Integrate: f(t) = \frac{2}{t}\sqrt{t^4+25} with respect to t .

Question:

Evaluate the integral {eq}\displaystyle \int \frac{2}{t}\sqrt{t^4+25} \, dt {/eq}.

Integration by Trigonometric Substitution:

When integrating a function which includes a quadratic expression, it is often helpful to make a trigonometric substitution. This substitution can be especially useful when the quadratic expression is under a square root. Some examples include:

1) To simplify the quadratic expression {eq}a^2-x^2 {/eq}, make the substitution {eq}x=a\sin u {/eq}.

2) To simplify the quadratic expression {eq}x^2+a^2 {/eq}, make the substitution {eq}x=a\tan u {/eq}.

3) To simplify the quadratic expression {eq}x^2-a^2 {/eq}, make the substitution {eq}x=a\sec u {/eq}.

Answer and Explanation:

To start evaluating the integral {eq}\displaystyle \int \frac{2}{t}\sqrt{t^4+25} \, dt {/eq}, we'll make a substitution to get rid of the fourth power of {eq}t {/eq}, as follows:

$$\begin{align*} \int \frac{2}{t}\sqrt{t^4+25} \, dt&=\int \frac{2t}{t^2}\sqrt{t^4+25} \, dt\\[0.3cm] &=\int \frac{\sqrt{t^4+25}}{t^2}(2t \, dt)&&\text{(preparing to make a substitution)}\\[0.3cm] &=\int \frac{\sqrt{u^2+25}}{u} \, du&&\left( \text{making the substitution }u=t^2,du=2t\, dt \right) \end{align*} $$

Next we'll use a tangent substitution:

$$\begin{align*} \int \frac{\sqrt{u^2+25}}{u} \, du&=\int \frac{\sqrt{(5\tan v)^2+25}}{5 \tan v}(5 \sec^2 v \, dv)&& \left( \text{making the substitution }u=5\tan v, du=5 \sec^2 v \, dv\text{ for }-\frac{\pi}{2}<v<\frac{\pi}{2} \right)\\[0.3cm] &=\int \frac{\sqrt{25\tan^2 v + 25}}{5 \tan v}(5 \sec^2 v \, dv)\\[0.3cm] &=\int \frac{\sqrt{25\sec^2 v}}{5 \tan v}(5 \sec^2 v \, dv)&&\text{(by the Pythagorean identity for the secant and tangent)}\\[0.3cm] &=\int \frac{5 \sec v}{5 \tan v}(5 \sec^2 v \, dv)&&\left(\text{since }\sec v > 0\text{ for }-\frac{\pi}{2}<v<\frac{\pi}{2}\right)\\[0.3cm] &=\int \frac{5\sec^3 v}{\tan v} \, dv\\[0.3cm] &=\int \frac{5\sin v}{\cos^2 v \sin^2 v} \, dv\\[0.3cm] &=\int \frac{5\sin v}{\cos^2 v(1-\cos^2 v)} \, dv&&\text{(by the Pythagorean identity for the sine and cosine)}\\[0.3cm] &=\int \left[\frac{5\sin v}{\cos^2 v}+\frac{5\sin v}{1-\cos^2 v}\right] \, dv&&\left(\text{using the identity }\frac{1}{w(1-w)}=\frac{1}{w}+\frac{1}{1-w}\right)\\[0.3cm] &=\int \frac{5\sin v}{\cos^2 v} \, dv + \int \frac{5 \sin v}{1-\cos^2 v} \, dv&&\text{(breaking the integral up)}\\ \end{align*} $$

Now, we can evaluate each of these two integrals separately:

$$\begin{align*} \int \frac{5\sin v}{\cos^2 v} \, dv&=\int 5\sec v \tan v \, dv\\[0.3cm] &=5 \sec v+C_1 \end{align*} $$

and

$$\begin{align*} \int \frac{5\sin v}{1-\cos^2 v} \, dv&=\int \frac{5\sin v}{\sin^2 v} \, dv&&\text{(by the Pythagorean identity for the sine and cosine)}\\[0.3cm] &=\int 5\csc v \, dv\\[0.3cm] &=-5 \ln|\csc v + \cot v|+C_2 \, \end{align*} $$

Substituting these two integrals back together, the entire integral is then equal to:

$$\begin{align*} 5\sec v - 5 \ln|\csc v + \cot v|+C&=5\sec v-5 \ln\left|\frac{\sec v - 1}{\tan v}\right|+C&&\text{(rewriting in terms of secant and tangent)}\\[0.3cm] &=5\sqrt{\tan^2 v + 1}-5 \ln\left|\frac{\sqrt{\tan^2 v + 1}-1}{\tan v}\right|+C&&\left( \text{rewriting in terms of tangent alone, using the fact that }\sec v>0\text{ for }-\frac{\pi}{2}<v<\frac{\pi}{2}\right)\\[0.3cm] &=5\sqrt{\left(\frac{u}{5}\right)^2+1}-5\ln\left|\frac{\sqrt{(u/5)^2+1}-1}{u}\right|+C&& \left( \text{rewriting in terms of the variable }u\text{, using the fact that }\frac{u}{5}=\tan v \right)\\[0.3cm] &=\sqrt{u^2+25}-5\ln\left|\frac{\sqrt{u^2+25}-5}{5u}\right|+C\\[0.3cm] &=\sqrt{u^2+25}-5\ln\left|\frac{\sqrt{u^2+25}-5}{u}\right|+C&&\text{(absorbing the constant }-5\ln 5\text{ into }C\text{)}\\[0.3cm] &=\sqrt{t^4+25}-5\ln\left|\frac{\sqrt{t^4+25}-5}{t^2}\right|+C&&\text{(rewriting in terms of the original variable }t\text{)}\\[0.3cm] &=\sqrt{t^4+25}-5\ln \frac{\sqrt{t^4+25}-5}{t^2}+C&&\text{(because the expression in absolute values is positive wherever it is defined)} \end{align*} $$

In summary, we've found that {eq}\boxed{\int \frac{2}{t}\sqrt{t^4+25}\,dt=\sqrt{t^4+25}-5\ln \frac{\sqrt{t^4+25}-5}{t^2}+C \,} {/eq}


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How to Use Trigonometric Substitution to Solve Integrals

from Math 104: Calculus

Chapter 13 / Lesson 11
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