Question

If the primitive of $f\left(x\right)=\frac{1}{3\sin x+{\sin }^{3}x}$ is equal to $\frac{1}{6}\log \left|\frac{t-1}{t+1}\right|+\frac{1}{12}\log \left|\frac{2+t}{2-t}\right|+C$ then

• A. $t=\cos x$
• B. $t=\tan x/2$
• C. $t=2\cos x$
• D. $t=\sin x$

Answer 1

The correct option is A $t=\cos x$

Let $a=3,x=3t$

Put in the equation

$\frac{1}{2\left(3\right)}\log \left|\frac{\frac{x}{3}-1}{\frac{x}{3}+1}\right|⟹\frac{1}{2\left(3\right)}\log \left|\frac{x-3}{x+3}\right|$ whose differential is $\frac{1}{x^2-9}$

$\frac{1}{2\left(6\right)}\log \left|\frac{2+\frac{x}{3}}{2-\frac{x}{3}}\right|⟹\frac{1}{2\left(6\right)}\log \left|\frac{6+x}{6-x}\right|$ whose differential is $\frac{1}{36-x^2}$

So the equation is $\frac{1}{x^2-9}+\frac{1}{36-x^2}$

Put $x=3t$

$⟹\frac{1}{-3t^4+15t^2-12}$