Engineering Mathematics Test 1

Explanation:

Given that, 1, 2, 3 are Eigen values of A_3×3

We can write as,

(λ – 1)(λ - 2)(λ - 3) = 0

(λ² – 2λ - λ + 2) (λ - 3) = 0

λ³ – 6λ² + 11λ – 6 = 0

This is a characteristic equation of A_3×3

By a Cayley-Hamilton theorem, matrix A satisfies its characteristic equation.

i.e.

Explanation:

\(\frac{{{d^2}y}}{{d{x^2}}} = 0\)

\(\frac{{dy}}{{dx}} = {C_1} = 5\)

y = C₁x + C₂

at x = 0, y = 3

3 = C₂

Hence, y = 5x + 3

∴ y(3) = 5 × 3 + 3 = 18

Explanation:

A and B are mutually exhaustive events

Therefore, P(A ∪ B) = 1

Now,

 P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

\(1=\frac{3}{4}+\frac{1}{2}-P\left( A\cap B \right)\)

\(\therefore P\left( A\cap B \right)=\frac{1}{4}\)

Now,

P(A’ ∪ B’) = P(A ∩ B)’

P(A ∩ B) + P(A ∩ B)’ = 1

\(\frac{1}{4}~+P{{\left( A\cap B \right)}^{'}}=1\)

\(P\left( {A}'\cup {B}' \right)=P{{\left( A\cap B \right)}^{'}}=\frac{3}{4}=0.75\)

Explanation:

For continuous function:

\(\mathop {\lim }\limits_{x \to {2^ - }} f\left( x \right) = f\left( 2 \right) = \mathop {\lim }\limits_{x \to {2^ + }} f\left( x \right)\)

\(\mathop {\lim }\limits_{x \to {2^ - }} \left( {4 - x} \right) = \mathop {\lim }\limits_{x \to 2} \left( {4 - x} \right) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {kx - 4} \right)\)

(4 - 2) = 2 k - 4

k = 3

Explanation:

\(curl\;\vec V = \bar 0\) (Given: It is irrorational)

Now,

\(\left| {\begin{array}{*{20}{c}} {\vec i}&{\vec j}&{\vec k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {x + 2y + az}&{bx - 3y - z}&{4x + cy + 2z} \end{array}} \right| = \bar 0\)

\( \Rightarrow \left( {c + 1} \right)\vec i - \left( {4 - a} \right)\vec j + \left( {b - 2} \right)\vec k = \bar 0\)

Verify from the options

∴ c = -1, a = 4, b = 2

Concept:

Laplace transform of cosh at is, \(L\left( {\cosh at} \right) = \frac{s}{{{s^2} - {a^2}}}\)

Laplace transform of cosh at is, \(L\left( {\sinh at} \right) = \frac{a}{{{s^2} - {a^2}}}\)

If L{f(t)} = F(s) then \(L\left\{ {{t^n}f\left( t \right)} \right\} = {\left( { - 1} \right)^n}\frac{{{d^n}}}{{d{s^n}}}\left[ {F\left( s \right)} \right]\)

Calculation:

\(L\left( {\cosh t} \right) = \frac{s}{{{s^2} - 1}}\)

\(L\left( {t\cosh t} \right) = - \frac{d}{{ds}}\left( {\frac{s}{{{s^2} - 1}}} \right)\)

\( = - \frac{{\left( {{s^2} - 1} \right)\left( 1 \right) - s\left( {2s} \right)}}{{{{\left( {{s^2} - 1} \right)}^2}}}\)

\(I = \mathop \smallint \limits_0^\infty \frac{{dx}}{{{{\left( {{x^2} + 1} \right)}^2}}}\)

Put x = tan θ

⇒ dx = sec² θ dθ

Limits will be: 0 to π/2

\(I = \mathop \smallint \limits_0^{\frac{\pi }{2}} \frac{{{{\sec }^2}\theta }}{{{{\left( {ta{n^2}\theta + 1} \right)}^2}}}d\theta\)

\(= \mathop \smallint \limits_0^{\frac{\pi }{2}} \frac{{{{\sec }^2}\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^2}}}d\theta\)

\(= \mathop \smallint \limits_0^{\frac{\pi }{2}} {\cos ^2}\theta d\theta\)

