To solve the integral ∫x2−1dx, we can use a trigonometric substitution. Here’s a step-by-step solution:
Trigonometric Substitution:
Let x=sec(θ). Then dx=sec(θ)tan(θ)dθ.
Also, we have:
x2−1=sec2(θ)−1=tan2(θ)=∣tan(θ)∣
Since θ ranges such that sec(θ)≥1, tan(θ) is positive, and ∣tan(θ)∣=tan(θ).
Substitute and Simplify:
Substitute x=sec(θ) and dx=sec(θ)tan(θ)dθ:
∫x2−1dx=∫tan(θ)⋅sec(θ)tan(θ)dθ=∫sec(θ)tan2(θ)dθ
Simplify the Integral:
Use the identity tan2(θ)=sec2(θ)−1:
∫sec(θ)tan2(θ)dθ=∫sec(θ)(sec2(θ)−1)dθ=∫sec3(θ)dθ−∫sec(θ)dθ
Integrate Separately:
The integral ∫sec(θ)dθ=ln∣sec(θ)+tan(θ)∣+C.
The integral ∫sec3(θ)dθ can be approached using integration by parts or other techniques, but for now, let's use a known result:
∫sec3(θ)dθ=21sec(θ)tan(θ)+21ln∣sec(θ)+tan(θ)∣+C
Combine Results:
Combine the integrals:
∫sec3(θ)dθ−∫sec(θ)dθ=(21sec(θ)tan(θ)+21ln∣sec(θ)+tan(θ)∣)−ln∣sec(θ)+tan(θ)∣+C=21sec(θ)tan(θ)−21ln∣sec(θ)+tan(θ)∣+C
Substitute Back:
Substitute back x=sec(θ) and tan(θ)=x2−1:
=21xx2−1−21ln∣x+x2−1∣+C
Therefore, the integral ∫x2−1dx is:
∫x2−1dx=21xx2−1−21ln∣x+x2−1∣+C
Would you like more details or have any questions?
Here are five more questions you might be interested in:
How do you solve ∫sin2(x)dx?
What is the derivative of ln(x2+1)?
Can you explain the method of partial fractions?
How do you find the limit limx→0xsin(x)?
How do you solve a second-order linear differential equation?
Tip: When solving integrals, always check if a trigonometric substitution or another method like integration by parts can simplify the problem.