CBSE Class 12 Mathematics Integrals Notes Set 04

Important Notes and Formulas

Definition

Let \( \phi(x) \) be the primitive or anti-derivative of a function f(x) defined on [a, b], i.e., \( \frac{d}{dx}[\phi(x)] = f(x) \). Then the definite integral of f(x) over [a, b] is denoted by \( \int_{a}^{b} f(x)\,dx \) and is defined as \( [\phi(b) - \phi(a)] \), i.e., \( \int_{a}^{b} f(x)\,dx = \phi(b) - \phi(a) \). This is also called the Newton–Leibnitz formula.
The numbers a and b are called the limits of integration, ‘a’ is called the lower limit and ‘b’ the upper limit. The interval [a, b] is called the interval of integration. The interval [a, b] is also known as the range of integration.

Important Tips
(i) In the above definition it does not matter which anti-derivative is used to evaluate the definite integral, because if \( \int f(x)\,dx = \phi(x) + c \), then \( \int_{a}^{b} f(x)\,dx = [\phi(x) + c]_{a}^{b} = (\phi(b) + c) - (\phi(a) + c) = \phi(b) - \phi(a) \).
In other words, to evaluate the definite integral there is no need to keep the constant of integration.
(ii) Every definite integral has a unique value.

Evaluation of Definite Integral by Substitution

When the variable in a definite integral is changed, the substitutions in terms of the new variable should be effected at three places:
(i) In the integrand    (ii) In the differential, say dx    (iii) In the limits
For example, if we put \( \phi(x) = t \) in the integral \( \int_{a}^{b} f\{\phi(x)\}\phi'(x)\,dx \), then \( \int_{a}^{b} f\{\phi(x)\}\phi'(x)\,dx = \int_{\phi(a)}^{\phi(b)} f(t)\,dt \).

Important Tips
(i) \( \int_{0}^{\pi} \frac{dx}{1 + \sin x} = 2 \)
(ii) \( \int_{0}^{\pi/2} \frac{dx}{\sin x + \cos x} = \sqrt{2}\,\log(\sqrt{2} + 1) \)
(iii) \( \int_{0}^{\pi/2} \log(\tan x)\,dx = 0 \)
(iv) \( \int_{0}^{a} \frac{dx}{1 + e^{f(x)}} = \frac{a}{2} \), where \( f(a - x) = -f(x) \)
(v) \( \int_{0}^{a} \frac{dx}{\sqrt{a^2 - x^2}} = \frac{\pi}{2} \)
(vi) \( \int_{0}^{a} \frac{dx}{x^2 + a^2} = \frac{\pi}{4a} \)
(vii) \( \int_{0}^{a} \sqrt{a^2 - x^2}\,dx = \frac{\pi a^2}{4} \)

Properties of Definite Integral

(1) \( \int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(t)\,dt \), i.e., the value of a definite integral remains unchanged if its variable is replaced by any other symbol.

(2) \( \int_{a}^{b} f(x)\,dx = -\int_{b}^{a} f(x)\,dx \), i.e., by the interchange in the limits of a definite integral, the sign of the integral is changed.

(3) \( \int_{a}^{b} f(x)\,dx = \int_{a}^{c} f(x)\,dx + \int_{c}^{b} f(x)\,dx \), (where a < c < b)
or \( \int_{a}^{b} f(x)\,dx = \int_{a}^{c_1} f(x)\,dx + \int_{c_1}^{c_2} f(x)\,dx + \ldots + \int_{c_n}^{b} f(x)\,dx \); (where \( a \lt c_1 \lt c_2 \lt \ldots \lt c_n \lt b \))
Generally this property is used when the integrand has two or more rules in the integration interval.

Important Tips
\( \int_{a}^{b} (|x - a| + |x - b|)\,dx = (b - a)^2 \)

Note: (i) Property (3) is useful when f(x) is not continuous in [a, b] because we can break up the integral into several integrals at the points of discontinuity so that the function is continuous in the sub-intervals.
(ii) The expression for f(x) changes at the end points of each of the sub-intervals. You might at times be puzzled about the end points as limits of integration. You may not be sure for x = 0 as the limit of the first integral or the next one. In fact, it is immaterial, as the area of the line is always zero. Therefore, even if you write \( \int_{-1}^{0} (1 - 2x)\,dx \), whereas 0 is not included in its domain, it is deemed to be understood that you are approaching x = 0 from the left in the first integral and from the right in the second integral. Similarly for x = 1.

