Which integration technique is applicable?

Question

Can the substitution method of integration be used for $\int_0^x\sqrt{t^2+1}dt$?

Answer 1

The first step in integrating sqrt(1 + t^2) is to make the trigonometric substitution t = tan(u) and dt = sec(u)^2 du. Then the integral of sqrt(t^2 + 1) becomes

∫ sqrt(t^2 + 1) dt

= ∫ sqrt(tan(u)^2 + 1) * sec(u)^2 du

= ∫ sqrt(sec(u)^2) * sec(u)^2 du

= ∫ sec(u)^3 du .

To integrate secant cubed, we use integration by parts. Recall the integration by parts formula,

∫ f * dg = fg - ∫ g * df

The task is to assign f, g, df, and df so that our integral becomes simpler. If we apply the assignment f = sec(u), dg = sec(u)^2 du, df = sec(u)tan(u) du and g = tan(u), we can transform the integral of secant cubed into the equation

∫ sec(u)^3 du

= tan(u)sec(u) - ∫ sec(u)tan(u)^2 du

= tan(u)sec(u) - ∫ sec(u)[sec(u)^2 - 1] du

= tan(u)sec(u) - ∫ sec(u)^3 du + ∫ sec(u) du

= tan(u)sec(u) - ∫ sec(u)^3 du + ln(sec(u) + tan(u))

This is an integral equation with the same integral term appearing on both sides. The integral equation we obtained in the previous step is

∫ sec(u)^3 du = tan(u)sec(u) + Ln(sec(u) + tan(u)) - ∫ sec(u)^3 du

which contains the same integral on both sides. Thus, we can rewrite it as 2*∫ sec(u)^3 du = tan(u)sec(u) + Ln(sec(u) + tan(u))

Dividing both sides by 2 gives us the antiderivative of secant cubed:

∫ sec(u)^3 du = 0.5*tan(u)sec(u) + 0.5*Ln(sec(u) + tan(u)) + c

Now we just need to make the reverse substitution with u = arctan(t) to find the antiderivative of the original function, sqrt(1 + t^2). Since tan(arctan(t)) = t, we get

0.5*tan(u)sec(u) + 0.5*ln(sec(u) + tan(u)) + c

= 0.5*t*sec(arctan(t)) + 0.5*ln[sec(arctan(t)) + t) + c

This can be simplified even further using the identity sec(arctan(x)) = sqrt(t^2 + 1). This gives us

0.5*t*sqrt(t^2 + 1) + 0.5*ln(sqrt(t^2 + 1) + t) + c

The above equation is the integral of the function sqrt(t^2 + 1). Then putting the limits we have the final answer as 0.5*x*sqrt(x^2 + 1) + 0.5*Ln(sqrt(x^2 + 1) + x) + c.

Answer 2

Hint: An easier approach: Try with $t=\sinh u$ and use that $\cosh^2 u-\sinh^2 u=1$.

Some details (hopefully without significant errors): Let $x=\sinh u_0$. Using $dt=\cosh u$ yields: $$ \int_0^x \sqrt{t^2+1} \; dt = \int_0^{u_0} \cosh^2 u \; du =\int_0^{u_0} \frac12(\cosh 2u + 1)\; du =\frac{1}{4}(\sinh 2u_0 + 2u_0)$$ Finally show (or use) that $u_0=\log (x+\sqrt{x^2+1})$ and $\sinh 2u_0 = 2\sinh u_0 \cosh u_0$.