\(= \mathop \smallint \limits_0^{\frac{\pi }{2}} \frac{{1 + \cos 2\theta }}{2}d\theta\)

Explanation:

\(f\left( z \right) = \frac{{{e^z}}}{{{{\left( {z - 3} \right)}^2}}}\)

C being |z| = 2

Poles of f(z) are z = 3, 3

Poles are lying outside the curve. Hence by Cauchy’s integral theorem,

\(\mathop \smallint \limits_c \frac{{{e^z}}}{{{{\left( {z - 3} \right)}^2}}} = dz = 0\)

Explanation:

Let f(x) = x – e^-x

f¹(x) = 1 + e^-x

Now,

From Newton – Raphson method

\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{{f^1}\left( {{x_n}} \right)}}\)

\( = {x_n} - \frac{{\left[ {{x_n} - {e^{ - xn}}} \right]}}{{\left[ {1 + {e^{ - xn}}} \right]}}\)

\({x_{n + 1}} = \frac{{\left( {{x_n} + {x_n}{e^{ - xn}} - {x_n} + {e^{ - xn}}} \right)}}{{\left( {1 + {e^{ - xn}}} \right)}}\)

\(\therefore {x_{n + 1}} = \frac{{\left( {1 + {x_n}} \right){e^{ - {x_n}}}}}{{\left( {1 + {e^{ - {x_n}}}} \right)}}\)

Given:

\(\frac{{dy}}{{dt}} = \left( {1 + y} \right)\)

\(\frac{{dy}}{{\left( {1 + y} \right)}} = dt\)

Integrating both sides and taking limits.

\(\mathop \smallint \limits_0^{999} \frac{{dy}}{{\left( {1 + y} \right)}} = \mathop \smallint \limits_0^t dt\)

\(\therefore \left[ {ln\;\left( {1 + y} \right)} \right]_0^{999} = t\)

In (1000) – In (1) = t

∴ t = In 1000 = 3 log_e 10

Concept:

Order: The order of a differential equation is the order of the highest derivative appearing in it.

Degree: The degree of a differential equation is the power of the highest derivative occurring in it, after the Equation has been expressed in a form free from radicals as far as the derivatives are concerned.

Calculation:

By definition of order and degree partial differential equation,

Order = 2

Degree = 1

Putting x = ∞ to the given limiting function, we get:

\({\left( {1 + \frac{1}{\infty }} \right)^\infty } = {1^\infty }\) 

1^∞ is one of the indeterminant forms.

We can modify the given limits as:

\(\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \frac{1}{x}} \right)^x} = \mathop {\lim }\limits_{x \to \infty } {e^{ln{{\left( {1 + \frac{1}{x}} \right)}^x}}}\) 

\( = {e^{\mathop {\lim }\limits_{x \to \infty } ln{{\left( {1 + \frac{1}{x}} \right)}^x}}}\) 

\( = {e^{\mathop {\lim }\limits_{x \to \infty } x\;ln\left( {1 + \frac{1}{x}} \right)}}\) 

The above can be written as:

e^f(x)    --(1)

where:

\(f\left( x \right) = \mathop {\lim }\limits_{x \to \infty } x\left( {1 + \frac{1}{x}} \right)\)       ---(2)

The above can be written as:

\(\mathop {\lim }\limits_{x \to \infty } \frac{{ln\left( {1 + \frac{1}{x}} \right)}}{{1/x}}\)  

Putting on the limit of x = ∞, we get:

\(\mathop {\lim }\limits_{x \to \infty } \frac{{In\left( {1 + \frac{1}{x}} \right)}}{{1/x}} = \frac{0}{0}\) 

Applying L-Hospitals rule, we get:

\( = \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {\frac{1}{{1 + \frac{1}{x}}}} \right)\left( {\frac{{ - 1}}{{{x^2}}}} \right)}}{{ - 1/{x^2}}}\) 

\( = \mathop {\lim }\limits_{x \to \infty } + \frac{1}{{1 + \frac{1}{x}}} = 1\) 

Putting this in equation (1), we can write:

\({e^{\mathop {\lim }\limits_{x \to \infty } x\;In\;\left( {1 + \frac{1}{x}} \right)}} = {e^1}\) 

Explanation:

\(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right) = 0\;\) 

