Question

Solve: $\int \frac{2\sin 2x-\cos x}{6-{\cos }^{2}x-4\sin x}d\theta$.

• A. $2\log ({\sin }^{2}x-\sin x+5)+7{\tan }^{-1}(x-2)+C$
• B. $2\log ({\sin }^{2}x-\sin 4x+5)-7{\tan }^{-1}(x-2)+C$
• C. $\log ({\sin }^{2}x-\sin 4x+5)-7{\tan }^{-1}(x-2)+C$
• D. $2\log ({\sin }^{2}x+\sin x+5)+7{\tan }^{-1}(x-2)+C$

Answer 1

The correct option is B $2\log ({\sin }^{2}x-\sin 4x+5)-7{\tan }^{-1}(x-2)+C$

$\int \frac{2\sin 2\theta -\cos \theta }{6-{\cos }^2\theta -4\sin \theta }d\theta$
$\int \frac{2\sin 2\theta -\cos \theta }{-4\sin \theta -{\cos }^2\theta +6}d\theta$

$=-\int \frac{2\sin 2\theta -\cos \theta }{4\sin \theta +{\cos }^2\theta -6}d\theta$

Substituting

$\sin 2\theta =2\sin \theta \cos \theta {\cos }^2\theta =1-{\sin }^2\theta$

$=\int \cos \theta (-\frac{4\sin \theta -1}{{\sin }^2\theta -4\sin \theta +5})d\theta$

Substituting $u=\sin \theta \to d\theta =\frac{1}{\cos \theta }du$

$=[-\int \frac{4u-1}{u^2-4u+5}du]$

$4u-1$ as $2(2u-u)+7$

$=\int (\frac{2(2u-4)}{u^2-4u+5}+\frac{7}{u^2-4u+5})du$

$=4\int \frac{u-2}{u^2-4u+5}du+7\int \frac{1}{u^2-4u+5}du$

$v=u^2-4u+5\to du=\frac{1}{2u-4}dv$

$=4\int \frac{1}{2v}dv$

$=\frac{1}{2}\int \frac{1}{v}dv$

$=\frac{lnv}{2}$

$=ln\frac{(u^2-4u+5)}{2}$

$=\int \frac{1}{(u-2{)}^2+1}du$

$v=u-2\to du=dv$

$=\int \frac{1}{v^2+1}dv$

$={\tan }^{-1}\left(v\right)$

$v=u-2$

$={\tan }^{-1}(u-2)$

$=4ln\frac{(u^2-4u+5)}{2}+7{\tan }^{-1}(u-2)$

$=-2ln(u^2-4u+5)-7{\tan }^{-1}(u-2)$

$=2ln({\sin }^2\theta -4\sin \theta +5)+7{\tan }^{-1}(\sin \theta -2)$

$\therefore \int \frac{2\sin 2\theta -\cos \theta }{6-{\cos }^2\theta -4\sin \theta }d\theta =3ln({\sin }^2\theta -4\sin \theta +5)+7{\tan }^{-1}(\sin \theta -2)$