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

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(A) ધારો કે $I = \int_0^{\pi / 4} \frac{\sec x}{1+2 \sin ^2 x} d x$. અંશ અને છેદને $\cos x$ વડે ગુણતા: $I = \int_0^{\pi / 4} \frac{\cos x}{\cos ^2 x(1

Vedclass · Accepted Answer

$\frac{1}{3} \log (\sqrt{2}+1)+\frac{\pi \sqrt{2}}{12}$

Vedclass · Answer

(A) ધારો કે $I = \int_0^{\pi / 4} \frac{\sec x}{1+2 \sin ^2 x} d x$. અંશ અને છેદને $\cos x$ વડે ગુણતા: $I = \int_0^{\pi / 4} \frac{\cos x}{\cos ^2 x(1+2 \sin ^2 x)} d x = \int_0^{\pi / 4} \frac{\cos x}{(1-\sin ^2 x)(1+2 \sin ^2 x)} d x$. ધારો કે $\sin x = t$, તેથી $\cos x d x = d t$. જ્યારે $x = 0, t = 0$. જ્યારે $x = \frac{\pi}{4}, t = \frac{1}{\sqrt{2}}$. $I = \int_0^{1 / \sqrt{2}} \frac{d t}{(1-t^2)(1+2 t^2)}$. આંશિક અપૂર્ણાંકનો ઉપયોગ કરતા: $\frac{1}{(1-t^2)(1+2 t^2)} = \frac{1}{3} \left( \frac{1}{1-t^2} + \frac{2}{1+2 t^2} ight)$. $I = \frac{1}{3} \int_0^{1 / \sqrt{2}} \left( \frac{1}{1-t^2} + \frac{2}{1+2 t^2} ight) d t$. $I = \frac{1}{3} \left[ \frac{1}{2} \log \left| \frac{1+t}{1-t} ight| + \frac{2}{\sqrt{2}} 	an ^{-1} (\sqrt{2} t) ight]_0^{1 / \sqrt{2}}$. $I = \frac{1}{3} \left[ \frac{1}{2} \log \left( \frac{1+1/\sqrt{2}}{1-1/\sqrt{2}} ight) + \sqrt{2} 	an ^{-1} (1) ight]$. $I = \frac{1}{3} \left[ \frac{1}{2} \log \left( \frac{\sqrt{2}+1}{\sqrt{2}-1} ight) + \sqrt{2} \cdot \frac{\pi}{4} ight]$. કારણ કે $\frac{\sqrt{2}+1}{\sqrt{2}-1} = (\sqrt{2}+1)^2$, તેથી $\log (\sqrt{2}+1)^2 = 2 \log (\sqrt{2}+1)$. $I = \frac{1}{3} \left[ \log (\sqrt{2}+1) + \frac{\pi \sqrt{2}}{4} ight] = \frac{1}{3} \log (\sqrt{2}+1) + \frac{\pi \sqrt{2}}{12}$.

$\int_0^{\frac{\pi}{4}} \frac{\sec x}{1+2 \sin ^2 x} d x=$

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