Question

Let $I_1 = \int_{\pi/4}^{\pi/3} \frac{1}{1-\cos x} dx$ and $I_2 = \int_{\pi/4}^{\pi/3} \frac{1}{1-\sin x} dx$. Then the value of $I_1 + I_2$ is

Answer 1

2

Answer 2

To find the value of $I_1 + I_2$, we will evaluate each integral separately using trigonometric identities.

Step 1: Evaluate $I_1 = \int_{\pi/4}^{\pi/3} \frac{1}{1-\cos x} dx$

We use the identity $1 - \cos x = 2\sin^2(x/2)$. So, $I_1 = \int_{\pi/4}^{\pi/3} \frac{1}{2\sin^2(x/2)} dx = \frac{1}{2} \int_{\pi/4}^{\pi/3} \csc^2(x/2) dx$.

Let $u = x/2$. Then $du = \frac{1}{2} dx$. When $x = \pi/4$, $u = \pi/8$. When $x = \pi/3$, $u = \pi/6$.

Substituting these into the integral: $I_1 = \int_{\pi/8}^{\pi/6} \csc^2 u \, du$. The integral of $\csc^2 u$ is $-\cot u$. $I_1 = [-\cot u]_{\pi/8}^{\pi/6} = -\cot(\pi/6) - (-\cot(\pi/8)) = -\sqrt{3} + \cot(\pi/8)$.

Alternatively, we can multiply the numerator and denominator by $1+\cos x$: $I_1 = \int_{\pi/4}^{\pi/3} \frac{1+\cos x}{(1-\cos x)(1+\cos x)} dx = \int_{\pi/4}^{\pi/3} \frac{1+\cos x}{1-\cos^2 x} dx = \int_{\pi/4}^{\pi/3} \frac{1+\cos x}{\sin^2 x} dx$ $I_1 = \int_{\pi/4}^{\pi/3} \left(\frac{1}{\sin^2 x} + \frac{\cos x}{\sin^2 x}\right) dx = \int_{\pi/4}^{\pi/3} (\csc^2 x + \cot x \csc x) dx$. The integral of $\csc^2 x$ is $-\cot x$, and the integral of $\cot x \csc x$ is $-\csc x$. $I_1 = [-\cot x - \csc x]_{\pi/4}^{\pi/3}$ $I_1 = (-\cot(\pi/3) - \csc(\pi/3)) - (-\cot(\pi/4) - \csc(\pi/4))$ $I_1 = \left(-\frac{1}{\sqrt{3}} - \frac{2}{\sqrt{3}}\right) - (-1 - \sqrt{2})$ $I_1 = -\frac{3}{\sqrt{3}} + 1 + \sqrt{2} = -\sqrt{3} + 1 + \sqrt{2}$.

Step 2: Evaluate $I_2 = \int_{\pi/4}^{\pi/3} \frac{1}{1-\sin x} dx$

We can use the identity $\sin x = \cos(\pi/2 - x)$. $I_2 = \int_{\pi/4}^{\pi/3} \frac{1}{1-\cos(\pi/2 - x)} dx$. Let $y = \pi/2 - x$. Then $dy = -dx$. When $x = \pi/4$, $y = \pi/2 - \pi/4 = \pi/4$. When $x = \pi/3$, $y = \pi/2 - \pi/3 = \pi/6$.

Substituting these into the integral: $I_2 = \int_{\pi/4}^{\pi/6} \frac{1}{1-\cos y} (-dy) = \int_{\pi/6}^{\pi/4} \frac{1}{1-\cos y} dy$. This integral is of the same form as $I_1$. Using the half-angle formula $1-\cos y = 2\sin^2(y/2)$: $I_2 = \frac{1}{2} \int_{\pi/6}^{\pi/4} \csc^2(y/2) dy$. Let $v = y/2$. Then $dv = \frac{1}{2} dy$. When $y = \pi/6$, $v = \pi/12$. When $y = \pi/4$, $v = \pi/8$.

$I_2 = \int_{\pi/12}^{\pi/8} \csc^2 v \, dv = [-\cot v]_{\pi/12}^{\pi/8} = -\cot(\pi/8) - (-\cot(\pi/12)) = -\cot(\pi/8) + \cot(\pi/12)$.

