If I=int frac{sin x+sin ^3 x}{cos 2 x} ,d x=P | vedclass

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(B) Given $I = \int \frac{\sin x + \sin^3 x}{\cos 2x} dx$. Using $\sin^2 x = 1 - \cos^2 x$,we have $\sin x(1 + \sin^2 x) = \sin x(1 + 1 - \cos^2 x) =

Vedclass · Accepted Answer

$\frac{1}{2}, \frac{-3}{4 \sqrt{2}}$

Vedclass · Answer

(B) Given $I = \int \frac{\sin x + \sin^3 x}{\cos 2x} dx$. Using $\sin^2 x = 1 - \cos^2 x$,we have $\sin x(1 + \sin^2 x) = \sin x(1 + 1 - \cos^2 x) = \sin x(2 - \cos^2 x)$. So,$I = \int \frac{\sin x(2 - \cos^2 x)}{2 \cos^2 x - 1} dx$. Let $\cos x = t$,then $-\sin x dx = dt$,so $\sin x dx = -dt$. $I = \int \frac{t^2 - 2}{2t^2 - 1} (-dt) = \int \frac{2 - t^2}{2t^2 - 1} dt$. $I = \frac{1}{2} \int \frac{4 - 2t^2}{2t^2 - 1} dt = \frac{1}{2} \int \frac{-(2t^2 - 1) + 3}{2t^2 - 1} dt$. $I = \frac{1}{2} \int (-1 + \frac{3}{2t^2 - 1}) dt = \frac{1}{2} [-t + \frac{3}{2\sqrt{2}} \log |\frac{\sqrt{2}t - 1}{\sqrt{2}t + 1}|] + c$. $I = -\frac{1}{2} \cos x + \frac{3}{4\sqrt{2}} \log |\frac{\sqrt{2} \cos x - 1}{\sqrt{2} \cos x + 1}| + c$. Comparing with $P \cos x + Q \log |\frac{\sqrt{2} \cos x - 1}{\sqrt{2} \cos x + 1}| + c$,we get $P = -\frac{1}{2}$ and $Q = \frac{3}{4\sqrt{2}}$. Note: The provided options seem to have a sign discrepancy for $P$. Based on the standard integral form,$P = -1/2$ and $Q = 3/(4\sqrt{2})$. If $P$ is $1/2$,the numerator in the integral must be $\sin x(\cos^2 x - 2)$.

If $I=\int \frac{\sin x+\sin ^3 x}{\cos 2 x} \,d x=P \cos x+Q \log \left|\frac{\sqrt{2} \cos x-1}{\sqrt{2} \cos x+1}\right|+c,$ (where $c$ is a constant of integration),then the values of $P$ and $Q$ are respectively

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