One of the fundamental formulas in geometry is for the area \(A\) of a circle of radius r: \(A = \pi r^2\). The calculus-based proof of that formula uses a definite integral evaluated by means of a trigonometric substitution, as will now be demonstrated.
Use the circle of radius \(r >0\) centered at the origin \((0,0)\) in the \(xy\)-plane, whose equation is \(x^2 + y^2 = r^2\) (see Figure [fig:circle](a) below).
By symmetry about the \(x\)-axis, the area \(A\) of the circle is twice the area of its upper hemisphere (see Figure [fig:circle](b) above), which is the area under the curve \(y =\sqrt{r^2 - x^2}\):
\[A ~=~ 2 \, \int_{-r}^{r} \sqrt{r^2 - x^2}~\dx \nonumber \]
To evaluate this integral, recall from trigonometry that any point \((x,y)\) on the circle can be written as \((x,y)=(r\cos\,\theta,r\sin\,\theta)\), where \(0 \le \theta < 2\pi\) (in radians) is the angle shown in Figure [fig:circle](a). Figure [fig:circle](b) shows that as \(x\) goes from \(x=-r\) to \(x=r\), the angle \(\theta\) goes from \(\theta = \pi\) to \(\theta = 0\). Now substitute \(x=r\,\cos\,\theta\) and \(dx=-r\,\sin\,\theta\,d\theta\) into the integral and change the limits of integration from \(x=-r\) and \(x=r\) to \(\theta =\pi\) and \(\theta=0\), respectively:
\(begin{aligned} A ~&=~ 2 \,\int_{\pi}^{0} \sqrt{r^2 - r^2\,\cos^2 \,\theta}~(-r\,\sin\,\theta)~,\dtheta ~=~ -2 \,\int_{0}^{\pi} \sqrt{r^2\,(1 - \cos^2 \,\theta)}~(-r\,\sin\,\theta)~,\dtheta\
\[6pt] &=~ 2 \,\int_{0}^{\pi} r\,\sqrt{\sin^2\,\theta}~r\,\sin\,\theta~\dtheta ~=~ 2r^2 \,\,\int_{0}^{\pi} \sin^2\,\theta~\dtheta ~=~ \cancel{2}r^2 \,\int_{0}^{\pi} \frac{1 - \cos\,2\theta}{\cancel{2}}~\dtheta\
\[6pt] &=~ r^2 \,\left(\theta ~-~ \frac{1}{2}\,\sin\,2\theta\right)~\Biggr|_0^{\pi} ~=~ r^2 \,\left(\pi ~-~ \frac{1}{2}\,\sin\,2\pi ~-~ \left(0 ~-~ \frac{1}{2}\,\sin\,0\right)\right)\
\[6pt] &=~ \pi r^2 \quad\checkmark\end{aligned} \nonumber \]
For an indefinite integral of the general form \(\int \sqrt{a^2 - u^2}\;\du\), the same calculation as above with the substitutions \(u=a \cos\,\theta\) and \(\du=-a \sin\,\theta\,\dtheta\) yields
\[\begin{aligned} \int \sqrt{a^2 - u^2}~\du ~&=~ \int \sqrt{a^2 - a^2\,\cos^2 \,\theta}~(-a\,\sin\,\theta)~\dtheta ~=~ -a^2 \,\int \sin^2\,\theta~\dtheta\
\[6pt] &=~ -a^2 \,\int \frac{1 - \cos\,2\theta}{2}~\dtheta ~=~ -\frac{a^2}{2}\,\theta ~+~ \frac{a^2\,\sin\,2\theta}{4} ~+~ C ~,\end{aligned} \nonumber \]
which is still in terms of \(\theta\). To put this back in terms of \(u\), use \(\theta = \cos^{-1} (\frac{u}{a})\), the double-angle formula \(\sin\,2\theta = 2\,\sin\,\theta\,\cos\,\theta\), and \(\sqrt{a^2 - u^2} = \sqrt{a^2\,\sin^2\,\theta} = a\,\sin\,\theta\). Then
\[\begin{aligned} \int \sqrt{a^2 - u^2}~\du ~&=~ -\frac{a^2}{2}\,\cos^{-1} \left(\frac{u}{a}\right) ~+~ \frac{2a^2\,\sin\,\theta\,\cos\,\theta}{4}\
\[6pt] &=~ -\frac{a^2}{2}\,\cos^{-1} \left(\frac{u}{a}\right) ~+~ \frac{(a\,\cos\,\theta)\,(a\,\sin\,\theta)}{2} ~+~ C \end{aligned} \nonumber \]
which results in the following formula:
It is left as an exercise to show that the substitution \(u=a \sin\,\theta\) gives: That these two seemingly different antiderivatives are equivalent follows immediately from the identity \(\sin^{-1} x \,+\, \cos^{-1} x \,=\, \frac{\pi}{2}\) for all \(-1 \le x \le 1\), which shows that the antiderivatives differ by the constant \(\frac{\pi a^2}{4}\) (absorbed in the generic constant \(C\)):
\[\begin{aligned} \frac{a^2}{2}\,\sin^{-1} \left(\frac{u}{a}\right) ~+~ \frac{1}{2}\,u\,\sqrt{a^2 - u^2} ~+~ C ~&=~ \frac{a^2}{2}\,\left(\frac{\pi}{2} ~-~ \cos^{-1} \left(\frac{u}{a}\right)\right) ~+~ \frac{1}{2}\,u\,\sqrt{a^2 - u^2} ~+~ C\
\[6pt] &=~ -\frac{a^2}{2}\,\cos^{-1} \left(\frac{u}{a}\right) ~+~ \frac{1}{2}\,u\,\sqrt{a^2 - u^2} ~+~ \underbracket[0.3pt]{C +\frac{\pi a^2}{4}}_{C}\end{aligned} \nonumber \]
Thus, either substitution—\(u=a \cos\,\theta\) or \(u=a \sin\,\theta\)—can be used when evaluating the integral \(\int \sqrt{a^2 - u^2}\,\du\). The latter choice is sometimes preferred, to avoid the negative sign in \(\du\) and the resulting formula.