⇒ f(x) is continuous at x = 0

\(f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{{x^2},\;\;x \ge 0}\\{ - {x^2},\;\;x < 0}\end{array}} \right.\)

\({f^1}\left( {0 - } \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( h \right)}}{h}\)

\(= \mathop {\lim }\limits_{h \to 0} \frac{{ - {h^2}}}{h} = 0\)

\({f^1}\left( {0 + } \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {0 + h} \right) - f\left( h \right)}}{h}\)

\( = \mathop {\lim }\limits_{h \to 0} \frac{{{h^2}}}{h} = 0\)

∴ f¹(0) = 0

Explanation:

u(x, y) = x³ + ax²y + bxy² + 2y³

The given function is harmonic function.

From CR equations,

u_x = v_y , u_y = -v_x

u_x = 3x² + 2axy + by², u_y = ax² + 2bxy + 6y²

⇒ v_y = 3x² + 2axy + by²

Now,

By integrating on both sides with respect to y.

\(\Rightarrow \smallint {v_y}dy = \smallint \left( {3{x^2} + 2axy + b{y^2}} \right)dy\)

⇒ v = 3x²y + axy² + \(\frac{{b{y^3}}}{3} + \phi \left( x \right)\)

⇒ v_x = 6xy + ay² + ϕ’(x) = -u_y

⇒ 6xy + ay² + ϕ’(x) = -ax² – 2bxy – 6y²

By comparing on both sides,

-2b = 6 ⇒ b = -3

a = -6

ϕ’(x) = -ax²

\(\Rightarrow \phi \left( x \right) = - \frac{{a{x^3}}}{3} + C = \frac{{6{x^3}}}{3} + C = 2{x^3} + C\)

Now, v = 3x²y – 6xy² – y³ + 2x³ + C

Given that, v (0, 0) = 1

⇒ C = 1

Now, v = 3x²y – 6xy² – y³ + 2x³ + 1

v(1, 1) = 3 – 6 – 1 + 2 – 1 = -1

Concept:

According Euler’s method

y_{n + 1} = y_n + h f(x_n, y_n)

Calculation:

Here,

h = 0.1, y₀ = y(0) = 0

y_(0.1) = y₀ + 0.1 (0 + 0) = 0

y_(0.2) = y_0.1 + h f(x_0.1, y_0.1)

y_(0.2) = 0 + (0.1) [0.1 + 0]

y_(0.2) = 0.01

y_(0.3) = y_(0.2) + f(x_0.2, y_0.2)

y_(0.3) = 0.01 + 0.1 [0.2 + 0.01]

∴ y_(0.3) = 0.031

Explanation:

Sample space if 4 appears on one of the dice:

S = {(1, 4), (2, 4), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 4), (6, 4)}

the sum of the numbers on dice is divisible by 2

E = {2, 4), (4, 2), (4, 4), (4, 6), (6, 4)}

probability that the sum of the numbers on dice is divisible by 2 is

\(p\left( E \right) = \frac{5}{{11}}\;\;\)

Alternate way:

Using conditional probability:

\({\rm{P}}\left( {\frac{{\rm{E}}}{{\rm{S}}}} \right) = \frac{{P\left( {E \cap S} \right)}}{{P\left( S \right)}} = \frac{{\frac{5}{{36}}}}{{\frac{{11}}{{36}}}} = \frac{5}{{11}}\)

Concept: 

Maxima and Minima:

Steps to finding maxima and minima

• find the stationary point at which f’(x) = 0.
• Then to find maxima and minima we will check the polarity of the second derivative.

 Case 1. If f”(xo) > 0 then x = xo will be minima where xo is stationary point on f(x).

 Case 2. If f”(xo) < 0 then x = xo will be maxima where xo is stationary point on f(x).

Calculation:

Given:

\(f\left( x \right) = \frac{1}{3}x\left( {{x^2} - 3} \right) = \frac{{{x^3}}}{3} - x\)

Now, we know that

For relative minimum and maximum;

putting \(\frac{{df}}{{dx}} = 0\) and solving for x, we get;

⇒ x² - 1 = 0

⇒ x = ±1

\(\frac{{{d^2}f}}{{d{x^2}}} = 2x\) which gives \(\frac{{{d^2}f}}{{d{x^2}}} = +~2\)  i.e. + Ve for x = +1

So, x = 1 is a point of relative minimum.

