JEE Advanced Level Test: Definite and Indefinite Integral - JEE MCQ

We have 

Putting xⁿ = 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 

∫_1/π^2/π [sin(1/x)] / x² dx

To simplify the integral, we use the substitution method.

Step 1 : Let u = 1/x

du/dx = -1/x²

dx = -1/u² dx

Step 2: Change the Limits of Integration

We need to adjust the limits of integration to match our substitution.

1. When x = 1/π:

   u = 1 / (1/π) = π

2. When x = 2/π:

   u = 1 / (2/π) = π/2

Step 3: Substitute and Simplify the Integral

Substitute u and dx into the integral:

∫_1/π^2/π [sin(1/x)] / x² dx = ∫_π^π/2 sin(u) * (-1/u²) du

Notice the negative sign. To simplify, reverse the limits of integration and remove the negative sign:

= ∫_π/2^π sin(u) du

Step 4: Integrate

Integrate sin(u) with respect to u:

∫ sin(u) du = -cos(u) + C

Apply the definite integral:

[ -cos(u) ]_π/2^π = [ -cos(π) ] - [ -cos(π/2) ]

Step 5: Compute the Values

Evaluate the cosine values:

• cos(π) = -1
• cos(π/2) = 0

Substitute these values into the expression:

= [ -(-1) ] - [ -0 ] = 1 - 0 = 1

Final Answer

∫_1/π^2/π [sin(1/x)] / x² dx = 1

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 tan²x = t, we get

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/x²
Thus, ∫(1 to 2)e^x(1/x - 1/x²)dx 
= [e^x/x](1 to 2) + c
= e²/2 - 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 = (sec² x)/(tan x) dx
=> (1/cos^2x) * (cosx /sinx) dx = dt
dt = dx/(cosx sinx)
I = ∫t dt
= [t²]/2 + c
= 1/2[ln(tanx)]² + c

ut sin x = t Þ cos x dx = dt, so that reduced integral is