∫sec(x)dx
lnIsec(x)+tan(x)I+c
To integrate ∫sec(x)dx, we can use the technique of u-substitution where u = tan(x/2). Here’s the step by step process:
1. Rewrite sec(x) in terms of sin(x) and cos(x):
sec(x) = 1/cos(x) = cos⁻¹(x)
2. Substitute u = tan(x/2) in the integral:
∫sec(x)dx = ∫cos⁻¹(x)dx
Let u = tan(x/2), then we have:
x = 2tan⁻¹(u), dx = (2/(1+u²))du
So, the integral becomes:
∫cos⁻¹(x)dx = 2∫cos⁻¹(tan(u/2)) • (2/(1+u²))du
3. Simplify using trigonometric identities:
Using the identity: cos⁻¹(x) = π/2 – sin⁻¹(x), we get:
2∫(π/2 – sin⁻¹(tan(u/2))) • (2/(1+u²))du
4. Integrate using the formula:
∫sin⁻¹(x)dx = x•sin⁻¹(x) + √(1-x²) + C
So, the integral becomes:
2[(π/2)•tan⁻¹(u) + u•sin⁻¹(u) – ln|u+√(1+u²)|] + C
5. Substitute back u = tan(x/2):
2[(π/2)•x/2 + tan(x/2)•sin⁻¹(tan(x/2)) – ln|tan(x/2)+√(1+tan²(x/2))|] + C
Finally, we simplify our answer by using the trigonometric identity:
sin⁻¹(tan(x/2)) = (1/2)•ln[(1+sin(x))/(1-sin(x))]
Thus, the final integral becomes:
ln|sec(x) + tan(x)| + C.
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