If I1 - -0 x sin x ecos 4x dx - I2 - -20cos x ecos 4xdx- then the value of -I1-I2- is where - denotes the greatest integer function

I_1 = ∫^π_0 x sin x e^cos 4x dx⋯ (i) I_1 = ∫^π_0 (π x) sin x e^cos 4x dx ⋯ (ii) On adding (i) and (ii), we get 2I_1 = π∫^π_0sin x. e^cos 4x dx 2I_1 = 2 π∫^π ...

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Byju's Answer

Standard XII

Mathematics

Property 4

If I1 = ∫π0 x...

Question

If $I_{1} = \int_{0}^{π} x sin x e^{cos 4 x} d x & I_{2} = \int_{0}^{\frac{π}{2}} cos x e^{cos 4 x} d x$ , then the value of $[\frac{I_{1}}{I_{2}}]$ is (where [.] denotes the greatest integer function)

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Solution

$I_{1} = \int_{0}^{π} x sin x e^{cos 4 x} d x \dots$ (i)
$I_{1} = \int_{0}^{π} (π - x) sin x e^{cos 4 x} d x \dots$ (ii)
On adding (i) and (ii), we get-
$2 I_{1} = π \int_{0}^{π} sin x . e^{cos 4 x} d x 2 I_{1} = 2 π \int_{0}^{\frac{π}{2}} sin x e^{cos 4 x} d x I_{1} = π \int_{0}^{\frac{π}{2}} sin x . e^{cos 4 x} d x I_{1} = π \int_{0}^{\frac{π}{2}} cos x . e^{cos 4 x} d x \frac{I_{1}}{I_{2}} = π \Rightarrow [\frac{I_{1}}{I_{2}}] = 3$

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If I1=∫π0x sinx ecos4x dx & I2=∫π20cosx ecos4xdx, then the value of [I1I2] is (where [.] denotes the greatest integer function)

I1=∫π0x sinx ecos4x dx⋯ (i) I1=∫π0(π−x)sinx ecos4x dx⋯ (ii) On adding (i) and (ii), we get- 2I1=π∫π0sinx.ecos4x dx2I1=2π∫π20sinx ecos4x dxI1=π∫π20sinx.ecos4x dxI1=π∫π20cosx.ecos4xdxI1I2=π⇒[I1I2]=3

If $I_{1} = \int_{0}^{π} x sin x e^{cos 4 x} d x & I_{2} = \int_{0}^{\frac{π}{2}} cos x e^{cos 4 x} d x$ , then the value of $[\frac{I_{1}}{I_{2}}]$ is (where [.] denotes the greatest integer function)