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

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Question

(B) Let $I = \int_0^{\frac{\pi}{4}} \frac{\sin^2 x}{1+\sin x \cos x} dx$. Multiply numerator and denominator by $2$ to get $I = \int_0^{\frac{\pi}{4}}

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Let $I = \int_0^{\frac{\pi}{4}} \frac{\sin^2 x}{1+\sin x \cos x} dx$. Multiply numerator and denominator by $2$ to get $I = \int_0^{\frac{\pi}{4}} \frac{2\sin^2 x}{2+2\sin x \cos x} dx = \int_0^{\frac{\pi}{4}} \frac{1-\cos 2x}{2+\sin 2x} dx$.This splits into $I = \int_0^{\frac{\pi}{4}} \frac{1}{2+\sin 2x} dx - \int_0^{\frac{\pi}{4}} \frac{\cos 2x}{2+\sin 2x} dx = I_1 - I_2$.For $I_1 = \int_0^{\frac{\pi}{4}} \frac{\sec^2 x}{2	an^2 x + 2	an x + 2} dx$. Let $t = 	an x$,$dt = \sec^2 x dx$. Limits change from $0$ to $1$.$I_1 = \frac{1}{2} \int_0^1 \frac{dt}{t^2+t+1} = \frac{1}{2} \int_0^1 \frac{dt}{(t+1/2)^2 + 3/4} = \frac{1}{2} \cdot \frac{2}{\sqrt{3}} [	an^{-1}(\frac{2t+1}{\sqrt{3}})]_0^1 = \frac{1}{\sqrt{3}} (\frac{\pi}{3} - \frac{\pi}{6}) = \frac{\pi}{6\sqrt{3}}$.For $I_2 = \int_0^{\frac{\pi}{4}} \frac{\cos 2x}{2+\sin 2x} dx$. Let $u = 2+\sin 2x$,$du = 2\cos 2x dx$.$I_2 = \frac{1}{2} \int_2^3 \frac{du}{u} = \frac{1}{2} \ln(\frac{3}{2}) = \frac{1}{2} \ln(3) - \frac{1}{2} \ln(2)$.Comparing $I = I_1 - I_2 = \frac{\pi}{6\sqrt{3}} - \frac{1}{2} \ln(3) + \frac{1}{2} \ln(2) = \frac{\pi}{6\sqrt{3}} + \frac{1}{2} \ln(2/3)$.Given form is $\frac{1}{a} \ln(a/3) + \frac{\pi}{b\sqrt{3}}$. With $a=2, b=6$,we get $\frac{1}{2} \ln(2/3) + \frac{\pi}{6\sqrt{3}}$.Thus $a=2, b=6$,so $a+b = 8$.

If $\int_0^{\frac{\pi}{4}} \frac{\sin^2 x}{1+\sin x \cos x} dx = \frac{1}{a} \log_e\left(\frac{a}{3}\right) + \frac{\pi}{b \sqrt{3}}$,where $a, b \in N$,then $a+b$ is equal to ....................

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