If a 0, then int_{-pi}^{pi} frac{sin^2 x}{1 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int_{-\pi}^{\pi} \frac{\sin^2 x}{1+a^x} dx$  $(i)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$, we replace $x$ w

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(A) Let $I = \int_{-\pi}^{\pi} \frac{\sin^2 x}{1+a^x} dx$  $(i)$Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$, we replace $x$ with $-\pi + \pi - x = -x$:$I = \int_{-\pi}^{\pi} \frac{\sin^2(-x)}{1+a^{-x}} dx = \int_{-\pi}^{\pi} \frac{\sin^2 x}{1+\frac{1}{a^x}} dx = \int_{-\pi}^{\pi} \frac{a^x \sin^2 x}{a^x+1} dx$  $(ii)$Adding $(i)$ and $(ii)$:$2I = \int_{-\pi}^{\pi} \frac{\sin^2 x + a^x \sin^2 x}{1+a^x} dx = \int_{-\pi}^{\pi} \frac{(1+a^x) \sin^2 x}{1+a^x} dx = \int_{-\pi}^{\pi} \sin^2 x dx$Since $\sin^2 x$ is an even function:$2I = 2 \int_{0}^{\pi} \sin^2 x dx = \int_{0}^{\pi} (1 - \cos 2x) dx$$2I = [x - \frac{\sin 2x}{2}]_{0}^{\pi} = (\pi - 0) - (0 - 0) = \pi$$I = \frac{\pi}{2}$

If $a > 0$, then $\int_{-\pi}^{\pi} \frac{\sin^2 x}{1+a^x} dx$ is equal to

Similar Questions

For $f(x) = x^4 + |x|$,let $I_1 = \int_{0}^{\pi} f(\cos x) dx$ and $I_2 = \int_{0}^{\frac{\pi}{2}} f(\sin x) dx$. Then $\frac{I_1}{I_2}$ is equal to

If $f(x)$ is an odd function of $x,$ then $\int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {f(\cos x)\,dx} $ is equal to

Let $[t]$ denote the greatest integer less than or equal to $t$. Then the value of $\int_1^2 |2x - [3x]| dx$ is

If $I = \frac{2}{\pi} \int_{-\pi / 4}^{\pi / 4} \frac{dx}{(1 + e^{\sin x})(2 - \cos 2x)}$,then $27 I^2$ equals . . . . . . . .

If $I_n = \int_0^{\pi / 2} \sin^n(x) dx$ and $I_n = (k) I_{n-2}$,then what will be the value of $k$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module