Question

Solve

$\int \frac{\sin 2x}{{\left(a+b\cos x\right)}^2}dx=?where\left(\lambda =a+b\cos x\right)$

• A. $-\frac{2}{b^2}\left[\log \left|\lambda \right|+\frac{a}{\lambda }\right]+C$
• B. $-\frac{2}{b^2}\left[\log \left|\lambda \right|-\frac{a}{\lambda }\right]+C$
• C. $-\frac{4}{b^2}\left[\log \left|\lambda \right|+\frac{a}{\lambda }\right]+C$
• D. $-\frac{4}{b^2}\left[\log \left|\lambda \right|-\frac{a}{\lambda }\right]+C$

Answer 1

The correct option is A $-\frac{2}{b^2}\left[\log \left|\lambda \right|+\frac{a}{\lambda }\right]+C$

$\int \frac{\sin 2x}{{(a+b\cos x)}^2}dx=\int \frac{2\sin x\cos x}{{(a+b\cos x)}^2}dxa+b\cos x=t⟹\cos x=\frac{t-a}{b}⟹-b\sin xdx=dt=-\frac{2}{b}\int \left(\frac{t-a}{b}\right)\frac{1}{t^2}dt=-\frac{2}{b^2}\int \left(\frac{t-a}{t^2}\right)dt=-\frac{2}{b^2}\left[\int \frac{1}{t}dt-\int \frac{a}{t^2}dt\right]=-\frac{2}{b^2}[\ln t-\frac{a{t}^{-2+1}}{-2+1}]+C=-\frac{2}{b^2}[\ln t+\frac{a}{t}]+C=-\frac{2}{b^2}[\ln (a+b\cos x)+\frac{a}{a+b\cos x}]+C=-\frac{2}{b^2}[\ln \left|\lambda \right|+\frac{a}{\lambda }]+C$