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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
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)
∫(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
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
∫(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}
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
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
∫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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