∫ limits0 e37(π sin(π log x)/x)dx is equal to

Question

∫ limits0 e37(π sin(π log x)/x)dx is equal to (A) 2 (B) - 2 (C) (2/π) (D) 2 π

Tardigrade Admin · Accepted Answer

Put π log x=z ∴ (π/x) dx=dz When x= 1, z = 0, When x =e37, π log e37=z ∴ z=37 π ∴ given integral =∫37π0 sin z dz =|-cos z|037 π =-cos(37 π)+cos 0=1-cos π =1+1=2

Q. $0 \int_{e} 37 \frac{π s i n ( π l o g x )}{x} d x$ is equal to

2

- 2

$\frac{2}{π}$

$2 π$

Put $π l o g x = z$
$∴ \frac{π}{x} d x = d z$
When $x = 1, z = 0$ , When $x = e^{37}, π l o g e^{37} = z$
$∴ z = 37 π$
$∴$ given integral $= \int_{0}^{37 π} s in z d z = ∣ - cos z ∣_{0}^{37 π}$
$= - cos (37 π) + cos 0 = 1 - cos π = 1 + 1 = 2$

Q. 0∫e​37​xπsin(πlogx)​dx is equal to

2

- 2

π2​

2π

Put πlogx=z ∴xπ​dx=dz When x=1,z=0, When x=e37,πloge37=z ∴z=37π ∴ given integral =∫037π​sinzdz=∣−cosz∣037π​ =−cos(37π)+cos0=1−cosπ=1+1=2

Q. $0 \int_{e} 37 \frac{π s i n ( π l o g x )}{x} d x$ is equal to

$\frac{2}{π}$

$2 π$

Put $π l o g x = z$
$∴ \frac{π}{x} d x = d z$
When $x = 1, z = 0$ , When $x = e^{37}, π l o g e^{37} = z$
$∴ z = 37 π$
$∴$ given integral $= \int_{0}^{37 π} s in z d z = ∣ - cos z ∣_{0}^{37 π}$
$= - cos (37 π) + cos 0 = 1 - cos π = 1 + 1 = 2$