### FINDING INTEGRALS USING THE METHOD OF TRIGONOMETRIC SUBSTITUTION

The following integration problems use the method of trigonometric (trig) substitution. It is a method for finding antiderivatives of functions which contain square roots of quadratic expressions or rational powers of the form $\displaystyle \frac{n}{2}$ (where $n$ is an integer) of quadratic expressions. Examples of such expressions are $$\displaystyle{ \sqrt{ 4-x^2 }} \ \ \ and \ \ \ \displaystyle{(x^2+1)^{3/2}}$$ The method of trig substitution may be called upon when other more common and easier-to-use methods of integration have failed. Trig substitution assumes that you are familiar with standard trigonometric identies, the use of differential notation, integration using u-substitution, and the integration of trigonometric functions. Recall that if $$x = f(\theta) \ ,$$ $$dx = f'(\theta) \ d\theta$$ For example, if $$x = \sec \theta \ ,$$ then $$dx = \sec \theta \tan \theta \ d\theta$$ The goal of trig substitution will be to replace square roots of quadratic expressions or rational powers of the form $\ \displaystyle \frac{n}{2} \$ (where $\ n \$ is an integer) of quadratic expressions, which may be impossible to integrate using other methods of integration, with integer powers of trig functions, which are more easily integrated. For example, if we start with the expression $$\displaystyle{ \sqrt{4-x^2} }$$ and let $$x = 2 \sin \theta \ ,$$ then $$\displaystyle{ \sqrt{4-x^2} } = \displaystyle{ \sqrt{4-(2 \sin \theta)^2 } }$$ $$= \displaystyle{ \sqrt{4-4 \sin^2 \theta } }$$ $$= \displaystyle{ \sqrt{4 (1- \sin^2 \theta ) } }$$ $$= \displaystyle{ \sqrt{4} \cdot \sqrt{1- \sin^2 \theta } }$$ (Recall that $\ \cos^2 \theta + \sin^2 \theta = 1 \$ so that $\ 1- \sin^2 \theta = \cos^2 \theta$.) $$= \displaystyle{2 \cdot \sqrt{ \cos^2 \theta } }$$ $$= \displaystyle{2 \cdot \Big|\cos \theta \Big| }$$ (Assume that $\ \displaystyle - \frac{\pi}{2} \le \theta \le \displaystyle \frac{\pi}{2} \$ so that $\ \cos \theta \ge 0$.) $$= \displaystyle{2 \cos \theta }$$ and $$dx = 2 \cos \theta \ d \theta$$ Thus, $$\displaystyle{ \int \sqrt{4-x^2} \, dx }$$ could be rewritten as $$\displaystyle{ \int \sqrt{4-x^2} \, dx } = \displaystyle{ \int 2 \cos \theta \cdot 2 \cos \theta \, d \theta } = \displaystyle{ 4 \int \cos^2 \theta \, d \theta }$$ (Recall that $\ \cos 2 \theta = 2 \cos^2 \theta -1 \$ so that $\ \cos^2 \theta = \displaystyle \frac{1}{2}(1+ \cos 2 \theta)$ .) $$= \displaystyle{ 4 \int \frac{1}{2}(1+ \cos 2 \theta) \, d \theta }$$ $$= \displaystyle{ 2 \int (1+ \cos 2 \theta) \, d \theta }$$ $$= \displaystyle{ 2 ( \theta + \frac{1}{2} \sin 2 \theta) } + C$$ $$= \displaystyle{ 2 \theta + \sin 2 \theta } + C$$ (Recall that $\ \sin 2 \theta = 2 \sin \theta \cos \theta$ .) $$= \displaystyle{ 2 \theta + 2 \sin \theta \cos \theta } + C$$ We need to write our final answer in terms of $\ x$. Since $\ x = 2 \sin \theta \$, it follows that $$\sin \theta = \displaystyle{ x \over 2} = \displaystyle{ opposite \over hypotenuse }$$ and $$\theta = \arcsin \Big(\displaystyle \frac{x}{2} \Big)$$ Using the given right triangle and the Pythagorean Theorem, we can determine any trig value of $\theta$.