For the range given the minimum value can also take place to x = -100

So, checking at x = 1 and x = -100, we get:

At x = 1,

\(f\left( 1 \right) = \frac{1}{3}\left( {1 - 3} \right) = - \frac{2}{3}\)

at, x = -100,

f(-100) \(= \frac{1}{3}\left( { - 100} \right)\left( {10000 - 3} \right)\)

f(-100) \(= - \frac{{100}}{3}\left( {9997} \right)\)

f(-100) = -333233.33

So, f(-100) is the minimum value in the interval given.

∴ the minimum value occurs at x = -100.

Explanation:

Given: \({x^2}{y^{11}} + 6x{y^I} + 6y = x\) 

Now,

D(D – 1)y + 6Dy + 6y = e^t, where \(D = \frac{d}{{dt}}\) 

Put x = e^t, t = ln x

⇒ (D² + 5D + 6) y = e^t

Auxillary Equation is D² + 5D + 6 = 0

Solving the above Equation we get roots as -2, -3

Now,

\(CF = {c_1}{e^{ - 2t}} + {c_2}{e^{ - 3t}}\)

\(CF = \frac{{{c_1}}}{{{x^2}}} + \frac{{{c_2}}}{{{x^3}}}\)

\(\therefore {\rm{The\;complementary\;function\;is\;}}\left( {C.F.} \right) = \frac{{{c_1}}}{{{x^2}}} + \frac{{{c_2}}}{{{x^3}}}\)

Now,

\(P.I = \frac{1}{{\left( {{D^2} + 5D + 6} \right)}}{e^t}\)

\(P.I. = \frac{{{e^t}}}{{\left( {{1^2} + 5 \times 1 + 6} \right)}} = \frac{{{e^t}}}{{12}}\)

\(\therefore P.I = \frac{x}{{12}}\)

\(\therefore {\rm{The\;particular\;integral\;is\;}}\frac{x}{{12}}\)

Now,

∴ The general solution is equal to C.F. + P.I.

\(\therefore y = \frac{{{c_1}}}{{{x^2}}} + \frac{{{c_2}}}{{{x^3}}} + \frac{x}{{12}}\)

Concept:

\({\rm{Fourier\;series\;of\;}}f\left( x \right) = \frac{{{a_0}}}{2} + \mathop \sum \limits_{n = 1}^\infty \left( {{a_n}\cos \frac{{n\pi x}}{l} + {b_n}\sin \frac{{n\pi x}}{l}} \right)\)

\({a_n} = \frac{1}{l}\mathop \smallint \limits_0^{2l} f\left( x \right)\cos \frac{{2\pi x}}{l}dx\)

Calculation:

\({a_2} = \frac{1}{3}\mathop \smallint \limits_0^6 f\left( x \right)\cos \frac{{2\pi x}}{3}dx\)

\({a_2} = \frac{1}{3}\mathop \smallint \limits_0^3 x\cos \frac{{2\pi }}{3}x\;dx + \frac{4}{3}\mathop \smallint \limits_3^6 \cos \frac{{2\pi }}{3}\;x\;dx\)

\({a_2} = \frac{1}{3}\left[ {x\mathop \smallint \limits_0^3 \cos \frac{{2\pi x}}{3}\;dx - \mathop \smallint \limits_0^3 \left( {\smallint \cos \frac{{2\pi x}}{3}dx} \right)dx} \right] + \frac{4}{3}\mathop \smallint \limits_3^6 \cos \frac{{2\pi x}}{3}dx\)

\({a_2} = \frac{1}{3}\left[ {\frac{{3x}}{{2\pi }}\left. {\sin \frac{{2\pi x}}{3}} \right|_0^3 + {{\left( {\frac{3}{{2\pi }}} \right)}^2}\left. {\cos \frac{{2\pi x}}{3}} \right|_0^3} \right] + \frac{4}{3} \times \frac{3}{{2\pi }}\left. {\sin \frac{{2\pi x}}{3}} \right|_3^6\)

\({a_2} = \frac{1}{3}\left[ {\frac{9}{{2\pi }}\sin 2\pi - \sin 0 + \frac{9}{{4{\pi ^2}}}\left\{ {\cos 2\pi - \cos 0} \right\}} \right] + 0\)

