If I - -4-4e x -sin x- 2 x dx-1 - esin x- then

I = ∫^π/4_ π/4e^ x |sin x|^2 x dx/(1 + e^sin x)... nbsp; (1) Using ∫^b_a f(x) dx = ∫^b_a f(a + b x)dx I = ∫^π/4_ π/4e^ x |sin x|^2 x dx/(1 + e^ sin x)... ...

If I = ∫π/4-π...

Question

If $I = \int_{\frac{- π}{4}}^{\frac{π}{4}} \frac{e^{sec x} | sin x | {sec}^{2} x d x}{(1 + e^{sin x})}$ , then

I

is an irrational number

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I = e^{\sqrt{2}} - e

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ln (I + e) = \sqrt{2}

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I

is a rational number

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Solution

The correct option is C $ln (I + e) = \sqrt{2}$
$I = \int_{\frac{- π}{4}}^{\frac{π}{4}} \frac{e^{sec x} | sin x | {sec}^{2} x d x}{(1 + e^{sin x})}$ ... (1)
Using $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$
$I = \int_{\frac{- π}{4}}^{\frac{π}{4}} \frac{e^{sec x} | sin x | {sec}^{2} x d x}{(1 + e^{- sin x})}$ ... (2)
Add (1) & (2)
$2 I = \int_{\frac{- π}{4}}^{\frac{π}{4}} e^{sec x} | sin x | {sec}^{2} x d x I = \int_{0}^{\frac{π}{4}} e^{sec x} sec x tan x d x$
$e^{sec x} = t I = \int_{e}^{e^{\sqrt{2}}} d t = e^{\sqrt{2}} - e$

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Standard XII Mathematics

If I=∫π4−π4esec x|sin x|sec2x dx(1+esin x), then

The correct option is C ln(I+e)=√2I=∫π4−π4esec x|sin x|sec2x dx(1+esin x)... (1) Using ∫baf(x)dx=∫baf(a+b−x)dx I=∫π4−π4esec x|sin x|sec2x dx(1+e−sin x)... (2) Add (1) & (2) 2I=∫π4−π4esec x|sin x|sec2x dxI=∫π40esec xsec x tan x dx esec x=tI=∫e√2edt=e√2−e

If $I = \int_{\frac{- π}{4}}^{\frac{π}{4}} \frac{e^{sec x} | sin x | {sec}^{2} x d x}{(1 + e^{sin x})}$ , then