Definite Integral Using Antiderivative Values

 

❓ Question

If

$$\int \frac{(1-5\cos^2 x)}{\sin^3 x \cos^2 x}\,dx = f(x)+c$$

then

$$\left|f\left(\frac{\pi}{3}\right)-f\left(\frac{\pi}{6}\right)\right|$$

is equal to ______.

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🖼 Question Image

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✍️ Short Explanation

This problem is based on:

👉 Indefinite integration
👉 Trigonometric identities
👉 Fundamental relation of antiderivative.

Main idea:

$$f(b)-f(a)=\int_a^b f'(x)\,dx$$

So directly evaluate definite integral.

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🔷 Step 1 — Convert into Simpler Form 💯

Given:

$$f'(x)=\frac{1-5\cos^2 x}{\sin^3 x\cos^2 x}$$

Split numerator:

$$= \frac{1}{\sin^3 x\cos^2 x} - \frac{5\cos^2 x}{\sin^3 x\cos^2 x}$$
$$= \frac{1}{\sin^3 x\cos^2 x} - \frac{5}{\sin^3 x}$$

Using:

$$1=\sin^2 x+\cos^2 x$$

Rewrite:

$$1-5\cos^2 x = \sin^2 x-4\cos^2 x$$

Thus:

$$f'(x) = \frac{\sin^2 x}{\sin^3 x\cos^2 x} - \frac{4\cos^2 x}{\sin^3 x\cos^2 x}$$
$$= \frac1{\sin x\cos^2 x} - \frac4{\sin^3 x}$$
$$= \sec^2 x\,\csc x - 4\csc^3 x$$

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🔷 Step 2 — Observe Derivative Pattern

Now use standard derivatives:

$$\frac{d}{dx}(\sec x)=\sec x\tan x$$
$$\frac{d}{dx}(\cot x)=-\csc^2 x$$

Try rewriting integrand as derivative of:

$$\frac{1}{\sin^2 x\cos x}$$

Differentiate:

$$\frac{d}{dx}(\sec x\csc^2 x) = \frac{1-5\cos^2 x}{\sin^3 x\cos^2 x}$$

Hence:

$$f(x)=\sec x\,\csc^2 x$$

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🔷 Step 3 — Evaluate at Limits

At:

$$x=\frac{\pi}{3}$$
$$f\left(\frac{\pi}{3}\right) = \sec\frac{\pi}{3}\, \csc^2\frac{\pi}{3}$$
$$= 2\times\left(\frac{2}{\sqrt3}\right)^2$$
$$= 2\times\frac43$$
$$= \frac83$$

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At:

$$x=\frac{\pi}{6}$$
$$f\left(\frac{\pi}{6}\right) = \sec\frac{\pi}{6}\, \csc^2\frac{\pi}{6}$$
$$= \frac2{\sqrt3}\times(2)^2$$
$$= \frac8{\sqrt3}$$

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🔷 Step 4 — Find Required Value

$$\left| f\left(\frac{\pi}{3}\right) - f\left(\frac{\pi}{6}\right) \right|$$
$$= \left| \frac83-\frac8{\sqrt3} \right|$$ $$= 8\left| \frac13-\frac1{\sqrt3} \right|$$
$$= \frac{8(\sqrt3-3)}{3\sqrt3}$$

Taking modulus:

$$= \frac{8(3-\sqrt3)}{3\sqrt3}$$

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🔷 Step 5 — Rationalise

$$= \frac{8(3-\sqrt3)\sqrt3}{9}$$
$$= \frac{24\sqrt3-24}{9}$$
$$= \frac{8(\sqrt3-1)}3$$

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🔷 Step 6 — JEE Trap Alert 🚨

❌ Indefinite integral me constant confuse kar lena

❌ Trigonometric simplification galat kar dena

❌ Modulus apply karna bhool jaana

Remember:

$$\boxed{ f(b)-f(a)=\int_a^b f'(x)\,dx }$$

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✅ Final Answer

$$\boxed{ \frac{8(\sqrt3-1)}3 }$$

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