Question

If $I=\int \frac{\sin x+{\sin }^{3}x}{\cos 2x}dx=P\cos x+Q\log \left|f\right(x\left)\right|+R,$ then
( $Q<0$)

• A. $P=1/2,Q=-\frac{3}{4\sqrt{2}}$
• B. $P=1/4,Q=-\frac{1}{\sqrt{2}}$
• C. $f\left(x\right)=\frac{\sqrt{2}\cos x+1}{\sqrt{2}\cos x-1}$
• D. $f\left(x\right)=\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}$

Answer 1

Answer: (A), (D)

$P=1/2,Q=-\frac{3}{4\sqrt{2}}$
$f\left(x\right)=\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}$

Given $I=\int \frac{\sin x+{\sin }^{3}x}{\cos 2x}dx=P\cos x+Q\log \left|f\right(x\left)\right|+R,$

$I=\int \frac{\sin x+{\sin }^{3}x}{\cos 2x}dx$

$=\int \frac{\sin x(1+{\sin }^{2}x)}{2{\cos }^{2}x-1}dx$

$=\int \frac{\sin x(2-{\cos }^{2}x)}{2{\cos }^{2}x-1}dx$

Let $\cos x=t\Rightarrow dt=-\sin xdx$.

$I=\int \frac{-(2-t^2)}{2t^2-1}dt=\int \frac{t^2-2}{2t^2-1}dt=\frac{1}{2}\int \frac{2t^2-4}{2t^2-1}dt$

$=\frac{1}{2}\int dt-\frac{3}{2}\int \frac{dt}{2t^2-1}$

$=\frac{1}{2}\cos x-\frac{3}{4\sqrt{2}}\log \left|\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}\right|+C$ ( applying integration formulas)

by comparing with the given we get

$P=\frac{1}{2},Q=-\frac{3}{4\sqrt{2}},f\left(x\right)=\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}$