Alternatively, multiply numerator and denominator by $1+\sin x$: $I_2 = \int_{\pi/4}^{\pi/3} \frac{1+\sin x}{(1-\sin x)(1+\sin x)} dx = \int_{\pi/4}^{\pi/3} \frac{1+\sin x}{1-\sin^2 x} dx = \int_{\pi/4}^{\pi/3} \frac{1+\sin x}{\cos^2 x} dx$ $I_2 = \int_{\pi/4}^{\pi/3} \left(\frac{1}{\cos^2 x} + \frac{\sin x}{\cos^2 x}\right) dx = \int_{\pi/4}^{\pi/3} (\sec^2 x + \tan x \sec x) dx$. The integral of $\sec^2 x$ is $\tan x$, and the integral of $\tan x \sec x$ is $\sec x$. $I_2 = [\tan x + \sec x]_{\pi/4}^{\pi/3}$ $I_2 = (\tan(\pi/3) + \sec(\pi/3)) - (\tan(\pi/4) + \sec(\pi/4))$ $I_2 = (\sqrt{3} + 2) - (1 + \sqrt{2})$ $I_2 = \sqrt{3} + 1 - \sqrt{2}$.

Step 3: Calculate $I_1 + I_2$

Using the direct integration results from the alternative method: $I_1 + I_2 = (-\sqrt{3} + 1 + \sqrt{2}) + (\sqrt{3} + 1 - \sqrt{2})$ $I_1 + I_2 = -\sqrt{3} + 1 + \sqrt{2} + \sqrt{3} + 1 - \sqrt{2}$ $I_1 + I_2 = (1+1) + (-\sqrt{3}+\sqrt{3}) + (\sqrt{2}-\sqrt{2})$ $I_1 + I_2 = 2$.

Using the results from the first method involving $\cot(\pi/8)$ and $\cot(\pi/12)$: $I_1 + I_2 = (-\sqrt{3} + \cot(\pi/8)) + (-\cot(\pi/8) + \cot(\pi/12))$ $I_1 + I_2 = -\sqrt{3} + \cot(\pi/12)$. To find $\cot(\pi/12)$, we use $\pi/12 = 15^\circ$. $\cot(15^\circ) = \cot(45^\circ - 30^\circ) = \frac{\cot 45^\circ \cot 30^\circ + 1}{\cot 30^\circ - \cot 45^\circ} = \frac{1 \cdot \sqrt{3} + 1}{\sqrt{3} - 1}$. Rationalizing the expression: $\cot(15^\circ) = \frac{(\sqrt{3}+1)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{(\sqrt{3}+1)^2}{3-1} = \frac{3+1+2\sqrt{3}}{2} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$. Substitute this value back into $I_1 + I_2$: $I_1 + I_2 = -\sqrt{3} + (2+\sqrt{3}) = 2$.

Both methods yield the same result.

The final answer is $\boxed{2}$.

Explanation of the solution: The integrals $I_1$ and $I_2$ are evaluated separately. For $I_1 = \int \frac{1}{1-\cos x} dx$, multiply numerator and denominator by $1+\cos x$ to get $\int \frac{1+\cos x}{\sin^2 x} dx = \int (\csc^2 x + \cot x \csc x) dx$. This integrates to $[-\cot x - \csc x]$. For $I_2 = \int \frac{1}{1-\sin x} dx$, multiply numerator and denominator by $1+\sin x$ to get $\int \frac{1+\sin x}{\cos^2 x} dx = \int (\sec^2 x + \tan x \sec x) dx$. This integrates to $[\tan x + \sec x]$. Evaluate these definite integrals from $\pi/4$ to $\pi/3$. $I_1 = (-\cot(\pi/3) - \csc(\pi/3)) - (-\cot(\pi/4) - \csc(\pi/4)) = (-\frac{1}{\sqrt{3}} - \frac{2}{\sqrt{3}}) - (-1 - \sqrt{2}) = -\sqrt{3} + 1 + \sqrt{2}$. $I_2 = (\tan(\pi/3) + \sec(\pi/3)) - (\tan(\pi/4) + \sec(\pi/4)) = (\sqrt{3} + 2) - (1 + \sqrt{2}) = \sqrt{3} + 1 - \sqrt{2}$. Finally, $I_1 + I_2 = (-\sqrt{3} + 1 + \sqrt{2}) + (\sqrt{3} + 1 - \sqrt{2}) = 2$.