Question

The value of $\int{\sqrt{\sec x-1}dx}$ is equal to

Answer 1

Hint:Simplify the expression $\int{\sqrt{\sec x-1}dx}$ using $\sec x=\dfrac{1}{\cos x}$ . We know the formula,
$\cos 2x=2{{\cos }^{2}}x-1$ and $\cos 2x=1-2si{{n}^{2}}x\Rightarrow 2si{{n}^{2}}x=1-\cos 2x$ .Then, replace x by $\dfrac{x}{2}$
in these two formulas. Use these two formulas and transform $\int{\sqrt{\dfrac{1-\cos x}{\cos x}}dx}$ . Now, assume $t=\sqrt{2}cos\dfrac{x}{2}$ and transform the equation $\int{\sqrt{\dfrac{1}{2{{\cos }^{2}}\dfrac{x}{2}-1}}\times \sqrt{2}\sin \dfrac{x}{2}dx}$ . We know the formula $\dfrac{d\left( \cos ax \right)}{dx}=\dfrac{1}{a}\left( -\sin ax \right)$ and
$\int{\sqrt{\dfrac{1}{{{t}^{2}}-1}}dt=\left[ \ln \left[ \sqrt{{{t}^{2}}-1}+t \right] \right]}+C$ . Use these formulas and solve it further.

Complete step-by-step answer:
According to the question, we have to integrate,
$\int{\sqrt{\sec x-1}dx}$ ……………….(1)
We know that cosine function is reciprocal of sec function.
$\sec x=\dfrac{1}{\cos x}$ …………………..(2)
Putting the value of $\sec x$ from equation (2) in equation (1), we get
$\int{\sqrt{\sec x-1}dx}$
$=\int{\sqrt{\dfrac{1}{\cos x}-1}dx}$
$=\int{\sqrt{\dfrac{1-\cos x}{\cos x}}dx}$ ……………………(3)
We know the formula, $\cos 2x=2{{\cos }^{2}}x-1$ .
Replacing x by $\dfrac{x}{2}$ in the above formula, we get
$\cos 2.\dfrac{x}{2}=2{{\cos }^{2}}\dfrac{x}{2}-1$
$\Rightarrow \cos x=2{{\cos }^{2}}\dfrac{x}{2}-1$ ………………………(4)
We know the formula, $\cos 2x=1-2si{{n}^{2}}x$ .
Replacing x by $\dfrac{x}{2}$ in the above formula, we get
$\cos 2.\dfrac{x}{2}=1-2si{{n}^{2}}\dfrac{x}{2}$
$\Rightarrow \cos x=1-2si{{n}^{2}}\dfrac{x}{2}$
$\Rightarrow 2si{{n}^{2}}\dfrac{x}{2}=1-\cos x$ ………………………(5)
Now, from equation (3), equation (4), and equation (5), we get
$=\int{\sqrt{\dfrac{1-\cos x}{\cos x}}dx}$
$=\int{\sqrt{\dfrac{2{{\sin }^{2}}\dfrac{x}{2}}{2{{\cos }^{2}}\dfrac{x}{2}-1}}dx}$
$=\int{\sqrt{\dfrac{1}{2{{\cos }^{2}}\dfrac{x}{2}-1}}\times \sqrt{2}\sin \dfrac{x}{2}dx}$ ………………..(6)
Let us assume, $t=\sqrt{2}cos\dfrac{x}{2}$ ……………………..(7)
Differentiating with respect to x in equation (7), we get
$\dfrac{dt}{dx}=\dfrac{d\left( \sqrt{2}cos\dfrac{x}{2} \right)}{dx}$
$\dfrac{dt}{dx}=\sqrt{2}\dfrac{d\left( \cos \dfrac{x}{2} \right)}{dx}$ ……………………….(8)
We know the formula, $\dfrac{d\left( \cos ax \right)}{dx}=\dfrac{1}{a}\left( -\sin ax \right)$ .
Replacing x by $\dfrac{x}{2}$ and a by $\dfrac{1}{2}$ in the above formula, we get
$\dfrac{d\left( \cos \dfrac{x}{2} \right)}{dx}=\dfrac{1}{2}\left( -\sin \dfrac{x}{2} \right)$ ……………………(9)
Now, from equation (8) and equation (9), we get
$\dfrac{dt}{dx}=\sqrt{2}\dfrac{d\left( \cos \dfrac{x}{2} \right)}{dx}$