(4) \( \int_{0}^{a} f(x)\,dx = \int_{0}^{a} f(a - x)\,dx \): This property can be used only when the lower limit is zero. It is generally used for those complicated integrals whose denominators are unchanged when x is replaced by (a − x).
The following integrals can be obtained with the help of the above property.
(i) \( \int_{0}^{\pi/2} \frac{\sin^n x}{\sin^n x + \cos^n x}\,dx = \int_{0}^{\pi/2} \frac{\cos^n x}{\cos^n x + \sin^n x}\,dx = \frac{\pi}{4} \)
(ii) \( \int_{0}^{\pi/2} \frac{\tan^n x}{1 + \tan^n x}\,dx = \int_{0}^{\pi/2} \frac{\cot^n x}{1 + \cot^n x}\,dx = \frac{\pi}{4} \)
(iii) \( \int_{0}^{\pi/2} \frac{1}{1 + \tan^n x}\,dx = \int_{0}^{\pi/2} \frac{1}{1 + \cot^n x}\,dx = \frac{\pi}{4} \)
(iv) \( \int_{0}^{\pi/2} \frac{\sec^n x}{\sec^n x + \mathrm{cosec}^n x}\,dx = \int_{0}^{\pi/2} \frac{\mathrm{cosec}^n x}{\mathrm{cosec}^n x + \sec^n x}\,dx = \frac{\pi}{4} \)
(v) \( \int_{0}^{\pi/2} f(\sin 2x) \sin x\,dx = \int_{0}^{\pi/2} f(\sin 2x) \cos x\,dx \)
(vi) \( \int_{0}^{\pi/2} f(\sin x)\,dx = \int_{0}^{\pi/2} f(\cos x)\,dx \)
(vii) \( \int_{0}^{\pi/2} f(\tan x)\,dx = \int_{0}^{\pi/2} f(\cot x)\,dx \)
(viii) \( \int_{0}^{1} f(\log x)\,dx = \int_{0}^{1} f[\log(1 - x)]\,dx \)
(ix) \( \int_{0}^{\pi/2} \log \tan x\,dx = \int_{0}^{\pi/2} \log \cot x\,dx \)
(x) \( \int_{0}^{\pi/4} \log(1 + \tan x)\,dx = \frac{\pi}{8} \log 2 \)
(xi) \( \int_{0}^{\pi/2} \log \sin x\,dx = \int_{0}^{\pi/2} \log \cos x\,dx = \frac{-\pi}{2} \log 2 = \frac{\pi}{2} \log \frac{1}{2} \)
(xii) \( \int_{0}^{\pi/2} \frac{a \sin x + b \cos x}{\sin x + \cos x}\,dx = \int_{0}^{\pi/2} \frac{a \sec x + b\,\mathrm{cosec}\,x}{\sec x + \mathrm{cosec}\,x}\,dx = \int_{0}^{\pi/2} \frac{a \tan x + b \cot x}{\tan x + \cot x}\,dx = \frac{\pi}{4}(a + b) \)

(5) \( \int_{-a}^{a} f(x)\,dx = \int_{0}^{a} [f(x) + f(-x)]\,dx \).
In special case: \( \int_{-a}^{a} f(x)\,dx = \begin{cases} 2\int_{0}^{a} f(x)\,dx, & \text{if } f(x) \text{ is an even function or } f(-x) = f(x) \\ 0, & \text{if } f(x) \text{ is an odd function or } f(-x) = -f(x) \end{cases} \)
This property is generally used when the integrand is either an even or an odd function of x.

(6) \( \int_{0}^{2a} f(x)\,dx = \int_{0}^{a} f(x)\,dx + \int_{0}^{a} f(2a - x)\,dx \)
In particular, \( \int_{0}^{2a} f(x)\,dx = \begin{cases} 0, & \text{if } f(2a - x) = -f(x) \\ 2\int_{0}^{a} f(x)\,dx, & \text{if } f(2a - x) = f(x) \end{cases} \)
It is generally used to make half the upper limit.

