If ≤ft({1-5² x}{^{5} x ² x}) dx = f(x)+C, where C is the constant of integration, then f≤ft(π/6)-f≤ft(π/4) is equal to [28 Jan 2026 Shift 1]

Correct Answer is : 43(8−6){4}{{3}}(8-{6})34(8−6)

Correct Answer is : 43(8−6){4}{{3}}(8-{6})34(8−6)

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Indefinite Integrals

Section: Mathematics

Question : 3 of 90

Marks: +1, -0

\int\left(\frac{1-5\cos^2 x}{\sin^{5} x \cos^{2} x}\right) dx = f(x)+C

f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)

Solution:

Concept:

The integrand is the derivative of

\frac{\tan x}{\sin^5 x}

, so the antiderivative is

f(x)=\frac{\tan x}{\sin^5 x}

Explanation:

Rewrite the integrand:

\frac{1-5\cos^2 x}{\sin^5 x \cos^2 x} = \frac{\sec^2 x}{\sin^5 x} - 5\csc^5 x

Notice that

\frac{d}{dx}\left(\frac{\tan x}{\sin^5 x}\right) = \frac{\sec^2 x - 5}{\sin^5 x} = \frac{1-5\cos^2 x}{\sin^5 x \cos^2 x}

Thus

f(x)=\frac{\tan x}{\sin^5 x} = \frac{1}{\sin^4 x \cos x}

Calculate

f\left(\frac{\pi}{6}\right)

\sin\frac{\pi}{6}=\frac12

\cos\frac{\pi}{6}=\frac{\sqrt3}{2}

, so

f\left(\frac{\pi}{6}\right)=\frac{1}{(1/2)^4\cdot (\sqrt3/2)} = \frac{1}{(1/16)(\sqrt3/2)} = \frac{32}{\sqrt3}

Calculate

f\left(\frac{\pi}{4}\right)

\sin\frac{\pi}{4}=\frac1{\sqrt2}

\cos\frac{\pi}{4}=\frac1{\sqrt2}

, so

f\left(\frac{\pi}{4}\right)=\frac{1}{(1/\sqrt2)^4\cdot (1/\sqrt2)} = \frac{1}{(1/4)(1/\sqrt2)} = 4\sqrt2

Therefore

f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right) = \frac{32}{\sqrt3} - 4\sqrt2

Factor

\frac{4}{\sqrt3}

\frac{4}{\sqrt3}\left(8 - \sqrt2\sqrt3\right) = \frac{4}{\sqrt3}(8 - \sqrt6)

Answer:

Option B:

\frac{4}{\sqrt3}(8-\sqrt6)

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