Let g(t)=_-/2^/2(()/(4)t+f(x))dx, where f(x)=_e(x+x^2+1), xR.

Question

Let $g(t)=\displaystyle\int_{-\pi/2}^{\pi/2}\cos\left(\frac{\pi}{4}t+f(x)\right)dx$, where $f(x)=\log_e\left(x+\sqrt{x^2+1}\right)$, $x\in\mathbb{R}$. Then which one of the following is correct?

Accepted Answer

$$\because$$ $$f(x) = \ln \left( {x + \sqrt {{x^2} + 1} } \right)$$

$$\therefore$$ $$f(x) + f( - x) = \ln \left( {\sqrt {{x^2} + 1}  + x} \right) + \ln \left( {\sqrt {{x^2} + 1}  - x} \right)$$

$$\therefore$$ $f(x) + f( - x) = 0$ .... (i)

$$\because$$ $$g(t) = \int_{ - \pi /2}^{\pi /2} {\cos \left( {{\pi  \over 4}t + f(x)} \right)} dx$$

$$ = \int_0^{\pi /2} {\left\{ {\cos \left( {{\pi  \over 4}t + f(x)} \right) + \cos \left( {{\pi  \over 4}t + f( - x)} \right)} \right\}} dx$$

$$ = \int_0^{\pi /2} {\left\{ {\cos \left( {{{\pi t} \over 4} + f(x)} \right) + \cos \left( {{{\pi t} \over 4} - f(x)} \right)} \right\}dx} $$

$$g(t) = 2\int_0^{\pi /2} {\cos {{\pi t} \over 4}.\cos (f(x))dx} $$

$$\therefore$$ $$g(1) = \sqrt 2 \int_0^{\pi /2} {\cos \left( {f(x)} \right)} dx$$ and

$$g(0) = 2\int_0^{\pi /2} {\cos \left( {f(x)} \right)} dx$$

$$\therefore$$ $$\sqrt 2 g(1) = g(0)$$

Solutions · 1