(7) \( \int_{a}^{b} f(x)\,dx = \int_{a}^{b} f(a + b - x)\,dx \)
Note: \( \int_{a}^{b} \frac{f(x)\,dx}{f(x) + f(a + b - x)} = \frac{1}{2}(b - a) \)

(8) \( \int_{0}^{a} x f(x)\,dx = \frac{1}{2} a \int_{0}^{a} f(x)\,dx \), if \( f(a - x) = f(x) \)

Important Tips
(i) Every continuous function defined on [a, b] is integrable over [a, b].
(ii) If f is continuous on [a, b], then the integral function g defined by \( g(x) = \int_{a}^{x} f(t)\,dt \) for x ∈ [a, b] is derivable on [a, b] and \( g'(x) = f(x) \) for all x ∈ [a, b].

Integration of Piecewise Continuous Functions

Any function f(x) which is discontinuous at a finite number of points in an interval [a, b] can be made continuous in sub-intervals by breaking the interval into these sub-intervals. If f(x) is discontinuous at points \( x_1, x_2, x_3, \ldots, x_n \) in (a, b), then we can define sub-intervals \( (a, x_1), (x_1, x_2), \ldots, (x_{n-1}, x_n), (x_n, b) \) such that f(x) is continuous in each of these sub-intervals. Such functions are called piecewise continuous functions. For integration of a piecewise continuous function, we integrate f(x) in these sub-intervals and finally add all the values.

Solved Examples

Example 1. \( \int_{0}^{\pi} \sin^2 x\,dx \) is equal to [MP PET 1999]
(a) \( \pi \)
(b) \( \pi/2 \)
(c) 0
(d) None of these
Answer:
Ans. (b) \( \pi/2 \)
\( I = \frac{1}{2}\int_{0}^{\pi} 2\sin^2 x\,dx = \frac{1}{2}\int_{0}^{\pi} [1 - \cos 2x]\,dx \)
⇒ \( I = \frac{1}{2}\left[x - \frac{\sin 2x}{2}\right]_{0}^{\pi} \) ⇒ \( I = \frac{1}{2}[\pi] = \frac{\pi}{2} \).

Example 2. If \( h(a) = h(b) \), then \( \int_{a}^{b} [f(g[h(x)])]^{-1} f'(g[h(x)])\,g'[h(x)]\,h'(x)\,dx \) is equal to [MP PET 2001]
(a) 0
(b) \( f(a) - f(b) \)
(c) \( f[g(a)] - f[g(b)] \)
(d) None of these
Answer:
Ans. (a) 0
Put \( f(g[h(x)]) = t \) ⇒ \( f'(g[h(x)])\,g'[h(x)]\,h'(x)\,dx = dt \)
∴ \( \int_{f(g[h(a)])}^{f(g[h(b)])} t^{-1}\,dt = [\log t]_{f(g[h(a)])}^{f(g[h(b)])} = 0 \)   [∵ h(a) = h(b)]

Example 3. The value of the integral \( \int_{e^{-1}}^{e^{2}} \left|\frac{\log_e x}{x}\right| dx \) is [IIT 2000]
(a) 3/2
(b) 5/2
(c) 3
(d) 5
Answer:
Ans. (b) 5/2
Put \( \log_e x = t \) ⇒ \( e^t = x \)
∴ \( dx = e^t\,dt \)
and the limits change to −1 to 2.
∴ \( I = \int_{-1}^{2} \left|\frac{t}{e^t}\right| e^t\,dt = \int_{-1}^{2} |t|\,dt \) ⇒ \( I = \int_{-1}^{0} -t\,dt + \int_{0}^{2} t\,dt \) ⇒ \( I = \left[\frac{-t^2}{2}\right]_{-1}^{0} + \left[\frac{t^2}{2}\right]_{0}^{2} \) ⇒ I = 5/2

