int_0^{pi /4} {frac{{sec x}}{{1 + 2{{sin }^2} | vedclass

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Question

(A) ધારો કે $I = \int_0^{\pi /4} {\frac{{\sec x}}{{1 + 2{{\sin }^2}x}}} dx = \int_0^{\pi /4} {\frac{{\cos x}}{{{{\cos }^2}x(1 + 2{{\sin }^2}x)}}} dx$$

Vedclass · Accepted Answer

$\frac{1}{3}\left[ {\log (\sqrt 2 + 1) + \frac{\pi }{{2\sqrt 2 }}} \right]$

Vedclass · Answer

(A) ધારો કે $I = \int_0^{\pi /4} {\frac{{\sec x}}{{1 + 2{{\sin }^2}x}}} dx = \int_0^{\pi /4} {\frac{{\cos x}}{{{{\cos }^2}x(1 + 2{{\sin }^2}x)}}} dx$$= \int_0^{\pi /4} {\frac{{\cos x}}{{(1 - {{\sin }^2}x)(1 + 2{{\sin }^2}x)}}} dx$ધારો કે $t = \sin x$,તેથી $dt = \cos x dx$. જ્યારે $x=0, t=0$ અને જ્યારે $x=\pi/4, t=1/\sqrt{2}$.$I = \int_0^{1/\sqrt{2}} \frac{dt}{(1-t^2)(1+2t^2)}$.આંશિક અપૂર્ણાંકનો ઉપયોગ કરતા: $\frac{1}{(1-t^2)(1+2t^2)} = \frac{1}{3} \left( \frac{1}{1-t^2} + \frac{2}{1+2t^2} ight)$.$I = \frac{1}{3} \left[ \int_0^{1/\sqrt{2}} \frac{dt}{1-t^2} + 2 \int_0^{1/\sqrt{2}} \frac{dt}{1+2t^2} ight]$$I = \frac{1}{3} \left[ \frac{1}{2} \log \left| \frac{1+t}{1-t} ight| + 2 \cdot \frac{1}{\sqrt{2}} 	an^{-1}(t\sqrt{2}) ight]_0^{1/\sqrt{2}}$$I = \frac{1}{3} \left[ \frac{1}{2} \log \left( \frac{1+1/\sqrt{2}}{1-1/\sqrt{2}} ight) + \sqrt{2} 	an^{-1}(1) ight]$$I = \frac{1}{3} \left[ \frac{1}{2} \log \left( \frac{\sqrt{2}+1}{\sqrt{2}-1} ight) + \sqrt{2} \cdot \frac{\pi}{4} ight]$કારણ કે $\frac{\sqrt{2}+1}{\sqrt{2}-1} = (\sqrt{2}+1)^2$,તેથી $\frac{1}{2} \log (\sqrt{2}+1)^2 = \log(\sqrt{2}+1)$.$I = \frac{1}{3} \left[ \log(\sqrt{2}+1) + \frac{\pi}{2\sqrt{2}} ight]$.

$\int_0^{\pi /4} {\frac{{\sec x}}{{1 + 2{{\sin }^2}x}}} dx$ ની કિંમત શોધો.

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