Evaluate Definite Integral Using tan inverse Trick

 

❓ Question

Evaluate the integral:

$${\int }_0^{\pi }\frac{(x+3)\sin x}{1+3{\cos }^{2}x}dx$$

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🖼️ Question Image

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✍️ Short Explanation

This problem is based on:

👉 Definite integrals
👉 Property of symmetry
👉 Standard trigonometric integration.

Main idea:

Use:

$$\boxed{ \int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx }$$

to simplify the $x$-term.

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🔷 Step 1 — Let the Integral be $I$ 💯

$$I= \int_0^\pi \frac{(x+3)\sin x}{1+3\cos^2 x}\,dx$$

Using property:

$$x\to\pi-x$$

we get:

$$I= \int_0^\pi \frac{(\pi-x+3)\sin x}{1+3\cos^2 x}\,dx$$

because:

$$\sin(\pi-x)=\sin x$$

and

$$\cos^2(\pi-x)=\cos^2 x$$

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🔷 Step 2 — Add Both Integrals

Adding:

$$2I= \int_0^\pi \frac{[(x+3)+(\pi-x+3)]\sin x}{1+3\cos^2 x}\,dx$$
$$= (\pi+6) \int_0^\pi \frac{\sin x}{1+3\cos^2 x}\,dx$$

Thus:

$$I= \frac{\pi+6}{2} \int_0^\pi \frac{\sin x}{1+3\cos^2 x}\,dx$$

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🔷 Step 3 — Evaluate Standard Integral

Let:

$$t=\cos x$$

Then:

$$dt=-\sin x\,dx$$

Limits:

$$x=0\Rightarrow t=1$$
$$x=\pi\Rightarrow t=-1$$

Hence:

$$\int_0^\pi \frac{\sin x}{1+3\cos^2 x}\,dx = \int_{-1}^{1} \frac{dt}{1+3t^2}$$

Now:

$$= 2\int_0^1\frac{dt}{1+3t^2}$$

Using:

$$\int\frac{dx}{1+a^2x^2} = \frac1a\tan^{-1}(ax)$$

we get:

$$= 2\cdot\frac1{\sqrt3} \left[ \tan^{-1}(\sqrt3 t) \right]_0^1$$
$$= \frac2{\sqrt3} \left( \frac\pi3 \right)$$
$$= \frac{2\pi}{3\sqrt3}$$

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🔷 Step 4 — Final Calculation

$$I= \frac{\pi+6}{2} \cdot \frac{2\pi}{3\sqrt3}$$
$$= \boxed{ \frac{\pi(\pi+6)}{3\sqrt3} }$$

or

$$\boxed{ \frac{\pi}{3\sqrt3}(\pi+6) }$$

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🔷 Step 5 — JEE Trap Alert 🚨

❌ Direct integration start kar dena

❌ Symmetry property miss kar dena

Remember:

When numerator contains:

$$x \text{ and limits } 0\to\pi$$

always try:

$$x\to\pi-x$$

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✅ Final Answer

$$\boxed{ \frac{\pi}{3\sqrt3}(\pi+6) }$$

(Option 4)

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