The value of int_{-frac{pi}{2}}^{frac{pi}{2}} | vedclass

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

Question

(C) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin^{2} x}{1+\pi^{\sin x}} \, dx$. Using the property $\int_{-a}^{a} f(x) \, dx = \int_{0}

Vedclass · Accepted Answer

$\frac{3 \pi}{4}$

Vedclass · Answer

(C) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin^{2} x}{1+\pi^{\sin x}} \, dx$. Using the property $\int_{-a}^{a} f(x) \, dx = \int_{0}^{a} [f(x) + f(-x)] \, dx$,we have: $I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1+\sin^{2} x}{1+\pi^{\sin x}} + \frac{1+\sin^{2}(-x)}{1+\pi^{\sin(-x)}} \right) \, dx$ Since $\sin^{2}(-x) = \sin^{2} x$ and $\sin(-x) = -\sin x$: $I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1+\sin^{2} x}{1+\pi^{\sin x}} + \frac{1+\sin^{2} x}{1+\pi^{-\sin x}} \right) \, dx$ $I = \int_{0}^{\frac{\pi}{2}} (1+\sin^{2} x) \left( \frac{1}{1+\pi^{\sin x}} + \frac{\pi^{\sin x}}{\pi^{\sin x}+1} \right) \, dx$ $I = \int_{0}^{\frac{\pi}{2}} (1+\sin^{2} x) \left( \frac{1+\pi^{\sin x}}{1+\pi^{\sin x}} \right) \, dx$ $I = \int_{0}^{\frac{\pi}{2}} (1+\sin^{2} x) \, dx = \int_{0}^{\frac{\pi}{2}} (1 + \frac{1-\cos 2x}{2}) \, dx = \int_{0}^{\frac{\pi}{2}} (\frac{3}{2} - \frac{1}{2}\cos 2x) \, dx$ $I = [\frac{3}{2}x - \frac{1}{4}\sin 2x]_{0}^{\frac{\pi}{2}} = \frac{3\pi}{4} - 0 = \frac{3\pi}{4}$.

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\frac{1+\sin^{2} x}{1+\pi^{\sin x}}\right) \, dx$ is

Similar Questions

If $f(x) = \int_0^{\sin^2 x} \sin^{-1} \sqrt{t} \, dt$ and $g(x) = \int_0^{\cos^2 x} \cos^{-1} \sqrt{t} \, dt$, then the value of $f(x) + g(x)$ is

By using the properties of definite integrals,evaluate the integral $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x \, dx$.

$\int_{-1}^{1} \frac{x^{3} + |x| + 3}{x^{2} + 4|x| + 3} dx$ is equal to -

The value of $\int_0^{n\pi + v} {|\sin x|\,dx} $ is

$\int_{-1}^{1} x^{17} \cos^{4} x \, dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module