NIMCET 2008 — Mathematics PYQ

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Question

NIMCET 2008 — Mathematics PYQ

NIMCET | Mathematics | 2008

The value of $\int_{0}^{\pi/2} \frac{dx}{1 + \tan^3 x}$ is:

Choose the correct answer:

Explanation

Step 1: Simplify the Integral

Let the given integral be $I$ :

I = \int_{0}^{\pi/2} \frac{1}{1 + \tan^3 x} dx

We can write $\tan x$ as $\frac{\sin x}{\cos x}$ :

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

I = \int_{0}^{\pi/2} \frac{\cos^3 x}{\cos^3 x + \sin^3 x} dx \quad \dots \text{(Equation 1)}

Step 2: Apply the Property $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$

Using this property, we replace $x$ with $(\frac{\pi}{2} - x)$ :

I = \int_{0}^{\pi/2} \frac{\cos^3(\frac{\pi}{2} - x)}{\cos^3(\frac{\pi}{2} - x) + \sin^3(\frac{\pi}{2} - x)} dx

Since $\cos(\frac{\pi}{2} - x) = \sin x$ and $\sin(\frac{\pi}{2} - x) = \cos x$ :

I = \int_{0}^{\pi/2} \frac{\sin^3 x}{\sin^3 x + \cos^3 x} dx \quad \dots \text{(Equation 2)}

Step 3: Add Equation 1 and Equation 2

I + I = \int_{0}^{\pi/2} \frac{\cos^3 x}{\cos^3 x + \sin^3 x} dx + \int_{0}^{\pi/2} \frac{\sin^3 x}{\sin^3 x + \cos^3 x} dx

2I = \int_{0}^{\pi/2} \frac{\cos^3 x + \sin^3 x}{\cos^3 x + \sin^3 x} dx

2I = \int_{0}^{\pi/2} 1 \cdot dx

Step 4: Integrate and Solve for $I$

2I = [x]_{0}^{\pi/2}

2I = \frac{\pi}{2} - 0

I = \frac{\pi}{4}

Final Answer:

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

The correct option is (c).

Choose the correct answer:

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