∴ a₂ = 0

Explanation:

Writing the equations in matrix form,

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 5&3&7\\ 3&{26}&2\\ 7&2&{10} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 4\\ 9\\ 5 \end{array}} \right]\)

R₁ → 3R₁

R₂ → 5R₂

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} {15}&9&{21}\\ {15}&{130}&{10}\\ 7&2&{10} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {12}\\ {45}\\ 5 \end{array}} \right]\)

R₂ → R₂ – R₁

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} {15}&9&{21}\\ 0&{121}&{ - 1}\\ 7&2&{10} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {12}\\ {33}\\ 5 \end{array}} \right]\)

\({R_1} \to \frac{7}{3}{R_1};{R_2} \to \frac{{{R_2}}}{{11}};{R_3} \to 5{R_3}\)

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} {35}&{21}&{49}\\ 0&{11}&{ - 1}\\ {35}&{10}&{50} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {28}\\ 3\\ {25} \end{array}} \right]\)

R₃ → R₃ + R₂ – R₁; R₁ → R₁|7

\( \Rightarrow \left[ {\begin{array}{*{20}{c}} 5&3&7\\ 0&{11}&{ - 1}\\ 0&0&0 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 4\\ 3\\ 0 \end{array}} \right]\)

Hence, Z = 0

5x + 3y = 4

\(11y = 3 \Rightarrow y = \frac{3}{{11}}\)

\(x = \frac{7}{{11}}\)

Thus,

\(\therefore {\rm{x\;}} + {\rm{\;y\;}} + {\rm{\;z\;}} = \frac{{10}}{{11}}\)

Concept:

Gauss Divergence Theorem:

The surface integral of the normal component of a vector function \(\vec F\) taken around a closed surface S is equal to the integral of the divergence of \(\vec F\) taken over the volume V enclosed by the surface S. Mathematically;

\(\oint_{S}^{}\vec{F}.\hat{n}ds=\int_{V}^{}div\vec{F}\;dV\)

Calculation:

Given:

\(\mathop \int\!\!\!\int \limits_S \left( {x + z} \right)dydz + \left( {y + z} \right)dzdx + \left( {x + y} \right)dxdy\) and x2 + y2 + z2 = a2

\(∴\;{\rm{\vec F}} = \left( {x + z} \right)\hat i + \left( {y + z} \right)\hat j + \left( {x + y} \right)\hat k\)

From the Gauss divergence theorem;

\(\oint_{S}^{}\vec{F}.\hat{n}ds=\int_{V}^{}div\vec{F}\;dV\)

\(∴ div\;\vec F=\nabla .{\rm{\vec F}} \Rightarrow 1 + 1 + 0 = 2\)

\(\int_{V}^{}\nabla.\vec F\;dV=\int_{V}^{}2\;dV\)

where \(\int_{V}^{}dV\) represents volume of sphere.

The equation for the surface of sphere is x² + y² + z² = a² , ∴  

∴ the radius of sphere is a.

The volume of the sphere \(= \frac{4}{3}\pi {a^3}\)

\(\Rightarrow \int_{V}^{}2\;dV=2\times \frac{4}{3}\pi {{a}^{3}}=\frac{8}{3}\pi {{a}^{3}}\)

Explanation:

If A_{3 × 3} has repeated eigen values, then also A can have 3 linearly independent eigen vectors. For example I₃ has only one distinct eigen value but there are 3 linearly independent eigen vectors.

∴ option (a) need not be true.

If A is 3 distinct eigen values then A has 3 linearly independent eigen vectors.

Therefore option (b) need not be true.

If A is non singular matrix, then A need not contain 3 linearly independent eigen vectors.

For example a diagonal matrix with 3 distinct eigen values 0, 1, 2 and 3 linearly independent eigen vectors. But the diagonal matrix is singular.

