int_0^{frac{pi}{2}} frac{sin left(frac{pi}{4} | vedclass

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

Question

(B) ધારો કે $I = \int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{4}+x\right)+\sin \left(\frac{3 \pi}{4}+x\right)}{\cos x+\sin x} d x$. નિત્યસમ $\si

Vedclass · Accepted Answer

$\frac{\pi}{2 \sqrt{2}}$

Vedclass · Answer

(B) ધારો કે $I = \int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{4}+x\right)+\sin \left(\frac{3 \pi}{4}+x\right)}{\cos x+\sin x} d x$. નિત્યસમ $\sin(A+B) = \sin A \cos B + \cos A \sin B$ નો ઉપયોગ કરતા: $\sin \left(\frac{\pi}{4}+x\right) = \frac{1}{\sqrt{2}} \cos x + \frac{1}{\sqrt{2}} \sin x = \frac{1}{\sqrt{2}} (\cos x + \sin x)$. $\sin \left(\frac{3 \pi}{4}+x\right) = \sin \left(\pi - (\frac{\pi}{4} - x)\right) = \sin (\frac{\pi}{4} - x) = \frac{1}{\sqrt{2}} \cos x - \frac{1}{\sqrt{2}} \sin x = \frac{1}{\sqrt{2}} (\cos x - \sin x)$. આ કિંમતોને સંકલનમાં મૂકતા: $I = \int_0^{\frac{\pi}{2}} \frac{\frac{1}{\sqrt{2}} (\cos x + \sin x) + \frac{1}{\sqrt{2}} (\cos x - \sin x)}{\cos x + \sin x} d x$. $I = \int_0^{\frac{\pi}{2}} \frac{\frac{2}{\sqrt{2}} \cos x}{\cos x + \sin x} d x = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x} d x$. ગુણધર્મ $\int_0^a f(x) d x = \int_0^a f(a-x) d x$ નો ઉપયોગ કરતા: $I = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\cos(\frac{\pi}{2}-x)}{\cos(\frac{\pi}{2}-x) + \sin(\frac{\pi}{2}-x)} d x = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} d x$. $I$ માટેના બંને પદોનો સરવાળો કરતા: $2I = \sqrt{2} \int_0^{\frac{\pi}{2}} \frac{\cos x + \sin x}{\cos x + \sin x} d x = \sqrt{2} \int_0^{\frac{\pi}{2}} 1 d x = \sqrt{2} [x]_0^{\frac{\pi}{2}} = \sqrt{2} \cdot \frac{\pi}{2} = \frac{\pi}{\sqrt{2}}$. તેથી,$I = \frac{\pi}{2 \sqrt{2}}$.

$\int_0^{\frac{\pi}{2}} \frac{\sin \left(\frac{\pi}{4}+x\right)+\sin \left(\frac{3 \pi}{4}+x\right)}{\cos x+\sin x} d x=$

Similar Questions

$n \in N$ માટે,જો $I_n = \int \frac{\sin nx}{\sin x} dx = \frac{2}{n-1} \sin(n-1)x + I_{n-2}$ અને $\int_0^\pi \frac{\sin nx}{\sin x} dx = \frac{k\pi}{2}$ હોય,તો $k =$

$\int_2^4 \frac{\log x^2}{\log x^2+\log (36-12x+x^2)} dx$ ની કિંમત શોધો.

$\frac{\int_{0}^{\pi/2} (x \cos x + 1) e^{\sin x} dx}{\int_{0}^{\pi/2} (x \sin x + 1) e^{\cos x} dx}$ નું નિરપેક્ષ મૂલ્ય - ની બરાબર છે.

$\int_0^{2a} f(x) dx - \int_a^{2a} f(x) dx =$

ધારો કે $u = \int\limits_0^1 {\frac{{\ln (x + 1)}}{{{x^2} + 1}}} \,dx$ અને $v = \int\limits_0^{\frac{\pi }{2}} {\ln (\sin 2x)} \,dx$,તો:

Vedclass Test Series

Exam Paper Generator

Online Exam Module