MHT CET · Maths · Indefinite Integration
\(\int \cos ^{\frac{-3}{7}} x \cdot \sin ^{\frac{-11}{7}} x d x=\)
- A \(\frac{-4}{7} \tan ^{\frac{-4}{7}} x+\mathrm{c}\), where c is a constant of integration.
- B \(\frac{4}{7} \tan ^{\frac{4}{7}} x+\mathrm{c}\), where c is a constant of integration.
- C \(\frac{-7}{4} \tan ^{\frac{-4}{7}} x+\mathrm{c}\), where c is a constant of integration.
- D \(\frac{7}{4} \tan ^{\frac{4}{7}} x+\mathrm{c}\), where c is a constant of integration.
Answer & Solution
Correct Answer
(C) \(\frac{-7}{4} \tan ^{\frac{-4}{7}} x+\mathrm{c}\), where c is a constant of integration.
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \int \cos ^{-\frac{3}{7}} x \cdot \sin ^{\frac{-11}{7}} x \cdot d x \\ & =\int \frac{\sin ^{-\frac{11}{7}} x}{\cos ^{\frac{3}{7}} x} \cdot \frac{1}{\cos ^{-2} x \cdot \cos ^2 x} \mathrm{~d} x \\ & =\int \frac{\sin ^{-\frac{11}{7}} x}{\cos ^{-\frac{11}{7}} x} \cdot \sec ^2 x \cdot \mathrm{~d} x \\ & =\int \tan ^{-\frac{11}{7}} x \cdot \sec ^2 x \mathrm{~d} x \\ & =\frac{\tan ^{-\frac{4}{7}} x}{-\frac{4}{7}}+\mathrm{c}\end{aligned}\)
\(\ldots\left[\because \int[\mathrm{f}(x)]^{\mathrm{n}} \cdot \mathrm{f}^{\prime}(x) \mathrm{d} x=\frac{[\mathrm{f}(x)]^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}\right]\)
\(=-\frac{7}{4} \tan ^{-\frac{4}{7}} x+c\)
\(\ldots\left[\because \int[\mathrm{f}(x)]^{\mathrm{n}} \cdot \mathrm{f}^{\prime}(x) \mathrm{d} x=\frac{[\mathrm{f}(x)]^{\mathrm{n}+1}}{\mathrm{n}+1}+\mathrm{c}\right]\)
\(=-\frac{7}{4} \tan ^{-\frac{4}{7}} x+c\)
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