Evaluate the integral: int frac{1}{(sin x + c | vedclass

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(B) Given integral $I = \int \frac{1}{(\sin x + \cos x + \sqrt{2} \sqrt{\sin 2x})^2} dx$. We can rewrite the denominator as $(\sqrt{\sin x} + \sqrt{\c

Vedclass · Accepted Answer

$\frac{-(1+3 \sqrt{	an x})}{3(1+\sqrt{	an x})^3}+C$

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(B) Given integral $I = \int \frac{1}{(\sin x + \cos x + \sqrt{2} \sqrt{\sin 2x})^2} dx$. We can rewrite the denominator as $(\sqrt{\sin x} + \sqrt{\cos x})^4$. Thus,$I = \int \frac{dx}{(\sqrt{\sin x} + \sqrt{\cos x})^4} = \int \frac{dx}{\cos^2 x (\sqrt{	an x} + 1)^4} = \int \frac{\sec^2 x}{(\sqrt{	an x} + 1)^4} dx$. Let $	an x = t^2$,then $\sec^2 x dx = 2t dt$. Substituting these into the integral: $I = \int \frac{2t dt}{(t+1)^4} = 2 \int \frac{t+1-1}{(t+1)^4} dt = 2 \int \left( \frac{1}{(t+1)^3} - \frac{1}{(t+1)^4} ight) dt$. Integrating term by term: $I = 2 \left[ \frac{(t+1)^{-2}}{-2} - \frac{(t+1)^{-3}}{-3} ight] + C = 2 \left[ \frac{1}{3(t+1)^3} - \frac{1}{2(t+1)^2} ight] + C$. Simplifying: $I = \frac{2}{3(t+1)^3} - \frac{1}{(t+1)^2} + C = \frac{2 - 3(t+1)}{3(t+1)^3} + C = \frac{2 - 3t - 3}{3(t+1)^3} + C = \frac{-(1+3t)}{3(1+t)^3} + C$. Substituting $t = \sqrt{	an x}$,we get $I = \frac{-(1+3\sqrt{	an x})}{3(1+\sqrt{	an x})^3} + C$.

Evaluate the integral: $\int \frac{1}{(\sin x + \cos x + \sqrt{2} \sqrt{\sin 2x})^2} dx$

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