int_0^{pi / 4} frac{1}{5 cos ^2 x+16 sin ^2 x | vedclass

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(C) ધારો કે $I = \int_0^{\pi / 4} \frac{1}{5 \cos ^2 x+16 \sin ^2 x+8 \sin x \cos x} d x$. અંશ અને છેદને $\cos^2 x$ વડે ભાગતા: $I = \int_0^{\pi / 4} \

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$\frac{1}{8} \operatorname{Tan}^{-1}\left(\frac{8}{9}\right)$

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(C) ધારો કે $I = \int_0^{\pi / 4} \frac{1}{5 \cos ^2 x+16 \sin ^2 x+8 \sin x \cos x} d x$. અંશ અને છેદને $\cos^2 x$ વડે ભાગતા: $I = \int_0^{\pi / 4} \frac{\sec^2 x}{5 + 16 	an^2 x + 8 	an x} d x$. ધારો કે $u = 	an x$, તેથી $du = \sec^2 x dx$. જ્યારે $x = 0, u = 0$. જ્યારે $x = \pi/4, u = 1$. $I = \int_0^1 \frac{du}{16u^2 + 8u + 5} = \int_0^1 \frac{du}{(4u+1)^2 + 4} = \frac{1}{4} \int_0^1 \frac{d(4u+1)}{(4u+1)^2 + 2^2}$. સૂત્ર $\int \frac{dx}{x^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$ નો ઉપયોગ કરતા: $I = \frac{1}{4} \left[ \frac{1}{2} 	an^{-1} \left( \frac{4u+1}{2} ight) ight]_0^1 = \frac{1}{8} [	an^{-1}(\frac{5}{2}) - 	an^{-1}(\frac{1}{2})]$. $	an^{-1} A - 	an^{-1} B = 	an^{-1} \left( \frac{A-B}{1+AB} ight)$ નો ઉપયોગ કરતા: $I = \frac{1}{8} 	an^{-1} \left( \frac{5/2 - 1/2}{1 + (5/2)(1/2)} ight) = \frac{1}{8} 	an^{-1} \left( \frac{2}{1 + 5/4} ight) = \frac{1}{8} 	an^{-1} \left( \frac{2}{9/4} ight) = \frac{1}{8} 	an^{-1} \left( \frac{8}{9} ight)$.

$\int_0^{\pi / 4} \frac{1}{5 \cos ^2 x+16 \sin ^2 x+8 \sin x \cos x} d x=$

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