Example 4. \( \int_{0}^{\pi/2} \frac{dx}{1 + \sin x} \) equals [MNR 1983; Rajasthan PET 1990; Kurukshetra CEE 1997]
(a) 0
(b) 1
(c) −1
(d) 2
Answer:
Ans. (b) 1
\( I = \int_{0}^{\pi/2} \frac{dx}{\sin^2 (x/2) + \cos^2 (x/2) + 2\sin (x/2) \cos (x/2)} \)
\( I = \int_{0}^{\pi/2} \frac{dx}{(\sin (x/2) + \cos (x/2))^2} = \int_{0}^{\pi/2} \frac{\sec^2 (x/2)}{(1 + \tan (x/2))^2}\,dx \)
Put \( 1 + \tan (x/2) = t \) ⇒ \( \frac{1}{2}\sec^2 (x/2)\,dx = dt \)
∴ \( I = 2\int_{1}^{2} \frac{dt}{t^2} = -2\left[\frac{1}{t}\right]_{1}^{2} = -2\left[\frac{1}{2} - \frac{1}{1}\right] = 1 \)

Example 5. \( \int_{3}^{6} \frac{1}{x + 1}\,dx \) is equal to
(a) \( [\log(x + 1)]_{3}^{6} \)
(b) \( [\log(t + 1)]_{3}^{6} \)
(c) Both (a) and (b)
(d) None of these
Answer:
Ans. (c) Both (a) and (b)
\( I = \int_{3}^{6} \frac{1}{x + 1}\,dx = [\log(x + 1)]_{3}^{6} \),   \( I = \int_{3}^{6} \frac{1}{t + 1}\,dt = [\log(t + 1)]_{3}^{6} \)

Example 6. Suppose f is such that f(−x) = −f(x) for every real x and \( \int_{0}^{1} f(x)\,dx = 5 \), then \( \int_{-1}^{0} f(t)\,dt = \) [MP PET 2000]
(a) 10
(b) 5
(c) 0
(d) −5
Answer:
Ans. (d) −5
Given, \( \int_{0}^{1} f(x)\,dx = 5 \)
Put x = −t ⇒ dx = −dt
∴ \( I = -\int_{0}^{-1} f(-t)\,dt = -\int_{-1}^{0} f(t)\,dt \) ⇒ I = −5

Example 7. \( \int_{-2}^{2} |1 - x^2|\,dx \) is equal to [IIT 1989; BIT Ranchi 1996; Kurukshetra CEE 1998]
(a) 2
(b) 4
(c) 6
(d) 8
Answer:
Ans. (b) 4
\( I = \int_{-2}^{2} |1 - x^2|\,dx = \int_{-2}^{-1} |1 - x^2|\,dx + \int_{-1}^{1} |1 - x^2|\,dx + \int_{1}^{2} |1 - x^2|\,dx \)
⇒ \( I = -\int_{-2}^{-1} (1 - x^2)\,dx + \int_{-1}^{1} (1 - x^2)\,dx - \int_{1}^{2} (1 - x^2)\,dx \) ⇒ \( I = \frac{4}{3} + \frac{4}{3} + \frac{4}{3} = 4 \).

Example 8. \( \int_{0}^{1.5} [x^2]\,dx \), where [.] denotes the greatest integer function, equals [DCE 2000, 2001; IIT 1988; AMU 1998]
(a) \( 2 + \sqrt{2} \)
(b) \( 2 - \sqrt{2} \)
(c) \( 1 + \sqrt{2} \)
(d) \( \sqrt{2} - 1 \)
Answer:
Ans. (b) \( 2 - \sqrt{2} \)
\( I = \int_{0}^{1.5} [x^2]\,dx = \int_{0}^{1} [x^2]\,dx + \int_{1}^{\sqrt{2}} [x^2]\,dx + \int_{\sqrt{2}}^{1.5} [x^2]\,dx \)
⇒ \( I = 0 + \int_{1}^{\sqrt{2}} 1\,dx + \int_{\sqrt{2}}^{1.5} 2\,dx = \sqrt{2} - 1 + 3 - 2\sqrt{2} \) ⇒ \( I = 2 - \sqrt{2} \)

Example 9. If \( f(x) = \begin{cases} e^{\cos x} \sin x, & |x| \leq 2 \\ 2, & \text{otherwise} \end{cases} \), then \( \int_{-2}^{3} f(x)\,dx = \) [IIT 2000]
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
Ans. (c) 2
|x| ≤ 2 ⇒ −2 ≤ x ≤ 2 and \( f(x) = e^{\cos x} \sin x \) is an odd function.
∴ \( I = \int_{-2}^{3} f(x)\,dx = \int_{-2}^{2} f(x)\,dx + \int_{2}^{3} f(x)\,dx \) ⇒ \( I = 0 + \int_{2}^{3} 2\,dx = [2x]_{2}^{3} = 2 \)
[∵ \( \int_{-a}^{a} f(x)\,dx = 0 \) if f(x) is odd, and in (2, 3], f(x) = 2]

