NIMCET 2012 — Mathematics PYQ

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Question

NIMCET 2012 — Mathematics PYQ

NIMCET | Mathematics | 2012

The value of $\int_{0}^{\sin^2 x} \sin^{-1} \sqrt{t} \, dt + \int_{0}^{\cos^2 x} \cos^{-1} \sqrt{t} \, dt$ is:

Choose the correct answer:

Explanation

1. Let the first integral be $I_1$ :

I_1 = \int_{0}^{\sin^2 x} \sin^{-1} \sqrt{t} \, dt

Let $t = \sin^2 \theta \implies dt = 2 \sin \theta \cos \theta \, d\theta = \sin 2\theta \, d\theta$ .

When $t=0, \theta=0$ . When $t=\sin^2 x, \theta=x$ .

I_1 = \int_{0}^{x} \theta \cdot \sin 2\theta \, d\theta

2. Let the second integral be $I_2$ :

I_2 = \int_{0}^{\cos^2 x} \cos^{-1} \sqrt{t} \, dt

Let $t = \cos^2 \theta \implies dt = -2 \cos \theta \sin \theta \, d\theta = -\sin 2\theta \, d\theta$ .

When $t=0, \theta=\pi/2$ . When $t=\cos^2 x, \theta=x$ .

I_2 = \int_{\pi/2}^{x} \theta \cdot (-\sin 2\theta) \, d\theta = \int_{x}^{\pi/2} \theta \sin 2\theta \, d\theta

3. Combine $I_1$ and $I_2$ :

I_1 + I_2 = \int_{0}^{x} \theta \sin 2\theta \, d\theta + \int_{x}^{\pi/2} \theta \sin 2\theta \, d\theta

Using the property $\int_{a}^{b} f + \int_{b}^{c} f = \int_{a}^{c} f$ :

I = \int_{0}^{\pi/2} \theta \sin 2\theta \, d\theta

4. Integration by Parts:

\int u v \, d\theta = u \int v \, d\theta - \int (u' \int v \, d\theta) \, d\theta

Let $u = \theta, v = \sin 2\theta$ :

I = \left[ \theta \left( \frac{-\cos 2\theta}{2} \right) \right]_{0}^{\pi/2} - \int_{0}^{\pi/2} (1) \left( \frac{-\cos 2\theta}{2} \right) d\theta

I = \left[ -\frac{\pi/2 \cos(\pi)}{2} - 0 \right] + \left[ \frac{\sin 2\theta}{4} \right]_{0}^{\pi/2}

I = \left[ -\frac{\pi/2 (-1)}{2} \right] + [0 - 0]

I = \frac{\pi}{4}

Correct Option:

(a) $\frac{\pi}{4}$

Choose the correct answer:

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