Since $$\displaystyle (adjacent)^2 + (opposite)^2 = (hypotenuse)^2 \ \ \longrightarrow$$ $$(adjacent)^2 + (x)^2 = (2)^2 \ \ \longrightarrow \ \ \ adjacent = \sqrt{4-x^2} \ \ \longrightarrow$$ $$\cos \theta = \displaystyle{ adjacent \over hypotenuse }= \displaystyle{ \sqrt{4-x^2} \over 2 }$$ Then $$\displaystyle{ 2 \theta + 2 \sin \theta \cos \theta } + C = 2 \arcsin \Big( \frac{x}{2} \Big) + 2 \cdot \displaystyle{ x \over 2} \cdot \displaystyle{ \sqrt{4-x^2} \over 2}$$ $$= \displaystyle 2 \arcsin \Big( \frac{x}{2} \Big) + \frac{1}{2} x \cdot \sqrt{4-x^2} + C$$ When using the method of trig substitution, we will always use one of the following three well-known trig identities :

• (I) $\ \ \ \ \ \ \ 1 - \sin^2 \theta = \cos^2 \theta$
• (II) $\ \ \ \ \ \ \ 1 + \tan^2 \theta = \sec^2 \theta$ and
• (III) $\ \ \ \ \ \ \ \sec^2 \theta - 1 = \tan^2 \theta$
For the expression $$\sqrt{a^2-x^2}$$ we use equation (I) and let $$x = a \sin \theta$$ (Assume that $\ \displaystyle - \frac{\pi}{2} \le \theta \le \displaystyle \frac{\pi}{2} \$ so that $\ \cos \theta \ge 0$. This allows for both positve and negative values of $\ x$.)
Then $$\sqrt{a^2-x^2} = \sqrt{a^2-a^2 \sin^2 \theta }$$ $$= \sqrt{a^2(1- \sin^2 \theta)}$$ $$= \sqrt{a^2 \cos^2 \theta}$$ $$= \sqrt{a^2} \sqrt{\cos^2 \theta}$$ $$= a \sqrt{\cos^2 \theta}$$ $$= a \big| \cos \theta \big|$$ $$= a \cos \theta$$ and $$dx = a \cos \theta \ d \theta$$ For the expression $$\sqrt{a^2+x^2}$$ we use equation (II) and let $$x = a \tan \theta$$ (Assume that $\ \displaystyle - \frac{\pi}{2} \le \theta \le \displaystyle \frac{\pi}{2} \$ so that $\ \cos \theta > 0 \$ and $\ \sec \theta \gt 0 \$. This allows for both positve and negative values of $\ x$.)
Then $$\sqrt{a^2+x^2} = \sqrt{a^2+a^2 \tan^2 \theta}$$ $$= \sqrt{a^2(1+ \tan^2 \theta)}$$ $$= \sqrt{a^2} \sqrt{\sec^2 \theta}$$ $$= a \sqrt{ \sec^2 \theta}$$ $$= a \big| \sec \theta \big|$$ $$= a \sec \theta$$ and $$dx = a \sec^2 \theta \ d \theta$$ For the expression $$\sqrt{x^2-a^2}$$ we use equation (III) and let $$x = a \sec \theta$$ (Assume that $\ 0 \le \theta \lt \displaystyle \frac{ \pi }{2} \ ,$ so that $\ \tan \theta \ge 0$. This allows for only positive values of $\ x$. If the integral includes negative values of $\ x$, you must use $\displaystyle \frac{\pi}{2} \lt \theta \le \pi \$ with $\ \sqrt{ \tan^2 \theta} = - \tan \theta$.)
Then $$\sqrt{x^2-a^2} = \sqrt{a^2 \sec^2 \theta - a^2}$$ $$= \sqrt{a^2(\sec^2 \theta - 1)}$$ $$= \sqrt{a^2} \sqrt{\sec^2 \theta - 1}$$ $$= a \sqrt{ \tan^2 \theta}$$ $$= a \big| \tan \theta \big|$$ $$= a \tan \theta$$ and $$dx = a \sec \theta \tan \theta\ d \theta$$ Recall the following well-known, basic indefinite trigonometric integral formulas :
• 1.) $\ \ \ \displaystyle{ \int \cos x \, \ dx } \ = \ \sin x + C$
• 2.) $\ \ \ \displaystyle{ \int \sin x \, \ dx } \ = \ - \cos x + C$
• 3.) $\ \ \ \displaystyle{ \int \sec^2 x \, \ dx } \ = \ \tan x + C$
• 4.) $\ \ \ \displaystyle{ \int \csc^2 x \, \ dx } \ = \ - \cot x + C$
• 5.) $\ \ \ \displaystyle{ \int \sec x \tan x \, \ dx } \ = \ \sec x + C$
• 6.) $\ \ \ \displaystyle{ \int \csc x \cot x \, \ dx } \ = \ - \csc x + C$
• 7.) $\ \ \ \displaystyle{ \int \tan x \, \ dx } \ = \ \ln | \sec x | + C$
• 8.) $\ \ \ \displaystyle{ \int \cot x \, \ dx } \ = \ \ln | \sin x | + C$
• 9.) $\ \ \ \displaystyle{ \int \sec x \, \ dx } \ = \ \ln | \sec x + \tan x| + C$
• 10.) $\ \ \ \displaystyle{ \int \csc x \, \ dx } \ = \ \ln | \csc x - \cot x | + C$

