Tough Calculus Question? Try This Shortcut for the π–Limit Integral! ⚡

 

❓ 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 Solution

We split the integral into two parts:

$$I = \int_{0}^{\pi} \frac{x\sin x}{1+3\cos^{2}x}\,dx + 3\int_{0}^{\pi} \frac{\sin x}{1+3\cos^{2}x}\,dx.$$

Let:

$$I_1=\int_0^\pi \frac{x\sin x}{1+3\cos^2x}\, dx,\qquad I_2=\int_0^\pi \frac{\sin x}{1+3\cos^2x}\, dx.$$

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🔹 Step 1 — Simplify $I_1$ using substitution symmetry

Let

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

Substitute:

$$x=\pi - t,\qquad dx=-dt,\qquad \sin(\pi - t)=\sin t,\qquad \cos(\pi - t)=-\cos t.$$

Thus:

$$J=\int_0^\pi \frac{(\pi - t)\sin t}{1+3\cos^2 t}\, dt.$$

Add original and transformed integrals:

$$2J = \int_0^\pi \frac{(\pi)\sin t}{1+3\cos^2 t}\, dt$$

because $x + (\pi-x) = \pi$.

Thus:

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

But this is just:

$$J=\frac{\pi }{2}{I}_2\mathrm{.}$$

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🔹 Step 2 — Solve $I_2$

$$I_2={\int }_0^{\pi }\frac{\sin x}{1+3{\cos }^{2}x}dx\mathrm{.}$$

Let $u = \cos x$, $du = -\sin x\,dx$.

When $x=0, u=1$.
When $x=\pi, u=-1$.

$$I_2={\int }_1^{-1}\frac{-du}{1+3u^2}={\int }_{-1}^1\frac{du}{1+3u^2}\mathrm{.}$$

This is a standard form:

$$\int \frac{du}{1+a^2{u}^2}=\frac{1}{a}{\tan }^{-1}(au)\mathrm{.}$$

Here $a = \sqrt{3}$:

$$I_2 = \frac{1}{\sqrt{3}}\left[\tan^{-1}(\sqrt{3}u)\right]_{-1}^{1} = \frac{1}{\sqrt{3}}\left(\tan^{-1}(\sqrt{3}) - \tan^{-1}(-\sqrt{3})\right).$$
$${\tan }^{-1}(\sqrt{3})=\frac{\pi }{3},{\tan }^{-1}(-\sqrt{3})=-\frac{\pi }{3}\mathrm{.}$$

Thus:

$$I_2=\frac{1}{\sqrt{3}}(\frac{\pi }{3}+\frac{\pi }{3})=\frac{2\pi }{3\sqrt{3}}\mathrm{.}$$

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🔹 Step 3 — Compute total integral

We already found:

$$I_1=\frac{\pi }{2}{I}_2\mathrm{.}$$

So:

$$I=I_1+3I_2=\frac{\pi }{2}{I}_2+3I_2=I_2(3+\frac{\pi }{2})\mathrm{.}$$

Insert value of $I_2 = \frac{2\pi}{3\sqrt{3}}$:

$$I=\frac{2\pi }{3\sqrt{3}}(3+\frac{\pi }{2})\mathrm{.}$$

Simplify:

$$I=\frac{2\pi }{3\sqrt{3}}\cdot \frac{6+\pi }{2}=\frac{\pi (6+\pi )}{3\sqrt{3}}\mathrm{.}$$

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🧮 Final Answer

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

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