Question

The value of $\int \frac{dx}{\sec x + \operatorname{cosec} x}$, is equal to

• A. $\left\{(\sin x + \cos x) + \frac{1}{\sqrt{2}}\log \left|\frac{\tan x/2 - 1 - \sqrt{2}}{\tan x/2 - 1 + \sqrt{2}}\right|\right\} + C$
• B. $2\left\{(\sin x + \cos x) + \frac{1}{\sqrt{2}}\log \left|\frac{\tan x/2 - 1 - \sqrt{2}}{\tan x/2 - 1 + \sqrt{2}}\right|\right\} + C$
• C. $\frac{1}{2}\left\{(\sin x - \cos x) + \frac{1}{\sqrt{2}}\log \left|\frac{\tan x/2 - 1 - \sqrt{2}}{\tan x/2 - 1 + \sqrt{2}}\right|\right\} + C$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{2}\left\{(\sin x - \cos x) + \frac{1}{\sqrt{2}}\log \left|\frac{\tan x/2 - 1 - \sqrt{2}}{\tan x/2 - 1 + \sqrt{2}}\right|\right\} + C$

Answer 2

To evaluate the integral $I = \int \frac{dx}{\sec x + \operatorname{cosec} x}$, we first rewrite the integrand in terms of $\sin x$ and $\cos x$:

$$\frac{1}{\sec x + \operatorname{cosec} x} = \frac{1}{\frac{1}{\cos x} + \frac{1}{\sin x}} = \frac{1}{\frac{\sin x + \cos x}{\sin x \cos x}} = \frac{\sin x \cos x}{\sin x + \cos x}$$

So the integral becomes:

$$I = \int \frac{\sin x \cos x}{\sin x + \cos x} dx$$

We use the identity $2 \sin x \cos x = (\sin x + \cos x)^2 - 1$. Let $u = \sin x + \cos x$. Then $\sin x \cos x = \frac{u^2 - 1}{2}$. Substituting this into the integral:

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

Substitute back $u = \sin x + \cos x$:

$$I = \frac{1}{2} \int \left( (\sin x + \cos x) - \frac{1}{\sin x + \cos x} \right) dx$$

This integral can be split into two parts:

$$I = \frac{1}{2} \int (\sin x + \cos x) dx - \frac{1}{2} \int \frac{1}{\sin x + \cos x} dx$$

Let's evaluate each part. Part 1: $\frac{1}{2} \int (\sin x + \cos x) dx = \frac{1}{2} (-\cos x + \sin x) + C_1 = \frac{1}{2} (\sin x - \cos x) + C_1$.

Part 2: $-\frac{1}{2} \int \frac{1}{\sin x + \cos x} dx$. To evaluate $\int \frac{1}{\sin x + \cos x} dx$, we use the half-angle substitution $t = \tan(x/2)$. Then $\sin x = \frac{2t}{1+t^2}$, $\cos x = \frac{1-t^2}{1+t^2}$, and $dx = \frac{2 dt}{1+t^2}$.

$$\int \frac{1}{\sin x + \cos x} dx = \int \frac{1}{\frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2}} \cdot \frac{2 dt}{1+t^2}$$ $$= \int \frac{1+t^2}{2t + 1 - t^2} \cdot \frac{2 dt}{1+t^2} = \int \frac{2 dt}{1+2t-t^2}$$

The denominator can be rewritten by completing the square:

$$1+2t-t^2 = -(t^2 - 2t - 1) = -((t-1)^2 - 1 - 1) = -((t-1)^2 - 2) = 2 - (t-1)^2$$

So the integral becomes:

$$\int \frac{2 dt}{2 - (t-1)^2}$$

This is of the form $\int \frac{k}{a^2 - x^2} dx = \frac{k}{2a} \log \left| \frac{a+x}{a-x} \right|$. Here, $k=2$, $a^2=2 \implies a=\sqrt{2}$, and $x=(t-1)$.

$$\int \frac{2 dt}{2 - (t-1)^2} = 2 \cdot \frac{1}{2\sqrt{2}} \log \left| \frac{\sqrt{2} + (t-1)}{\sqrt{2} - (t-1)} \right| + C_2$$ $$= \frac{1}{\sqrt{2}} \log \left| \frac{t-1+\sqrt{2}}{-t+1+\sqrt{2}} \right| + C_2$$

Now, substitute this back into Part 2:

$$-\frac{1}{2} \left[ \frac{1}{\sqrt{2}} \log \left| \frac{t-1+\sqrt{2}}{-t+1+\sqrt{2}} \right| \right] + C_3$$

Using the property $\log|A/B| = -\log|B/A|$:

$$= \frac{1}{2\sqrt{2}} \log \left| \frac{-t+1+\sqrt{2}}{t-1+\sqrt{2}} \right| + C_3$$

To match the options, we need to manipulate the argument of the logarithm. Let's rewrite the argument in the options:

$$\frac{\tan x/2 - 1 - \sqrt{2}}{\tan x/2 - 1 + \sqrt{2}} = \frac{t - (1+\sqrt{2})}{t - (1-\sqrt{2})}$$

Let's see if we can get this form from our result. Our result is $\frac{1}{\sqrt{2}} \log \left| \frac{t-1+\sqrt{2}}{-t+1+\sqrt{2}} \right|$. Let's multiply the numerator and denominator by $-1$:

$$\frac{1}{\sqrt{2}} \log \left| \frac{-(1-t-\sqrt{2})}{-(t-1-\sqrt{2})} \right| = \frac{1}{\sqrt{2}} \log \left| \frac{1-t-\sqrt{2}}{t-1-\sqrt{2}} \right|$$

This is the integral of $\frac{1}{\sin x + \cos x}$. So, the total integral is:

$$I = \frac{1}{2} (\sin x - \cos x) - \frac{1}{2} \left[ \frac{1}{\sqrt{2}} \log \left| \frac{1-t-\sqrt{2}}{\tan(x/2)-1-\sqrt{2}} \right| \right] + C$$ $$I = \frac{1}{2} (\sin x - \cos x) - \frac{1}{2\sqrt{2}} \log \left| \frac{1-\tan(x/2)-\sqrt{2}}{\tan(x/2)-1-\sqrt{2}} \right| + C$$

The options use $\log \left|\frac{\tan x/2 - 1 - \sqrt{2}}{\tan x/2 - 1 + \sqrt{2}}\right|$.

Combining the two parts:

$$I = \frac{1}{2} (\sin x - \cos x) + \frac{1}{2\sqrt{2}} \log \left| \frac{\tan x/2 - 1 - \sqrt{2}}{\tan x/2 - 1 + \sqrt{2}} \right| + C$$

This can be written as:

$$I = \frac{1}{2} \left\{ (\sin x - \cos x) + \frac{1}{\sqrt{2}} \log \left| \frac{\tan x/2 - 1 - \sqrt{2}}{\tan x/2 - 1 + \sqrt{2}} \right| \right\} + C$$

This matches option (c).