Question

$\int_{}^{}\frac{\mathbf{3}\mathbf{\cos}\mathbf{x}\mathbf{+ 3}\mathbf{\sin}\mathbf{x}}{\mathbf{4}\mathbf{\sin}\mathbf{x}\mathbf{+ 5}\mathbf{\cos}\mathbf{x}}\mathbf{dx =}$

• A. $\frac{27}{41}x - \frac{3}{41}\log(4\sin x + 5\cos x) + c$
• B. $\frac{27}{41}x + \frac{3}{41}\log(4\sin x + 5\cos x) + c$
• C. $\frac{27}{41}x - \frac{3}{41}\log(4\sin x - 5\cos x) + c$
• D. None of these

Answer 1

Answer: (A)

$\frac{27}{41}x - \frac{3}{41}\log(4\sin x + 5\cos x) + c$

Answer 2

$\mathbf{3}\mathbf{\cos}\mathbf{x}\mathbf{+ 3}\mathbf{\sin}\mathbf{x}\mathbf{= M}\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{(4}\mathbf{\sin}\mathbf{x}\mathbf{+ 5}\mathbf{\cos}\mathbf{x}\mathbf{) + N(4}\mathbf{\sin}\mathbf{x}\mathbf{+ 5}\mathbf{\cos}\mathbf{x}\mathbf{) \Rightarrow 3}\mathbf{\cos}\mathbf{x}\mathbf{+ 3}\mathbf{\sin}\mathbf{x}\mathbf{= M(4}\mathbf{\cos}\mathbf{x}\mathbf{- 5}\mathbf{\sin}\mathbf{x}\mathbf{) + N(4}\mathbf{\sin}\mathbf{x}\mathbf{+ 5}\mathbf{\cos}\mathbf{x}\mathbf{)} \Rightarrow$ Comparing the coefficient of $\sin x$ and $\cos x$on both sides.

$\Rightarrow$ $- 5M + 4N = 3$ and $4M + 5N = 3$ $\Rightarrow$ $M = \frac{- 3}{41},N = \frac{27}{41}$

$\therefore$ $I = \int_{}^{}\frac{\frac{- 3}{41}(4\cos x - 5\sin x) + \frac{27}{41}(4\sin x + 5\cos x)}{4\sin x + 5\cos x}dx$

$\Rightarrow I = \int_{}^{}{\frac{27}{41}dx + \left( \frac{- 3}{41} \right)}\int_{}^{}\frac{4\cos x - 5\sin x}{4\sin x + 5\cos x}dx$

$\Rightarrow I = \frac{27}{41}x - \frac{3}{41}\log(4\sin x + 5\cos x) + c$.