Example 10. \( \int_{0}^{\pi} e^{\sin^2 x} \cos^3 x\,dx = \)
OR
For any integer n, \( \int_{0}^{\pi} e^{\sin^2 x} \cos^3 (2n + 1)x\,dx = \) [MP PET 2002; MNR 1992, 98]
(a) −1
(b) 0
(c) 1
(d) None of these
Answer:
Ans. (b) 0
Let \( f_1(x) = e^{\sin^2 x}\cos^3 x = -f_1(\pi - x) \)
and \( f_2(x) = e^{\sin^2 x}\cos^3 (2n + 1)x = -f_2(\pi - x) \)
∴ I = 0.

Example 11. \( \int_{0}^{2a} \frac{f(x)}{f(x) + f(2a - x)}\,dx \) is equal to [IIT 1988; Kurukshetra CEE 1999; Karnataka CET 2000]
(a) a
(b) a/2
(c) 2a
(d) 0
Answer:
Ans. (a) a
\( I = \int_{0}^{2a} \frac{f(x)}{f(x) + f(2a - x)}\,dx = \int_{0}^{2a} \frac{f(2a - x)}{f(2a - x) + f(x)}\,dx \)
\( 2I = \int_{0}^{2a} \frac{f(x) + f(2a - x)}{f(x) + f(2a - x)}\,dx = \int_{0}^{2a} dx = [x]_{0}^{2a} = 2a \)
∴ I = a.

Example 12. \( \int_{0}^{\pi/2} \frac{\sqrt{\tan x}}{1 + \sqrt{\tan x}}\,dx = \int_{0}^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}\,dx \) is equal to [MNR 1989, 90; Rajasthan PET 1991, 95]
(a) \( \pi/4 \)
(b) \( \infty \)
(c) −1
(d) 1
Answer:
Ans. (a) \( \pi/4 \)
We know, \( \int_{0}^{\pi/2} \frac{\tan^n x\,dx}{1 + \tan^n x} = \frac{\pi}{4} \) for any value of n.
∴ \( I = \pi/4 \).

Example 13. The integral \( \int_{-1/2}^{1/2} \left([x] + \ln\left(\frac{1 + x}{1 - x}\right)\right) dx \) equals
(a) \( \frac{-1}{2} \)
(b) 0
(c) 1
(d) \( 2 \ln\left(\frac{1}{2}\right) \)
Answer:
Ans. (a) \( \frac{-1}{2} \)
\( \log\left(\frac{1 + x}{1 - x}\right) \) is an odd function of x as f(−x) = −f(x).
\( I = \int_{-1/2}^{1/2} [x]\,dx + 0 \) ⇒ \( I = \int_{-1/2}^{0} [x]\,dx + \int_{0}^{1/2} [x]\,dx \) ⇒ \( I = \int_{-1/2}^{0} -1\,dx + 0 \) ⇒ \( -[x]_{-1/2}^{0} = \frac{-1}{2} \).

Example 14. The value of the integral \( \int_{-1}^{1} \log\left[x + \sqrt{x^2 + 1}\right] dx \) is [MP PET 2001]
(a) 0
(b) log 2
(c) log 1/2
(d) None of these
Answer:
Ans. (a) 0
\( f(x) = \log[x + \sqrt{x^2 + 1}] \) is an odd function, i.e., f(−x) = −f(x) ⇒ f(x) + f(−x) = 0 ⇒ I = 0.

