If I_1=∫0(π)/2e^(sin⁡x)1+xcos⁡xdx and I_2=∫0(π)/2e^(cos⁡x)1-xsin⁡xdx, then I_1/I_2 is equal to (where x denotes the greatest integer less than or…

Question

If I_1=∫0(π)/2e^(sin⁡x)1+xcos⁡xdx and I_2=∫0(π)/2e^(cos⁡x)1-xsin⁡xdx, then I_1/I_2 is equal to (where x denotes the greatest integer less than or equal to x)

MARKS Content Team · Accepted Answer

Correct answer: 2. I_1=∫0(π)/2e^(sin⁡x)+xe^(sin⁡x).cos⁡xdx=xe^(sin⁡x) (π)/2 0 =(πe)/2 I_2=∫0(π)/2e^(cos⁡x)+x-e^(cos⁡x).sin⁡xdx=xe^(cos⁡x) (π)/2 0 =(π)/2 Hence, I_1I_2=e⇒e=2

If $I_{1} = \int_{0}^{\frac{π}{2}} e^{\sin x} (1 + x \cos x) d x$ and $I_{2} = \int_{0}^{\frac{π}{2}} e^{\cos x} (1 - x \sin x) d x,$ then $[\frac{I_{1}}{I_{2}}]$ is equal to (where $[x]$ denotes the greatest integer less than or equal to $x$ )

Solution

More Definite Integration Questions

If I1=∫0π2esin⁡x1+xcos⁡xdx and I2=∫0π2ecos⁡x1-xsin⁡xdx, then I1I2 is equal to (where x denotes the greatest integer less than or equal to x)

Solution

More Definite Integration Questions

If $I_{1} = \int_{0}^{\frac{π}{2}} e^{\sin x} (1 + x \cos x) d x$ and $I_{2} = \int_{0}^{\frac{π}{2}} e^{\cos x} (1 - x \sin x) d x,$ then $[\frac{I_{1}}{I_{2}}]$ is equal to (where $[x]$ denotes the greatest integer less than or equal to $x$ )