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We have
Putting x^{n} = t so that n x^{n–1} dx = dt
Comparing it with the given value, we get
put t = 1/x ⇒ dt = 1/x2 as t = π/2 and π
Put x = 2 cos θ ⇒ dx =  2 sin θ dθ, then
Integrate it by parts taking log (1+ x/2 )as first function
Since sinq is positive in interval (0, π)
By adding (i) and (ii), we get
Now, Put tan^{2}x = t, we get
denotes the greater integer less than or equal to x
If [x] denotes the greater integer less than or equal to x, then the value of
If f(x) = tan x  tan3 x + tan5 x  …… to ∞ with 0 < x < π/4, then
I = ∫0 π2 log(tan x).dx
I = ∫0 π2 log(cot x).dx
Adding both the equations, we get
2I = ∫0 π2 log(tanx) + log(cot x) dx
2I = ∫0 π2 log(1).dx
= 0
f’(x) = 1/x^{2}
Thus, ∫(1 to 2)e^{x}(1/x  1/x^{2})dx
= [e^{x}/x](1 to 2) + c
= e^{2}/2  e
Here on adding we get
Differentiating both sides, we get
Comparing the coefficient of like terms on both sides, we get
Differentiating both sides, we get
Comparing the like powers of x in both sides, we get
t = ln(tan x)
dt = (sec^{2} x)/(tan x) dx
=> (1/cos^2x) * (cosx /sinx) dx = dt
dt = dx/(cosx sinx)
I = ∫t dt
= [t^{2}]/2 + c
= 1/2[ln(tanx)]^{2} + c
ut sin x = t Þ cos x dx = dt, so that reduced integral is
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