$\int \cos ^{\frac{-3}{7}} x \cdot \sin ^{\frac{-11}{7}} x d x=$

$\int \cos ^{\frac{-3}{7}} x \cdot \sin ^{\frac{-11}{7}} x d x=$
  1. $\frac{-4}{7} \tan ^{\frac{-4}{7}} x+\mathrm{c}$, where c is a constant of integration.
  2. $\frac{4}{7} \tan ^{\frac{4}{7}} x+\mathrm{c}$, where c is a constant of integration.
  3. $\frac{-7}{4} \tan ^{\frac{-4}{7}} x+\mathrm{c}$, where c is a constant of integration.
  4. $\frac{7}{4} \tan ^{\frac{4}{7}} x+\mathrm{c}$, where c is a constant of integration.

Solution

$\begin{aligned} & \int \cos ^{-\frac{3}{7}} x \cdot \sin ^{\frac{-11}{7}} x \cdot d x \\ & =\int \frac{\sin ^{-\frac{11}{7}} x}{\cos ^{\frac{3}{7}} x} \cdot \frac{1}{\cos ^{-2} x \cdot \cos ^2 x} \mathrm{~d} x \\ & =\int \frac{\sin ^{-\frac{11}{7}} x}{\cos ^{-\frac{11}{7}} x} \cdot \sec ^2 x \cdot \mathrm{~d} x \\ & =\int \tan ^{-\frac{11}{7}} x \cdot \sec ^2 x \mathrm{~d} x \\ & =\frac{\tan ^{-\frac{4}{7}} x}{-\frac{4}{7}}+\mathrm{c}\end{aligned}$ $\ldots\left[\because \int[\mathrm{f}(x)]^{\mathrm{n}} \cdot \mathrm{f}^{\prime}(x) \mathrm{d} x=\frac{[\mathrm{f}(x)]^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}\right]$ $=-\frac{7}{4} \tan ^{-\frac{4}{7}} x+c$

Asked in: MHT CET 2024 (16 May Shift 2)

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