\,x}\,dx \)    (iii) \( \int_{0}^{\pi} \frac{x}{1 + \cos^2 x}\,dx \)
(iv) \( \int_{0}^{\pi} x\log(\sin x)\,dx \)    (v) \( \int_{0}^{2\pi} \sin^5 x\,dx \)

Answers for Do Yourself

Do Yourself 1
(i) (a) \( \tan x - \sec x + C \)    (b) \( \frac{\pi}{2}x - \frac{x^2}{2} + C \)
(ii) (a) \( -\frac{(3e)^{-2x}}{2\log(3e)} + C \)    (b) \( \frac{\left(\frac{a}{b}\right)^x}{\log(a/b)} + \frac{\left(\frac{b}{a}\right)^x}{\log(b/a)} + 2x + C,\ (a \neq b) \)
(iii) \( \frac{x^3}{3} - \frac{x^2}{2} + x - \log|x + 1| + C \)

Do Yourself 2
(i) \( \log|\log(\log x)| + C \)    (ii) \( \log|\cos x + \sin x| + C \)
(iii) \( \frac{1}{3}(\tan^{-1} x)^3 + C \)    (iv) \( \frac{1}{e}\log|e^x + x^e| + C \)

Do Yourself 3
(i) \( \frac{1}{\cos(a - b)} \cdot \log\left|\frac{\sin(x + a)}{\cos(x + b)}\right| + C \)    (ii) \( 2[\log|\sin x| - \cot x - x] + C \)
(iii) \( \frac{1}{4}\log|\sec 2x + \tan 2x| + C \)    (iv) \( \frac{1}{6}\sec^3 2x - \frac{1}{2}\sec 2x + C \)

Do Yourself 4
(i) \( \frac{1}{2}\log\left|\frac{e^x - 1}{e^x + 1}\right| + C \)    (ii) \( -\frac{1}{24}\log\left|\frac{4 + 3\cos x}{4 - 3\cos x}\right| + C \)
(iii) \( \frac{1}{2}\log(x^2 + 4x + 5) - 2\tan^{-1}(x + 2) + C \)    (iv) \( \frac{1}{\sqrt{2}}\log\left(x + \sqrt{x^2 + \frac{1}{2}}\right) + C \)

Do Yourself 5
(i) \( \log\left(\frac{x^2 + 1}{x^2 + 2}\right) + C \)    (ii) \( \log\left(\frac{1 + \sin x}{2 + \sin x}\right) + C \)
(iii) \( \frac{x}{2} + \log|x| - \frac{3}{4}\log|1 - 2x| + C \)

Do Yourself 6
(i) \( \frac{1}{2}\left[\sqrt{x - x^2} - (1 - 2x)\sin^{-1}\sqrt{x}\right] + C \)    (ii) \( \log\left|\frac{x}{x + 1}\right| - \frac{\log|x|}{x + 1} + C \)

Do Yourself 7
(i) \( \frac{e^{2x}}{4x} + C \)    (ii) \( \frac{e^x}{1 + x^2} + C \)

Do Yourself 8
\( \frac{x\sqrt{x^2 + 6}}{2} - 3\log\left(x + \sqrt{x^2 + 6}\right) + C \)

Do Yourself 9
(i) \( \frac{\pi}{2} \)    (ii) \( \frac{\pi}{6\sqrt{3}} + \log\frac{\sqrt{3}}{2} \)    (iii) \( 1 - \log 2 \)

Do Yourself 10
(i) 2    (ii) \( \frac{\pi^2}{4} \)    (iii) \( \frac{\pi^2}{2\sqrt{2}} \)    (iv) \( \frac{\pi^2}{2}\log\frac{1}{2} \)    (v) 0