Example 15. The value of \( \int_{-\pi}^{\pi} (1 - x^2) \sin x \cos^2 x\,dx \) is [Roorkee 1992; MNR 1998]
(a) 0
(b) 1
(c) 2
(d) 3
Answer:
Ans. (a) 0
Let \( f_1(x) = (1 - x^2) \), \( f_2(x) = \sin x \) and \( f_3(x) = \cos^2 x \)
Now, \( f_1(x) = f_1(-x) \), \( f_2(x) = -f_2(-x) \) and \( f_3(x) = f_3(-x) \)
∴ \( I = \int_{-\pi}^{\pi} f(x)\,dx = \int_{-\pi}^{\pi} [f_1(x) f_2(x) f_3(x)]\,dx = -\int_{-\pi}^{\pi} [f_1(-x) f_2(-x) f_3(-x)]\,dx \)
∴ I = 0

Example 16. If n is any integer, then \( \int_{0}^{\pi} e^{\cos^2 x} \cos^3 (2n + 1)x\,dx \) is equal to [IIT 1985; Rajasthan PET 1995]
(a) \( \pi \)
(b) 1
(c) 0
(d) None of these
Answer:
Ans. (c) 0
\( I = \int_{0}^{\pi} e^{\cos^2 (\pi - x)} \cdot \cos^3 (2n + 1)(\pi - x)\,dx \)
⇒ \( I = -\int_{0}^{\pi} e^{\cos^2 x} \cdot \cos^3 (2n + 1)x\,dx \) ⇒ I = −I
⇒ 2I = 0 ⇒ I = 0.

Example 17. \( \int_{\pi/4}^{3\pi/4} \frac{dx}{1 + \cos x} \) is equal to [IIT 1999]
(a) 2
(b) −2
(c) 1/2
(d) −1/2
Answer:
Ans. (a) 2
\( I = \int_{\pi/4}^{3\pi/4} \frac{1}{1 - \cos x}\,dx \)   \( \left[∵ \cos\left(\frac{\pi}{4} + \frac{3\pi}{4} - x\right) = -\cos x\right] \)
∴ \( 2I = \int_{\pi/4}^{3\pi/4} \frac{2}{1 - \cos^2 x}\,dx \)
⇒ \( 2I = 2\int_{\pi/4}^{3\pi/4} \mathrm{cosec}^2 x\,dx \) ⇒ \( 2I = -2[\cot x]_{\pi/4}^{3\pi/4} = 4 \) ⇒ I = 2.

Example 18. The value of \( \int_{2}^{3} \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}}\,dx \) is [IIT 1994; Kurukshetra CEE 1998]
(a) 1
(b) 0
(c) −1
(d) 1/2
Answer:
Ans. (d) 1/2
\( I = \int_{2}^{3} \frac{\sqrt{x}}{\sqrt{5 - x} + \sqrt{x}}\,dx \)
Put x = 2 + 3 − t ⇒ dx = −dt
∴ \( I = \int_{3}^{2} \frac{\sqrt{5 - t}}{\sqrt{t} + \sqrt{5 - t}}(-dt) = \int_{2}^{3} \frac{\sqrt{5 - x}}{\sqrt{5 - x} + \sqrt{x}}\,dx \) and \( 2I = \int_{2}^{3} \frac{\sqrt{x} + \sqrt{5 - x}}{\sqrt{5 - x} + \sqrt{x}}\,dx = \int_{2}^{3} 1\,dx \)
⇒ \( 2I = [x]_{2}^{3} = 1 \) ⇒ I = 1/2

Example 19. If f(a + b − x) = f(x), then \( \int_{a}^{b} x f(x)\,dx \) is equal to [Kurukshetra CEE 1993; AIEEE 2003]
(a) \( \frac{a + b}{2}\int_{a}^{b} f(b - x)\,dx \)
(b) \( \frac{a + b}{2}\int_{a}^{b} f(x)\,dx \)
(c) \( \frac{b - a}{2}\int_{a}^{b} f(x)\,dx \)
(d) None of these
Answer:
Ans. (b) \( \frac{a + b}{2}\int_{a}^{b} f(x)\,dx \)
\( I = \int_{a}^{b} x f(x)\,dx \) and \( I = \int_{a}^{b} (a + b - x) f(a + b - x)\,dx \)
⇒ \( I = \int_{a}^{b} (a + b - x) f(x)\,dx \) ⇒ \( I = \int_{a}^{b} (a + b) f(x)\,dx - \int_{a}^{b} x f(x)\,dx \) ⇒ \( 2I = (a + b)\int_{a}^{b} f(x)\,dx \) ⇒ \( I = \frac{a + b}{2}\int_{a}^{b} f(x)\,dx \)