Concept:

By Green’s theorem

\(\mathop \oint \nolimits_C Mdx + Ndy = \mathop \int\!\!\!\int \limits_R^{} \left[ {\frac{{\partial N}}{{\partial x}} - \frac{{\partial M}}{{\partial y}}} \right]dx\;dy\)

Calculation:

\(\frac{{\partial N}}{{\partial x}} - \frac{{\partial M}}{{\partial y}} = - 2y + 3x{y^2}\)

\(\frac{{\partial N}}{{\partial x}} - \frac{{\partial M}}{{\partial y}} = \mathop \smallint \limits_{x = 0}^2 \mathop \smallint \limits_{y = 0}^2 \left[ { - 2y + 3x{y^2}} \right]dydx\)

\(\frac{{\partial N}}{{\partial x}} - \frac{{\partial M}}{{\partial y}} = \mathop \smallint \limits_0^2 \left[ { - 2\left( {\frac{{{y^2}}}{2}} \right) + 3x\left( {\frac{{{y^3}}}{3}} \right)} \right]_0^2dx\)

\(\frac{{\partial N}}{{\partial x}} - \frac{{\partial M}}{{\partial y}} = \mathop \smallint \limits_0^2 \left( { - 4 + 8x} \right)dx\)

\(\frac{{\partial N}}{{\partial x}} - \frac{{\partial M}}{{\partial y}} = \left[ { - 4x + 4{x^2}} \right]_0^2\)

\(\frac{{\partial N}}{{\partial x}} - \frac{{\partial M}}{{\partial y}} = - 8 + 16\)

\(\therefore \frac{{\partial N}}{{\partial x}} - \frac{{\partial M}}{{\partial y}} = 8\)

Explanation:

Given:

Mean number of defectives = 2 = np

⇒ 2 = 20p

Calculation:

\({\rm{Hence}},{\rm{\;probability\;of\;defective\;part\;}} = p = \frac{2}{{20}} = 0.1\)

Probability of non-defective = 0.9

Now,

∴ Probability of at least three defectives in a sample of 20 = 1 – [P(0) + P(1) + P(2)]

\( = 1 - \left[ {{}_{}^{20}{C_0}{{\left( {0.9} \right)}^{20}} + {}_{}^{20}{C_1}\left( {0.1} \right){{\left( {0.9} \right)}^{19}} + {}_{}^{20}{C_2}{{\left( {0.1} \right)}^2}{{\left( {0.9} \right)}^{18}}} \right]\) 

= 0.323

Thus, the number of samples having at least three defective parts out of 1000 samples

= 1000 × 0.323

= 323

\({\rm{let\;I}} = {\rm{\;}}\mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \sec {\rm{x\;dx}}\)

multiply and divide by (sec x + tan x)

\({\rm{I}} = \mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \sec {\rm{x\;}} \times \frac{{\sec {\rm{x}} + \tan {\rm{x}}}}{{{\rm{secx}} + \tan {\rm{x}}}}{\rm{\;dx\;}}\)

\({\rm{I}} = \mathop \smallint \limits_0^{\frac{{\rm{\pi }}}{2}} \frac{{{{\sec }^2}x + {\rm{secx}}.\tan {\rm{x}}}}{{{\rm{secx}} + \tan {\rm{x}}}}{\rm{\;dx}}\)

Let t= sec x + tan x

Differentiation on both sides w.r.t x

dt = (sec x tan x + sec² x) dx

Substitute these values in above equation,

\({\rm{I}} = {\rm{\;}}\mathop \smallint \nolimits_0^{\frac{{\rm{\pi }}}{{2{\rm{\;}}}}} \frac{{{\rm{dt}}}}{{\rm{t}}} = {\log _{\rm{e}}}{\rm{t\;}}\)

\({\rm{I}} = \left[ {{{\log }_{\rm{e}}}\left( {\sec {\rm{x}} + \tan {\rm{x}}} \right)} \right]_0^{\frac{\pi }{2}}\)

\({\rm{I}} = \left[ {{{\log }_{\rm{e}}}\left( {\sec \frac{\pi }{2} + \tan \frac{\pi }{2}} \right) - \;{{\log }_e}\left( {\sec 0 - \tan 0} \right)} \right]_{}^{}\)

\({\rm{I}} = \left[ {{{\log }_{\rm{e}}}\left( {\sec \frac{\pi }{2} + \tan \frac{\pi }{2}} \right) - \;{{\log }_e}\left( {1\;} \right)} \right]_{}^{}\)

As, it does not produce any exact result, because tan 90 and sec 90 are ND (not defined). This is a divergent