Q113 / 160 · AP EAMCET 2024 (19 May Shift 2) · Mathematics

$\int\left(\frac{x}{x \cos x-\sin x}\right)^2 d x=$

  • A
    $\frac{x \operatorname{cosec} x}{x \cos x-\sin x}+\cot x+c$
  • Correct answer
    B
    $\frac{x \operatorname{cosec} x}{x \cos x-\sin x}-\cot x+c$
  • C
    $\frac{x \operatorname{cosec} x}{x \cos x+\sin x}+\cot x+c$
  • D
    $\frac{x}{x \cos x-\sin x}-\cot x+c$

Solution

Let $I=\int\left(\frac{x}{x \cos x-\sin x}\right)^2 d x$ $\begin{aligned} & =\int \frac{x^2}{(x \cos x-\sin x)^2} d x \\ & =-\int \frac{x}{\sin x} \times \frac{(-x \sin x)}{(x \cos x-\sin x)^2} d x \\ & =\frac{-x}{\sin x} \times \frac{(-1)}{(x \cos x-\sin x)}-\int \frac{\sin x-x \cos x}{\sin ^2 x(x \cos x-\sin x)} d x \\ & =\frac{x}{\sin x(x \cos x-\sin x)}+\int \operatorname{cosec}^2 x d x \\ & =\frac{x \operatorname{cosec} x}{x \cos x-\sin x}-\cot x+c .\end{aligned}$

Hence, the correct answer is option: 'B'.

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