Example 20. If \( \int_{0}^{\pi} x f(\sin x)\,dx = k\int_{0}^{\pi} f(\sin x)\,dx \), then the value of k will be [IIT 1982]
(a) \( \pi \)
(b) \( \pi/2 \)
(c) \( \pi/4 \)
(d) 1
Answer:
Ans. (b) \( \pi/2 \)
Given, \( \int_{0}^{\pi} x f(\sin x)\,dx = k\int_{0}^{\pi} f(\sin x)\,dx \)
⇒ \( \int_{0}^{\pi} (\pi - x) f(\sin(\pi - x))\,dx = k\int_{0}^{\pi} f(\sin(\pi - x))\,dx \) ⇒ \( \pi\int_{0}^{\pi} f(\sin x)\,dx - \int_{0}^{\pi} x f(\sin x)\,dx = k\int_{0}^{\pi} f(\sin x)\,dx \)
⇒ \( \pi\int_{0}^{\pi} f(\sin x)\,dx - 2k\int_{0}^{\pi} f(\sin x)\,dx = 0 \) ⇒ \( (\pi - 2k)\int_{0}^{\pi} f(\sin x)\,dx = 0 \)
∴ \( \pi - 2k = 0 ⇒ k = \pi/2 \).

Example 21. \( \int_{-10}^{20} [\cot^{-1} x]\,dx \), where [.] denotes the greatest integer function, is
(a) \( 30 + \cot 1 + \cot 3 \)
(b) \( 30 + \cot 1 + \cot 2 + \cot 3 \)
(c) \( 30 + \cot 1 + \cot 2 \)
(d) None of these
Answer:
Ans. (b) \( 30 + \cot 1 + \cot 2 + \cot 3 \)
Let \( I = \int_{-10}^{20} [\cot^{-1} x]\,dx \),
we know \( \cot^{-1} x \in (0, \pi) \) ∀ x ∈ R
thus, \( [\cot^{-1} x] = \begin{cases} 3, & x \in (-\infty, \cot 3) \\ 2, & x \in (\cot 3, \cot 2) \\ 1, & x \in (\cot 2, \cot 1) \\ 0, & x \in (\cot 1, \infty) \end{cases} \)
Hence, \( I = \int_{-10}^{\cot 3} 3\,dx + \int_{\cot 3}^{\cot 2} 2\,dx + \int_{\cot 2}^{\cot 1} 1\,dx + \int_{\cot 1}^{20} 0\,dx = 30 + \cot 1 + \cot 2 + \cot 3 \)

Example 22. \( \int_{0}^{2} [x^2 - x + 1]\,dx \), where [.] denotes the greatest integer function, is
(a) \( \frac{7 - \sqrt{5}}{2} \)
(b) \( \frac{7 + \sqrt{5}}{2} \)
(c) \( \frac{\sqrt{5} - 3}{2} \)
(d) None of these
Answer:
Ans. (d) None of these
Here \( x^2 - x + 1 \lt 1 \) for 0 < x < 1, \( 1 \leq x^2 - x + 1 \lt 2 \) for \( 1 \leq x \lt \frac{1 + \sqrt{5}}{2} \) and \( 2 \leq x^2 - x + 1 \leq 3 \) for \( \frac{1 + \sqrt{5}}{2} \leq x \leq 2 \).
Let \( I = \int_{0}^{2} [x^2 - x + 1]\,dx = \int_{0}^{1} [x^2 - x + 1]\,dx + \int_{1}^{\frac{1 + \sqrt{5}}{2}} [x^2 - x + 1]\,dx + \int_{\frac{1 + \sqrt{5}}{2}}^{2} [x^2 - x + 1]\,dx \)
\( = \int_{0}^{1} 0\,dx + \int_{1}^{\frac{1 + \sqrt{5}}{2}} 1\,dx + \int_{\frac{1 + \sqrt{5}}{2}}^{2} 2\,dx = \frac{\sqrt{5} - 1}{2} + (3 - \sqrt{5}) = \frac{5 - \sqrt{5}}{2} \)
Since \( \frac{5 - \sqrt{5}}{2} \) is not among options (a), (b), (c), the answer is (d).