NIMCET 2010 — Mathematics PYQ

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Question

NIMCET 2010 — Mathematics PYQ

NIMCET | Mathematics | 2010

The value of the integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin x + \cos x}{3 + \sin 2x} dx$ is:

Choose the correct answer:

Explanation

Step 1: Use Substitution

We know that the derivative of $(\sin x - \cos x)$ is $(\cos x + \sin x)$ .

Let $t = \sin x - \cos x$

Then, $dt = (\cos x + \sin x) dx$

Step 2: Change the Limits

When $x = 0$ : $t = \sin 0 - \cos 0 = 0 - 1 = -1$
When $x = \frac{\pi}{4}$ : $t = \sin \frac{\pi}{4} - \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$

Step 3: Express the denominator in terms of $t$

Squaring $t = \sin x - \cos x$ :

t^2 = (\sin x - \cos x)^2

t^2 = \sin^2 x + \cos^2 x - 2\sin x \cos x

t^2 = 1 - \sin 2x

\sin 2x = 1 - t^2

Substituting this into the denominator:

3 + \sin 2x = 3 + (1 - t^2) = 4 - t^2

Step 4: Substitute everything into the integral

I = \int_{-1}^{0} \frac{dt}{4 - t^2}

Step 5: Integrate using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \log \left| \frac{a+x}{a-x} \right|$

Here $a = 2$ :

I = \left[ \frac{1}{2(2)} \log \left| \frac{2 + t}{2 - t} \right| \right]_{-1}^{0}

I = \frac{1}{4} \left[ \log \left| \frac{2 + 0}{2 - 0} \right| - \log \left| \frac{2 + (-1)}{2 - (-1)} \right| \right]

I = \frac{1}{4} \left[ \log(1) - \log \left( \frac{1}{3} \right) \right]

I = \frac{1}{4} [0 - (\log 1 - \log 3)]

I = \frac{1}{4} \log 3

Conclusion:

The value of the integral is $\frac{1}{4} \log 3$ .

Correct Option:

(c) $\frac{1}{4} \log 3$