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
\[\dfrac{dt}{dx}=\sqrt{2}\dfrac{d\left( \cos \dfrac{x}{2} \right)}{dx}\]
\[\begin{align}
  & \dfrac{dt}{dx}=-\sqrt{2}\dfrac{1}{2}(-\sin \dfrac{x}{2}) \\
 & \dfrac{dt}{dx}=-\dfrac{1}{\sqrt{2}}\left( \sin \dfrac{x}{2} \right) \\
\end{align}\]
\[\Rightarrow -\sqrt{2}.dt=\left( \sin \dfrac{x}{2} \right)dx\] ………………..(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
\[\begin{align}
  & =\left( -2 \right)\left[ \ln \left[ \sqrt{{{\left( \sqrt{2}\cos \dfrac{x}{2} \right)}^{2}}-1}+t \right] \right]+C \\
 & =\left( -2 \right)\left[ \ln \left[ \sqrt{2{{\cos }^{2}}\dfrac{x}{2}-1}+\sqrt{2}cos\dfrac{x}{2} \right] \right]+C \\
\end{align}\]
\[=\left( -2 \right)\left[ \ln \left[ \sqrt{2}\times \left( \sqrt{{{\cos }^{2}}\dfrac{x}{2}-1}+cos\dfrac{x}{2} \right) \right] \right]+C\] ……………………….(13)
We know the formula, \[\ln (pqr)=ln(p)+ln(qr)\] .
Using this formula in equation (13), we get
\[=-2\ln \sqrt{2}-2\ln \left( \sqrt{{{\cos }^{2}}\dfrac{x}{2}-1}+cos\dfrac{x}{2} \right)+C\]
\[\begin{align}
  & ={{C}_{1}}-2\ln \left( \sqrt{{{\cos }^{2}}\dfrac{x}{2}-1}+cos\dfrac{x}{2} \right)+C \\
 & =-2\ln \left( \sqrt{{{\cos }^{2}}\dfrac{x}{2}-1}+cos\dfrac{x}{2} \right)+C+{{C}_{1}} \\
 & =-2\ln \left( \sqrt{{{\cos }^{2}}\dfrac{x}{2}-1}+cos\dfrac{x}{2} \right)+A \\
\end{align}\]
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.