NIMCET 2009 — Mathematics PYQ

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Question

NIMCET 2009 — Mathematics PYQ

NIMCET | Mathematics | 2009

The value of $\int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} dx$ is:

Choose the correct answer:

Explanation

To solve this definite integral, we use the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$ (often called the King's property).

1. Let the integral be $I$ :

I = \int_{0}^{\pi} \frac{x \sin x}{1 + \cos^2 x} dx \quad \dots (1)

2. Applying the property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$ :

I = \int_{0}^{\pi} \frac{(\pi - x) \sin(\pi - x)}{1 + \cos^2(\pi - x)} dx

Since $\sin(\pi - x) = \sin x$ and $\cos(\pi - x) = -\cos x$ (so $\cos^2(\pi - x) = \cos^2 x$ ):

I = \int_{0}^{\pi} \frac{(\pi - x) \sin x}{1 + \cos^2 x} dx \quad \dots (2)

3. Adding equations (1) and (2):

I + I = \int_{0}^{\pi} \frac{x \sin x + (\pi - x) \sin x}{1 + \cos^2 x} dx

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

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

4. Solving the integral using substitution:

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

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

I = \frac{\pi}{2} \int_{1}^{-1} \frac{-du}{1 + u^2}

I = \frac{\pi}{2} \int_{-1}^{1} \frac{du}{1 + u^2}

5. Integrating:

I = \frac{\pi}{2} [\tan^{-1} u]_{-1}^{1}

I = \frac{\pi}{2} [\tan^{-1}(1) - \tan^{-1}(-1)]

I = \frac{\pi}{2} \left[ \frac{\pi}{4} - \left( -\frac{\pi}{4} \right) \right]

I = \frac{\pi}{2} \left[ \frac{\pi}{4} + \frac{\pi}{4} \right]

I = \frac{\pi}{2} \cdot \frac{\pi}{2} = \frac{\pi^2}{4}

Conclusion

The final value of the integral is $\frac{\pi^2}{4}$ .

Correct Option:

(b) $\frac{\pi^2}{4}$

Choose the correct answer:

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