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QUESTION: 1

Evaluate:

Solution:

Let I = ∫(0 to pi/2)x^{3} sin(tan−1x^{4})/(1+x^{8})dx

Put tan−1 x^{4} = t

4x^{3}/(1+x^{8})dx = dt

⇒x^{3}dx/(1+x^{8}) = dt/4

Now, I = 1/4∫sin t dt

=−[1/4 cos t]0 to pi/2 + c

= -1/4[cos(0) - cos(pi/2)] +c

= 1/4[-cos(pi/2) + cos(0)] +c

= ¼[-0+1]

= 1/4

QUESTION: 2

Solution:

The given form of integral function (say ∫f(x)) can be transformed into another by changing the independent variable x to t,

Substituting x = g(t) in the function ∫f(x), we get;

dx/dt = g'(t)

or dx = g'(t).dt

Thus, I = ∫f(x).dx = f(g(t)).g'(t).dt

Therefore, ∫f(g(x)).g'(x).dx = f(b) - f(a)

QUESTION: 3

Solution:

∫(π/8 to π/4)1/(sin^{2}x cos^{2}x)

= ∫(π/8 to π/4)(sin^{2}x + cos^2x)/(sin^{2}x cos^{2}x) dx

= ∫(π/8 to π/4)(1/cos^{2}x +1/sin^{2}x) dx

= ∫(π/8 to π/4)(sec^{2} x dx + cosec^{2} x dx)

= [tan x - cot x](π/8 to π/4) + c

= [tan(π/8) - tan(π/4)] - [cot((π/8) - cot(π/4)]

= ((2)^{½} +1 - 1) - (1/(2)^{½} +1 - 1)

= 2

QUESTION: 4

The value of is:

Solution:

QUESTION: 5

If is

Solution:

In the question, it should be f’(2) instead of f”(2)

Explanation:- f(x) = ∫(0 to x) log(1+x^{2})

f’(x) = 2xdx/(1+x^{2})

f’(2) = 2(2)/(1+(2)^{2})

= 4/5

QUESTION: 6

Solution:

∫(1 to 3) x/(1+x^{2}) dx

Put t = 1+x^{2}

dt = 2x dx

= 1/2∫( 1 to 3) dt/t

= ½ log[t](1 to 3)

= ½[log(1+x^{2})](1 to 3)

= ½[log 10 - log 2]

= ½ (log(10/2)]

= ½[log 5]

= log(5)^{½}

QUESTION: 7

If then the value of k is:

Solution:

Let I=∫(0 to k) 1/[1 + 4x^{2}]dx = π8

Now, ∫(0 to k) 1/[4(1/4 + x^{2})]dx

= 2/4[tan^{−1} 2x]0 to k

= 1/2tan^{-1} 2k − 0 = π/8

1/2tan^{−1} 2k = π8

⇒ tan^{−1} 2k = π/4

⇒ 2k = 1

∴ k = 1/2

QUESTION: 8

Evaluate:

Solution:

QUESTION: 9

The value of is:

Solution:

1 + tan^{2} x = sec^{2} x

∫(π/2 to π/4)(sec^{2} x)/(1 + tan^{2} x)

=> ∫(π/2 to π/4)(sec^{2} x)/(sec^{2} x) dx

= ∫1(π/2 to π/4) dx

= [x](π/2 to π/4)

= π/2 - π/4 = π/4

QUESTION: 10

Solution:

∫x(5^{x}) dx

u = x, dv = 5^{x} dx

du = dx, v = (5^{x}/ln 5)

∫x(5^{x}) dx = [x(5^{x})/ln 5](0 to 1) - ∫(0 to 1)(5^{x})/ln 5 dx

= 5/ ln 5 - 0 - 1/ln 5[5^{x}/ln 5](0 to 1)

= 5/ln 5 - 5/(ln 5)^{2} + 1/(ln 5)^{2}

=5/ln 5 - 4/(ln 5)^{2}

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