Question

$\int \frac{3+2\cos x}{{(2+3\cos x)}^2}dx$ is equal to

• A. $\left(\frac{\sin x}{2+3\cos x}\right)+c$
• B. $\left(\frac{\sin x}{2+3\sin x}\right)+c$
• C. Both a and b
• D. None of the above

Answer 1

Answer: (A)

$\left(\frac{\sin x}{2+3\cos x}\right)+c$

I$=\int \frac{3+2\cos x}{{(2+3\cos x)}^2}dx$
Multiply numerator and denominator by ${\csc }^{2}x$
I$=\int \frac{{\csc }^{2}x(3+2\cos x)}{{\csc }^{2}x{(2+3\cos x)}^2}$
I$=\int \frac{3{\csc }^{2}x+2\cos x{\csc }^{2}x}{{(2\csc x+3\cos x\csc x)}^2}$
I$=\int \frac{3{\csc }^{2}x+2\csc x\cot x}{{(3\cot x+2\csc x)}^2}dx$
I$=-\int \frac{-3{\csc }^{2}x+2\csc x\cot x}{{(3\cot x+2\csc x)}^2}dx$
I$=-\int {(3\cot x+2\csc x)}^{-2}(-3{\csc }^{2}x-2\csc x\cot x)dx$
$\int {\left[f\left(x\right)\right]}^n{f}^{{}^{\prime }}\left(x\right)dx=\frac{{f\left(x\right)}^{n+1}}{n+1}$
I$=-\frac{{[3\cot x+2\csc x]}^{-2+1}}{-2+1}+C$
I$=\frac{1}{3\cot x+2\csc x}+C$
I$=\frac{\sin x}{3\cos x+2}+C$