JEE Mains · Maths · STD 12 - 7.1 indefinite integral
If \(\int(\frac{1-5~cos^{2}x}{sin^{5}x~cos^{2}x})dx=f(x)+C\) where C is the constant of integration, then \(f(\frac{\pi}{6})-f(\frac{\pi}{4})\) is equal to
- A \(\frac{1}{\sqrt{3}}(26+\sqrt{3})\)
- B \(\frac{4}{\sqrt{3}}(8-\sqrt{6})\)
- C \(\frac{1}{\sqrt{3}}(26-\sqrt{3})\)
- D \(\frac{2}{\sqrt{3}}(4+\sqrt{6})\)
Answer & Solution
Correct Answer
(B) \(\frac{4}{\sqrt{3}}(8-\sqrt{6})\)
Step-by-step Solution
Detailed explanation
\(\int\frac{dx}{sin^{5}x~cos^{2}x}-5\int\frac{dx}{sin^{5}x}\) \(\int\frac{sec^{2}xdx}{sin^{5}x}-5\int\frac{dx}{sin^{5}x}\) By IBP \(=\frac{tan~x}{sin^{5}x}-\int-\frac{5}{sin^{6}x} \cdot cos~x \cdot tan~xdx-5\int\frac{dx}{sin^{5}x}\) \(=\frac{tan~x}{sin^{5}x}+c\)…
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