int_{-frac{pi}{2}}^{frac{pi}{2}} left(frac{1+ | vedclass

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

Question

(C) माना $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin^{2} x}{1+\pi^{\sin x}} \, dx$. गुणधर्म $\int_{-a}^{a} f(x) \, dx = \int_{0}^{a} [f(x)

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) माना $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin^{2} x}{1+\pi^{\sin x}} \, dx$. गुणधर्म $\int_{-a}^{a} f(x) \, dx = \int_{0}^{a} [f(x) + f(-x)] \, dx$ का उपयोग करने पर: $I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1+\sin^{2} x}{1+\pi^{\sin x}} + \frac{1+\sin^{2}(-x)}{1+\pi^{\sin(-x)}} \right) \, dx$ चूंकि $\sin^{2}(-x) = \sin^{2} x$ और $\sin(-x) = -\sin x$: $I = \int_{0}^{\frac{\pi}{2}} \left( \frac{1+\sin^{2} x}{1+\pi^{\sin x}} + \frac{1+\sin^{2} x}{1+\pi^{-\sin x}} \right) \, dx$ $I = \int_{0}^{\frac{\pi}{2}} (1+\sin^{2} x) \left( \frac{1}{1+\pi^{\sin x}} + \frac{\pi^{\sin x}}{\pi^{\sin x}+1} \right) \, dx$ $I = \int_{0}^{\frac{\pi}{2}} (1+\sin^{2} x) \left( \frac{1+\pi^{\sin x}}{1+\pi^{\sin x}} \right) \, dx$ $I = \int_{0}^{\frac{\pi}{2}} (1+\sin^{2} x) \, dx = \int_{0}^{\frac{\pi}{2}} (1 + \frac{1-\cos 2x}{2}) \, dx = \int_{0}^{\frac{\pi}{2}} (\frac{3}{2} - \frac{1}{2}\cos 2x) \, dx$ $I = [\frac{3}{2}x - \frac{1}{4}\sin 2x]_{0}^{\frac{\pi}{2}} = \frac{3\pi}{4} - 0 = \frac{3\pi}{4}$.

$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left(\frac{1+\sin^{2} x}{1+\pi^{\sin x}}\right) \, dx$ का मान ज्ञात कीजिए।

Similar Questions

$\int_0^{2 \pi} \sin ^6 x \cos ^5 x \, dx$ का मान ज्ञात कीजिए।

$\frac{\int_{0}^{\pi/2} (x \cos x + 1) e^{\sin x} dx}{\int_{0}^{\pi/2} (x \sin x + 1) e^{\cos x} dx}$ का निरपेक्ष मान - के बराबर है।

$\int\limits_0^1 {\sqrt[3]{{2{x^3} - 3{x^2} - x + 1}}\,dx} $ का मान ज्ञात कीजिए।

$\int_{0}^{\frac{\pi}{2}} \frac{\sin^{\frac{2}{3}} x}{\sin^{\frac{2}{3}} x + \cos^{\frac{2}{3}} x} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module