निश्चित समाकलन int_{-frac{pi}{4}}^{frac{pi}{4 | vedclass

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(C) माना $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left(1+e^{x \cos x}\right)\left(\sin ^{4} x+\cos ^{4} x\right)} \cdots(1)$गुणधर्म $\in

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

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(C) माना $I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left(1+e^{x \cos x}ight)\left(\sin ^{4} x+\cos ^{4} xight)} \cdots(1)$गुणधर्म $\int_{a}^{b} f(x) d x = \int_{a}^{b} f(a+b-x) d x$ का उपयोग करते हुए,जहाँ $a = -\frac{\pi}{4}$ और $b = \frac{\pi}{4}$,हमें $a+b = 0$ प्राप्त होता है। अतः,$f(x)$ का मान $f(-x)$ हो जाता है:$I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left(1+e^{-x \cos(-x)}ight)\left(\sin ^{4}(-x)+\cos ^{4}(-x)ight)} = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left(1+e^{-x \cos x}ight)\left(\sin ^{4} x+\cos ^{4} xight)} \cdots(2)$$(1)$ और $(2)$ को जोड़ने पर:$2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \frac{1}{1+e^{x \cos x}} + \frac{1}{1+e^{-x \cos x}} ight) \frac{d x}{\sin ^{4} x+\cos ^{4} x}$चूँकि $\frac{1}{1+e^{x \cos x}} + \frac{1}{1+e^{-x \cos x}} = 1$,इसलिए:$2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\sin ^{4} x+\cos ^{4} x} = 2 \int_{0}^{\frac{\pi}{4}} \frac{d x}{\sin ^{4} x+\cos ^{4} x}$$I = \int_{0}^{\frac{\pi}{4}} \frac{d x}{\sin ^{4} x+\cos ^{4} x} = \int_{0}^{\frac{\pi}{4}} \frac{(1+	an^2 x) \sec^2 x}{	an^4 x + 1} dx$माना $	an x = u$,तो $\sec^2 x dx = du$। जब $x 	o 0, u 	o 0$ और $x 	o \frac{\pi}{4}, u 	o 1$:$I = \int_{0}^{1} \frac{1+u^2}{u^4+1} du = \int_{0}^{1} \frac{1+\frac{1}{u^2}}{u^2+\frac{1}{u^2}} du = \int_{0}^{1} \frac{1+\frac{1}{u^2}}{(u-\frac{1}{u})^2 + 2} du$माना $u-\frac{1}{u} = t$,तो $(1+\frac{1}{u^2}) du = dt$। जब $u 	o 0, t 	o -\infty$ और $u 	o 1, t 	o 0$:$I = \int_{-\infty}^{0} \frac{dt}{t^2+2} = \left[ \frac{1}{\sqrt{2}} 	an^{-1} \left( \frac{t}{\sqrt{2}} ight) ight]_{-\infty}^{0} = \frac{1}{\sqrt{2}} (0 - (-\frac{\pi}{2})) = \frac{\pi}{2\sqrt{2}}$

निश्चित समाकलन $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{\left(1+e^{x \cos x}\right)\left(\sin ^{4} x+\cos ^{4} x\right)}$ का मान किसके बराबर है?

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