Question

If $I=\int \frac{\sin x+{\sin }^{3}x}{\cos 2x}dx=P\cos x+Q\log \left|\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}\right|+R,$then

• A. $P=\frac{1}{2}$
• B. $P=\frac{1}{4}$
• C. $Q=-\frac{3}{2\sqrt{2}}$
• D. $Q=-\frac{3}{4\sqrt{2}}$

Answer 1

Answer: (A), (D)

$Q=-\frac{3}{4\sqrt{2}}$

Let $\cos x=t\Rightarrow \cos 2x=2t^2-1\text{and}dt=-\sin xdx$. Thus
$I=\int \frac{t^2-2}{2t^2-1}dt=\frac{1}{2}\int \frac{2t^2-4}{2t^2-1}dt$
$=\frac{1}{2}\int dt-\frac{3}{2}\int \frac{dt}{2t^2-1}$
$=\frac{1}{2}t-\frac{3}{2\sqrt{2}}\times \frac{1}{2}\log \left|\frac{\sqrt{2}t-1}{\sqrt{2}t+1}\right|+C$
$=\frac{1}{2}\cos x-\frac{3}{4\sqrt{2}}\log \left|\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}\right|+C$
So. $P=\frac{1}{2},$ $Q=-\frac{3}{4\sqrt{2}}$