Integrals Resulting in Inverse Trigonometric Functions: Learn It 1

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Integrals Resulting in Inverse Trigonometric Functions: Learn It 1

Calculate integrals that lead to inverse trigonometric function solutions

In this section we focus on integrals that result in inverse trigonometric functions. We have worked with these functions before.

Recall that trigonometric functions are not one-to-one unless the domains are restricted.

When working with inverses of trigonometric functions, we always need to be careful to take these restrictions into account. Also, we developed formulas for derivatives of inverse trigonometric functions. The formulas developed there give rise directly to integration formulas involving inverse trigonometric functions.

Integrals Resulting in Inverse Trigonometric Functions

Let us begin with the three formulas. Along with these formulas, we use substitution to evaluate the integrals. We prove the formula for the inverse sine integral.

integration formulas resulting in inverse trigonometric Functions

The following integration formulas yield inverse trigonometric functions:

$\displaystyle\int \frac{du}{\sqrt{{a}^{2}-{u}^{2}}}={ \sin }^{-1}\frac{u}{|a|}+C$
$\displaystyle\int \frac{du}{{a}^{2}+{u}^{2}}=\frac{1}{a}\phantom{\rule{0.05em}{0ex}}{ \tan }^{-1}\frac{u}{a}+C$
$\displaystyle\int \frac{du}{u\sqrt{{u}^{2}-{a}^{2}}}=\frac{1}{|a|}\phantom{\rule{0.05em}{0ex}}{ \sec }^{-1}\frac{|u|}{a}+C$

Proof

Let $y={ \sin }^{-1}\dfrac{x}{a}.$ Then $a \sin y=x.$

Now let’s use implicit differentiation. We obtain,

$\begin{array}{ccc}\hfill \frac{d}{dx}(a \sin y)& =\hfill & \frac{d}{dx}(x)\hfill \\ \\ \hfill a \cos y\frac{dy}{dx}& =\hfill & 1\hfill \\ \hfill \frac{dy}{dx}& =\hfill & \frac{1}{a \cos y}.\hfill \end{array}$

For $-\frac{\pi }{2}\le y\le \frac{\pi }{2}, \cos y\ge 0.$

Thus, applying the Pythagorean identity ${ \sin }^{2}y+{ \cos }^{2}y=1,$ we have $\cos y=\sqrt{1={ \sin }^{2}y}.$

This gives,

$\begin{array}{cc}\frac{1}{a \cos y}\hfill & =\frac{1}{a\sqrt{1-{ \sin }^{2}y}}\hfill \\ \\ & =\frac{1}{\sqrt{{a}^{2}-{a}^{2}{ \sin }^{2}y}}\hfill \\ & =\frac{1}{\sqrt{{a}^{2}-{x}^{2}}}.\hfill \end{array}$

Then for $\text{−}a\le x\le a,$ we have,

$\displaystyle\int \frac{1}{\sqrt{{a}^{2}-{u}^{2}}}du={ \sin }^{-1}\left(\frac{u}{a}\right)+C.$

$_\blacksquare$

Evaluate the definite integral ${\displaystyle\int }_{0}^{\frac{1}{2}}\dfrac{dx}{\sqrt{1-{x}^{2}}}.$

We can go directly to the formula for the antiderivative in the rule on integration formulas resulting in inverse trigonometric functions, and then evaluate the definite integral.

We have,

$\begin{array}{}\\ {\displaystyle\int }_{0}^{\frac{1}{2}}\dfrac{dx}{\sqrt{1-{x}^{2}}}\hfill & ={ \sin }^{-1}x{|}_{0}^{\frac{1}{2}}\hfill \\ & ={ \sin }^{-1}{\frac{1}{2}}-{ \sin }^{-1}0\hfill \\ & =\frac{\pi }{6}-0\hfill \\ & =\frac{\pi }{6}.\hfill \end{array}$

Evaluate the integral $\displaystyle\int \frac{dx}{\sqrt{4-9{x}^{2}}}.$

Substitute

$u=3x.$ Then

$du=3dx$ and we have,

$\displaystyle\int \frac{dx}{\sqrt{4-9{x}^{2}}}=\frac{1}{3}\displaystyle\int \frac{du}{\sqrt{4-{u}^{2}}}.$

Applying the formula with $a=2,$ we obtain,

$\begin{array}{cc}{\displaystyle\int} \dfrac{dx}{\sqrt{4-9{x}^{2}}}\hfill & =\frac{1}{3}{\displaystyle\int} \dfrac{du}{\sqrt{4-{u}^{2}}}\hfill \\ & =\frac{1}{3}{ \sin }^{-1}\left(\frac{u}{2}\right)+C\hfill \\ & =\frac{1}{3}{ \sin }^{-1}\left(\frac{3x}{2}\right)+C.\hfill \end{array}$

Evaluate the definite integral ${\displaystyle\int }_{0}^{\sqrt{3}\text{/}2}\dfrac{du}{\sqrt{1-{u}^{2}}}.$

The format of the problem matches the inverse sine formula. Thus,

$\begin{array}{}\\ {\displaystyle\int }_{0}^{\sqrt{3}\text{/}2}\dfrac{du}{\sqrt{1-{u}^{2}}}\hfill & ={ \sin }^{-1}u{|}_{0}^{\sqrt{3}\text{/}2}\hfill \\ & =\left[{ \sin }^{-1}(\frac{\sqrt{3}}{2})\right]-\left[{ \sin }^{-1}(0)\right]\hfill \\ & =\frac{\pi }{3}.\hfill \end{array}$