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

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(B) ધારો કે $I = \int_0^{\frac{\pi}{4}} \frac{\sin^2 x}{1+\sin x \cos x} dx$. અંશ અને છેદને $2$ વડે ગુણતા $I = \int_0^{\frac{\pi}{4}} \frac{2\sin^2 x}

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$8$

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(B) ધારો કે $I = \int_0^{\frac{\pi}{4}} \frac{\sin^2 x}{1+\sin x \cos x} dx$. અંશ અને છેદને $2$ વડે ગુણતા $I = \int_0^{\frac{\pi}{4}} \frac{2\sin^2 x}{2+2\sin x \cos x} dx = \int_0^{\frac{\pi}{4}} \frac{1-\cos 2x}{2+\sin 2x} dx$.આને $I = \int_0^{\frac{\pi}{4}} \frac{1}{2+\sin 2x} dx - \int_0^{\frac{\pi}{4}} \frac{\cos 2x}{2+\sin 2x} dx = I_1 - I_2$ તરીકે લખી શકાય.$I_1 = \int_0^{\frac{\pi}{4}} \frac{\sec^2 x}{2	an^2 x + 2	an x + 2} dx$. ધારો કે $t = 	an x$,$dt = \sec^2 x dx$. સીમાઓ $0$ થી $1$ બદલાય છે.$I_1 = \frac{1}{2} \int_0^1 \frac{dt}{t^2+t+1} = \frac{1}{2} \int_0^1 \frac{dt}{(t+1/2)^2 + 3/4} = \frac{1}{2} \cdot \frac{2}{\sqrt{3}} [	an^{-1}(\frac{2t+1}{\sqrt{3}})]_0^1 = \frac{1}{\sqrt{3}} (\frac{\pi}{3} - \frac{\pi}{6}) = \frac{\pi}{6\sqrt{3}}$.$I_2 = \int_0^{\frac{\pi}{4}} \frac{\cos 2x}{2+\sin 2x} dx$. ધારો કે $u = 2+\sin 2x$,$du = 2\cos 2x dx$.$I_2 = \frac{1}{2} \int_2^3 \frac{du}{u} = \frac{1}{2} \ln(\frac{3}{2}) = \frac{1}{2} \ln(3) - \frac{1}{2} \ln(2)$.સરખામણી કરતા $I = I_1 - I_2 = \frac{\pi}{6\sqrt{3}} - \frac{1}{2} \ln(3) + \frac{1}{2} \ln(2) = \frac{\pi}{6\sqrt{3}} + \frac{1}{2} \ln(2/3)$.આપેલ સ્વરૂપ $\frac{1}{a} \ln(a/3) + \frac{\pi}{b\sqrt{3}}$ છે. $a=2, b=6$ લેતા,આપણને $\frac{1}{2} \ln(2/3) + \frac{\pi}{6\sqrt{3}}$ મળે છે.આમ $a=2, b=6$,તેથી $a+b = 8$.

જો $\int_0^{\frac{\pi}{4}} \frac{\sin^2 x}{1+\sin x \cos x} dx = \frac{1}{a} \log_e\left(\frac{a}{3}\right) + \frac{\pi}{b \sqrt{3}}$,જ્યાં $a, b \in N$,તો $a+b$ ની કિંમત .................... છે.

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