& \dfrac{dt}{dx}=-\sqrt{2}\dfrac{1}{2}(-\sin \dfrac{x}{2}) \\\ & \dfrac{dt}{dx}=-\dfrac{1}{\sqrt{2}}\left( \sin \dfrac{x}{2} \right) \\\ \end{aligned}$$ $$\Rightarrow -\sqrt{2}.dt=\left( \sin \dfrac{x}{2} \right)dx$$ ………………..(10) From equation (6), equation (7), and equation (10), we get $$=\int{\sqrt{\dfrac{1}{2{{\cos }^{2}}\dfrac{x}{2}-1}}\times \sqrt{2}\left( \sin \dfrac{x}{2} \right)dx}$$ $$=\int{\sqrt{\dfrac{1}{{{t}^{2}}-1}}\times \sqrt{2}\left( -\sqrt{2} \right)dt}$$ $$=-2\int{\sqrt{\dfrac{1}{{{t}^{2}}-1}}dt}$$ ………………………….(11) We know the formula, $$\int{\sqrt{\dfrac{1}{{{t}^{2}}-1}}dt=\left[ \ln \left[ \sqrt{{{t}^{2}}-1}+t \right] \right]}+C$$ . Now, using this formula and simplifying equation (11), we get $$=-2\int{\sqrt{\dfrac{1}{{{t}^{2}}-1}}dt}$$ $$=(-2)\left[ \ln \left[ \sqrt{{{t}^{2}}-1}+t \right] \right]+C$$ ………………..(12) From equation (7), we have $$t=\sqrt{2}cos\dfrac{x}{2}$$ . Now, using equation (7) and transforming equation (12), we get $$\begin{aligned} & =\left( -2 \right)\left[ \ln \left[ \sqrt{{{\left( \sqrt{2}\cos \dfrac{x}{2} \right)}^{2}}-1}+t \right] \right]+C \\\ & =\left( -2 \right)\left[ \ln \left[ \sqrt{2{{\cos }^{2}}\dfrac{x}{2}-1}+\sqrt{2}cos\dfrac{x}{2} \right] \right]+C \\\ \end{aligned}$$ $$=\left( -2 \right)\left[ \ln \left[ \sqrt{2}\times \left( \sqrt{{{\cos }^{2}}\dfrac{x}{2}-1}+cos\dfrac{x}{2} \right) \right] \right]+C$$ ……………………….(13) We know the formula, $$\ln (pqr)=ln(p)+ln(qr)$$ . Using this formula in equation (13), we get $$=-2\ln \sqrt{2}-2\ln \left( \sqrt{{{\cos }^{2}}\dfrac{x}{2}-1}+cos\dfrac{x}{2} \right)+C$$ $$\begin{aligned} & ={{C}_{1}}-2\ln \left( \sqrt{{{\cos }^{2}}\dfrac{x}{2}-1}+cos\dfrac{x}{2} \right)+C \\\ & =-2\ln \left( \sqrt{{{\cos }^{2}}\dfrac{x}{2}-1}+cos\dfrac{x}{2} \right)+C+{{C}_{1}} \\\ & =-2\ln \left( \sqrt{{{\cos }^{2}}\dfrac{x}{2}-1}+cos\dfrac{x}{2} \right)+A \\\ \end{aligned}$$ Here, A is a constant. Hence, the value of $$\int{\sqrt{\sec x-1}dx}$$ is $$-2\ln \left( \sqrt{{{\cos }^{2}}\dfrac{x}{2}-1}+cos\dfrac{x}{2} \right)+A$$ . Note: To solve this question, one might think to take $$\cos x$$ equal to t in $$\int{\sqrt{\left( \dfrac{1-\cos x}{\cos x} \right)dx}}$$ . If we do so then, we will not be able to transform it into a simpler form. $$\cos x=t$$ ………………….(1) We know the identity, $${{\sin }^{2}}x+{{\cos }^{2}}x=1$$ . Using this identity to find the value of $$\sin x$$ . $$\sin x=\sqrt{1-{{\cos }^{2}}x}=\sqrt{1-{{t}^{2}}}=\sqrt{1-t}.\sqrt{1+t}$$ ………………..(2) On differentiating equation (1), we get $$\Rightarrow \dfrac{d(\cos x)}{dx}=\dfrac{dt}{dx}$$ $$\Rightarrow dx=\dfrac{-dt}{\sin x}$$ ……………………(3) Using equation (1), equation (2), and equation (3), transforming $$\int{\sqrt{\left( \dfrac{1-\cos x}{\cos x} \right)dx}}$$ , we get $$\int{\left( \dfrac{\sqrt{1-t}}{t} \right)\left( \dfrac{-dt}{\sin x} \right)}$$ $$=\int{\left( \dfrac{\sqrt{1-t}}{t} \right)\left( \dfrac{-dt}{\sqrt{1-t}.\sqrt{1+t}} \right)}$$ The above equation is complex to be solved. Therefore, we should not approach this question by this method.