Integration July 2017 Board Questions Solution

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Integration: July 2017 Board Questions & Solutions

Question 1: Evaluate: $I = \int \frac{1}{x \log x} dx$

Solution:

Let $I = \int \frac{1}{x \log x} dx$

Put $\log x = t$
Differentiating both sides with respect to $x$:
$\frac{1}{x} dx = dt$

Substituting these values in the integral:
$I = \int \frac{1}{t} dt$

$I = \log |t| + c$

Resubstituting $t = \log x$:
$I = \log |\log x| + c$

Question 2: Evaluate: $I = \int e^x \left(\frac{x-1}{x^2}\right) dx$

Solution:

Let $I = \int e^x \left(\frac{x}{x^2} - \frac{1}{x^2}\right) dx$

$I = \int e^x \left(\frac{1}{x} - \frac{1}{x^2}\right) dx$

This integral is of the standard form:
$$\int e^x [f(x) + f'(x)] dx = e^x f(x) + c$$

Here, let $f(x) = \frac{1}{x}$
Then, its derivative is $f'(x) = -\frac{1}{x^2}$

Therefore, applying the standard formula:
$I = e^x \left(\frac{1}{x}\right) + c$

$I = \frac{e^x}{x} + c$

Question 3: Evaluate the definite integral: $I = \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} dx$

Solution:

Let $I = \int_0^{\frac{\pi}{2}} \frac{\sin x}{\sin x + \cos x} dx$     --- (Equation 1)

Using the definite integral property: $\int_0^a f(x) dx = \int_0^a f(a-x) dx$

$$I = \int_0^{\frac{\pi}{2}} \frac{\sin(\frac{\pi}{2} - x)}{\sin(\frac{\pi}{2} - x) + \cos(\frac{\pi}{2} - x)} dx$$

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

$I = \int_0^{\frac{\pi}{2}} \frac{\cos x}{\cos x + \sin x} dx$     --- (Equation 2)

Adding Equation 1 and Equation 2:

$$2I = \int_0^{\frac{\pi}{2}} \left( \frac{\sin x}{\sin x + \cos x} + \frac{\cos x}{\cos x + \sin x} \right) dx$$

$$2I = \int_0^{\frac{\pi}{2}} \frac{\sin x + \cos x}{\sin x + \cos x} dx$$

$$2I = \int_0^{\frac{\pi}{2}} 1 \cdot dx$$

$$2I = [x]_0^{\frac{\pi}{2}}$$

$2I = \left(\frac{\pi}{2} - 0\right)$

$I = \frac{\pi}{4}$

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