Most of the following problems are average. A few are somewhat challenging. Make careful and precise use of the differential notation $dx$ and $d \theta$ and be careful when arithmetically and algebraically simplifying expressions. You should be proficient integrating various powers and rational functions involving trig functions. You may need to use the following additional well-known trig identities.
• A.) $\ \ \ \sin 2x = 2 \sin x \cos x$
• B.) $\ \ \ \cos 2x = 2 \cos^2 x - 1 \ \ \ \$ so that $\ \ \ \ \cos^2 x = \displaystyle{ \frac{1}{2}(1+\cos 2x)}$
• C.) $\ \ \ \cos 2x = 1 - 2 \sin^2 x \ \ \ \$ so that $\ \ \ \ \sin^2 x = \displaystyle{ \frac{1}{2} (1-\cos 2x) }$
• D.) $\ \ \ \cos 2x = \cos^2 x - \sin^2 x$

• E.) $\ \ \ 1 + \cot^2 x = \csc^2 x \ \ \ \$ so that $\ \ \ \ \cot^2 x = \csc^2 x - 1$

• PROBLEM 1 : Integrate $\displaystyle{ \int { \sqrt{1-x^2} } \,dx }$

• PROBLEM 2 : Integrate $\displaystyle{ \int { (x^2-1)^{3/2} \over x } \, dx }$ .

• PROBLEM 3 : Integrate $\displaystyle{ \int { 1 \over (1-x^2)^{3/2} } \, dx }$ .

• PROBLEM 4 : Integrate $\displaystyle{ \int { \sqrt{x^2+1} \over x } \, dx }$ .

• PROBLEM 5 : Integrate $\displaystyle{ \int x^3 \sqrt{4-9x^2} \, dx }$ .

• PROBLEM 6 : Integrate $\displaystyle{ \int { \sqrt{1-x^2} \over x } \, dx }$ .

• PROBLEM 7 : Integrate $\displaystyle{ \int { \sqrt{x^2-9} \over x^2 } \, dx }$ .

• PROBLEM 8 : Integrate $\displaystyle{ \int { \sqrt{x^2+1} \over x^2 } \,dx }$ .

• PROBLEM 9 : Integrate $\displaystyle{ \int \sqrt{x^2+25 } \,dx }$ .

• PROBLEM 10 : Integrate $\displaystyle{ \int { \sqrt{x^2-4} } \,dx }$ .

• PROBLEM 11 : Integrate $\displaystyle{ \int { x \over \sqrt{x^4-16} } \, dx }$ .

• PROBLEM 12 : Integrate $\displaystyle{ \int { 1 \over \sqrt{x^2-4x} } \, dx }$ .

• PROBLEM 13 : Integrate $\displaystyle{ \int { x \over \sqrt{x^2+4x+5} } \, dx }$ .

• PROBLEM 14 : Integrate $\displaystyle{ \int { x \cdot \sqrt{10x-x^2} } \, dx }$ .

• PROBLEM 15 : Integrate $\displaystyle{ \int { \sqrt{ {x-1} \over x } } \, dx }$ .

• PROBLEM 16 : Integrate $\displaystyle{ \int { \sqrt{1-x} \cdot \sqrt{x+3} } \, dx }$ .

Click HERE or HERE to see a detailed solution to problem 16.

• PROBLEM 17 : A motorboat is resting at the origin, $(0, 0)$, and a skier tethered to the boat with a 20-foot long rope, is resting at the point $(20, 0)$. The boat then begins moving along the positive $y$-axis, pulling the skier along the unknown path $y=f(x)$. Use integration to find an equation for this path.