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Question: Integration of sec inversex...

Integration of sec inversex

Answer

x sec^{-1}(x) - sgn(x) ln|x + sqrt(x^2-1)| + C

Explanation

Solution

To integrate sec⁡−1(x)\sec^{-1}(x), we use the method of integration by parts.

Let I=∫sec⁡−1(x)dxI = \int \sec^{-1}(x) dx.
We can write this as I=∫sec⁡−1(x)⋅1dxI = \int \sec^{-1}(x) \cdot 1 dx.

Choose u=sec⁡−1(x)u = \sec^{-1}(x) and dv=1dxdv = 1 dx.
Then, we find dudu and vv:
du=ddx(sec⁡−1(x))dx=1∣x∣x2−1dxdu = \frac{d}{dx}(\sec^{-1}(x)) dx = \frac{1}{|x|\sqrt{x^2-1}} dx
v=∫1dx=xv = \int 1 dx = x

Now, apply the integration by parts formula ∫udv=uv−∫vdu\int u dv = uv - \int v du:
I=sec⁡−1(x)⋅x−∫x⋅1∣x∣x2−1dxI = \sec^{-1}(x) \cdot x - \int x \cdot \frac{1}{|x|\sqrt{x^2-1}} dx
I=xsec⁡−1(x)−∫x∣x∣x2−1dxI = x \sec^{-1}(x) - \int \frac{x}{|x|\sqrt{x^2-1}} dx

The term x∣x∣\frac{x}{|x|} is equal to sgn(x)\text{sgn}(x) (the sign function), which is 11 for x>0x>0 and −1-1 for x<0x<0.
Since the domain of sec⁡−1(x)\sec^{-1}(x) is (−∞,−1]∪[1,∞)(-\infty, -1] \cup [1, \infty), we consider two cases for xx:

Case 1: x>1x > 1
In this case, ∣x∣=x|x| = x, so x∣x∣=1\frac{x}{|x|} = 1.
I=xsec⁡−1(x)−∫1x2−1dxI = x \sec^{-1}(x) - \int \frac{1}{\sqrt{x^2-1}} dx
The integral ∫1x2−a2dx=ln⁡∣x+x2−a2∣+C\int \frac{1}{\sqrt{x^2-a^2}} dx = \ln|x + \sqrt{x^2-a^2}| + C. Here a=1a=1.
So, ∫1x2−1dx=ln⁡∣x+x2−1∣\int \frac{1}{\sqrt{x^2-1}} dx = \ln|x + \sqrt{x^2-1}|.
Since x>1x > 1, x+x2−1x + \sqrt{x^2-1} is always positive, so ∣x+x2−1∣=x+x2−1|x + \sqrt{x^2-1}| = x + \sqrt{x^2-1}.
Thus, for x>1x > 1:
I=xsec⁡−1(x)−ln⁡(x+x2−1)+C1I = x \sec^{-1}(x) - \ln(x + \sqrt{x^2-1}) + C_1

Case 2: x<−1x < -1
In this case, ∣x∣=−x|x| = -x, so x∣x∣=−1\frac{x}{|x|} = -1.
I=xsec⁡−1(x)−∫−1x2−1dxI = x \sec^{-1}(x) - \int \frac{-1}{\sqrt{x^2-1}} dx
I=xsec⁡−1(x)+∫1x2−1dxI = x \sec^{-1}(x) + \int \frac{1}{\sqrt{x^2-1}} dx
I=xsec⁡−1(x)+ln⁡∣x+x2−1∣+C2I = x \sec^{-1}(x) + \ln|x + \sqrt{x^2-1}| + C_2
For x<−1x < -1, x+x2−1x + \sqrt{x^2-1} is negative (e.g., if x=−2x=-2, x+x2−1=−2+3≈−0.268x+\sqrt{x^2-1} = -2+\sqrt{3} \approx -0.268).
So, ∣x+x2−1∣=−(x+x2−1)|x + \sqrt{x^2-1}| = -(x + \sqrt{x^2-1}).
Thus, for x<−1x < -1:
I=xsec⁡−1(x)+ln⁡(−(x+x2−1))+C2I = x \sec^{-1}(x) + \ln(-(x + \sqrt{x^2-1})) + C_2

Combining the cases:
We can express the result in a single general form using the sign function, sgn(x)\text{sgn}(x):
The integral ∫x∣x∣x2−1dx=∫sgn(x)x2−1dx\int \frac{x}{|x|\sqrt{x^2-1}} dx = \int \frac{\text{sgn}(x)}{\sqrt{x^2-1}} dx.
If x>1x > 1, this is ∫1x2−1dx=ln⁡∣x+x2−1∣\int \frac{1}{\sqrt{x^2-1}} dx = \ln|x + \sqrt{x^2-1}|.
If x<−1x < -1, this is ∫−1x2−1dx=−ln⁡∣x+x2−1∣\int \frac{-1}{\sqrt{x^2-1}} dx = -\ln|x + \sqrt{x^2-1}|.
So, ∫sgn(x)x2−1dx=sgn(x)ln⁡∣x+x2−1∣\int \frac{\text{sgn}(x)}{\sqrt{x^2-1}} dx = \text{sgn}(x) \ln|x + \sqrt{x^2-1}|.

Substituting this back into the main integral:
I=xsec⁡−1(x)−sgn(x)ln⁡∣x+x2−1∣+CI = x \sec^{-1}(x) - \text{sgn}(x) \ln|x + \sqrt{x^2-1}| + C

This form correctly covers both cases.
For x>1x>1, sgn(x)=1\text{sgn}(x)=1, so I=xsec⁡−1(x)−ln⁡(x+x2−1)+CI = x \sec^{-1}(x) - \ln(x + \sqrt{x^2-1}) + C.
For x<−1x<-1, sgn(x)=−1\text{sgn}(x)=-1, so I=xsec⁡−1(x)−(−1)ln⁡∣x+x2−1∣+C=xsec⁡−1(x)+ln⁡∣x+x2−1∣+CI = x \sec^{-1}(x) - (-1) \ln|x + \sqrt{x^2-1}| + C = x \sec^{-1}(x) + \ln|x + \sqrt{x^2-1}| + C.

The final answer is xsec⁡−1(x)−sgn(x)ln⁡∣x+x2−1∣+Cx \sec^{-1}(x) - \text{sgn}(x) \ln|x + \sqrt{x^2-1}| + C.

Explanation of the solution:

The integral of sec⁡−1(x)\sec^{-1}(x) is found using integration by parts. Let u=sec⁡−1(x)u = \sec^{-1}(x) and dv=dxdv = dx. This gives du=1∣x∣x2−1dxdu = \frac{1}{|x|\sqrt{x^2-1}} dx and v=xv = x. Applying the formula ∫udv=uv−∫vdu\int u dv = uv - \int v du, we get xsec⁡−1(x)−∫x∣x∣x2−1dxx \sec^{-1}(x) - \int \frac{x}{|x|\sqrt{x^2-1}} dx. The term x∣x∣\frac{x}{|x|} is sgn(x)\text{sgn}(x). The remaining integral ∫sgn(x)x2−1dx\int \frac{\text{sgn}(x)}{\sqrt{x^2-1}} dx evaluates to sgn(x)ln⁡∣x+x2−1∣\text{sgn}(x) \ln|x + \sqrt{x^2-1}|. Combining these, the result is xsec⁡−1(x)−sgn(x)ln⁡∣x+x2−1∣+Cx \sec^{-1}(x) - \text{sgn}(x) \ln|x + \sqrt{x^2-1}| + C.