int_{-pi / 2}^{pi / 2} sin ^4 x cos ^6 x , dx | vedclass

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

Question

(B) અહીં વિધેય $f(x) = \sin^4 x \cos^6 x$ એ યુગ્મ વિધેય હોવાથી: $\int_{-\pi / 2}^{\pi / 2} \sin ^4 x \cos ^6 x \, dx = 2 \int_0^{\pi / 2} \sin ^4 x \c

Vedclass · Accepted Answer

$\frac{3 \pi}{256}$

Vedclass · Answer

(B) અહીં વિધેય $f(x) = \sin^4 x \cos^6 x$ એ યુગ્મ વિધેય હોવાથી: $\int_{-\pi / 2}^{\pi / 2} \sin ^4 x \cos ^6 x \, dx = 2 \int_0^{\pi / 2} \sin ^4 x \cos ^6 x \, dx$ વોલિસના સૂત્ર $\int_0^{\pi/2} \sin^m x \cos^n x \, dx = \frac{\Gamma(\frac{m+1}{2}) \Gamma(\frac{n+1}{2})}{2 \Gamma(\frac{m+n+2}{2})}$ નો ઉપયોગ કરતા: $I = 2 	imes \frac{\Gamma(\frac{4+1}{2}) \Gamma(\frac{6+1}{2})}{2 \Gamma(\frac{4+6+2}{2})} = \frac{\Gamma(\frac{5}{2}) \Gamma(\frac{7}{2})}{\Gamma(6)}$ $\Gamma(n+1) = n\Gamma(n)$ અને $\Gamma(\frac{1}{2}) = \sqrt{\pi}$ નો ઉપયોગ કરતા: $\Gamma(\frac{5}{2}) = \frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi} = \frac{3}{4} \sqrt{\pi}$ $\Gamma(\frac{7}{2}) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \sqrt{\pi} = \frac{15}{8} \sqrt{\pi}$ $\Gamma(6) = 5! = 120$ $I = \frac{(\frac{3}{4} \sqrt{\pi}) (\frac{15}{8} \sqrt{\pi})}{120} = \frac{45 \pi}{32 	imes 120} = \frac{3 \pi}{256}$

$\int_{-\pi / 2}^{\pi / 2} \sin ^4 x \cos ^6 x \, dx$ ની કિંમત શોધો.

Similar Questions

$\int_0^1 x^{3/2} \sqrt{1-x} \, dx$ ની કિંમત શોધો.

$\int_{-2}^2 x^4(4-x^2)^{\frac{7}{2}} dx=$

વિધેય $f(x) = \int_{x^2}^{x^2+1} e^{-t^2} dt$ એ કયા અંતરાલમાં વધતું વિધેય છે?

જો $\int_{\pi /2}^x \sqrt{3 - 2\sin^2 u} \,du + \int_0^y \cos t \,dt = 0$ હોય,તો $\frac{dy}{dx} = $

આપેલ છે કે $\frac{d}{d x} \int_0^{\phi(x)} f(t) d t=f(\phi(x)) \phi^{\prime}(x)$. બધા $x \in \left(0, \frac{\pi}{2}\right)$ માટે,જો $\int_1^{\cos x} t^2 f(t) d t=\cos 2 x$ હોય,તો $f\left(\frac{1}{\sqrt{2}}\right)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module