Example \(\PageIndex{1}\): trigsub1
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Solution
Evaluate \(~\displaystyle\int \sqrt{9-4x^2}~\dx~\).
Solution: The integrand is of the form \(\sqrt{a^2 - u^2}\) with \(a=3\) and \(u=2x\), so that \(\du = 2\dx\). Then \(\dx = \frac{1}{2}\du\) and so:
\[\begin{aligned} \int \sqrt{9-4x^2}~\dx ~&=~ \frac{1}{2}\,\int \sqrt{a^2 - u^2}~\du\
\[6pt] &=~ \frac{1}{2}\,\left(\frac{a^2}{2}\,\sin^{-1} \left(\frac{u}{a}\right) ~+~ \frac{1}{2}\,u\,\sqrt{a^2 - u^2}\;\right) ~+~ C\
\[6pt] &=~ \frac{9}{4}\,\sin^{-1} \left(\frac{2x}{3}\right) ~+~ \frac{1}{2}\,x\,\sqrt{9 - 4x^2} ~+~ C\end{aligned} \nonumber \]
In general, when other methods fail, use the table below as a guide for certain types of integrals, making use of the specified substitution and trigonometric identity:
For example, the substitution \(u = a\,\tan\,\theta\) leads to the following formula:
Similarly, the substitution \(u = a\,\sec\,\theta\) yields this formula:
The proof of each formula requires this result from Example
Example \(\PageIndex{1}\): intparts7
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Solution
in Section 6.1:
The above substitutions can be used even if no square roots are present.
Example \(\PageIndex{1}\): trigsub2
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Solution
Evaluate \(~\displaystyle\int \frac{\dx}{(1 + x^2)^2}~\).
Solution: Notice that this integral cannot be evaluated by using the Power Formula with the substitution \(u=1+x^2\) (why?). Integration by parts does not look promising, either. So try a trigonometric substitution. The integrand contains a term of the form \(a^2 + u^2\) (with \(a=1\) and \(u=x\)), so use the substitution \(x=\tan\,\theta\). Then \(\dx = \sec^2 \theta\,\dtheta\) and so
\[\begin{aligned} \int \frac{\dx}{(1 + x^2)^2} ~&=~ \int \frac{\sec^2 \theta~\dtheta}{(1 + \tan^2 \theta)^2}\
\[6pt] &=~ \int \frac{\sec^2 \theta\,\dtheta}{(\sec^2 \theta)^2}\
\[6pt] &=~ \int \frac{\dtheta}{\sec^2 \theta^2}\
\[6pt] &=~ \int \cos^2 \theta~\dtheta\
\[6pt] &=~ \int \frac{1 + \cos\,2\theta}{2}\,\dtheta\
\[6pt] &=~ \frac{\theta}{2} ~+~ \frac{1}{4}\,\sin\,2\theta ~+~ C\
\[6pt] &=~ \frac{\theta}{2} ~+~ \frac{1}{2}\,\sin\,\theta\;\cos\,\theta ~+~ C\end{aligned} \nonumber \]
by the trigonometric double-angle identity \(\sin\,2\theta = 2\,\sin\,\theta\;\cos\,\theta\). The simplest way to get expressions for \(\sin\,\theta\) and \(\cos\,\theta\) in terms of \(x\) is to draw a right triangle with an angle \(\theta\) such that \(\tan\,\theta = x = \frac{x}{1}\), as in the drawing on the right. The hypotenuse must then be \(\sqrt{1+x^2}\) (by the Pythagorean Theorem), which makes it easy to read off the values of \(\sin\,\theta\) and \(\cos\,\theta\):
\[\sin\,\theta ~=~ \frac{x}{\sqrt{1+x^2}} \qquad\text{and}\qquad \cos\,\theta ~=~ \frac{1}{\sqrt{1+x^2}} \nonumber \]
Since \(\theta = \tan^{-1} x\), putting the integral back in terms of \(x\) yields:
\[\begin{aligned} \int \frac{\dx}{(1 + x^2)^2} ~&=~ \frac{1}{2}\,\tan^{-1} x ~+~ \frac{1}{2}\,\frac{x}{\sqrt{1+x^2}}\;\frac{1}{\sqrt{1+x^2}} ~+~ C\
\[6pt] &=~ \frac{1}{2}\,\tan^{-1} x ~+~ \frac{x}{2\,(1 + x^2)} ~+~ C\end{aligned} \nonumber \]
Note: An alternative method for getting \(\sin\,\theta\) and \(\cos\,\theta\) in terms of \(x\) would be to put \(\tan\,\theta = x\) in the identity \(\sec^2 \theta = 1 + \tan^2 \theta\) to solve for \(\cos\,\theta\), then use the identity \(\sin^2 \theta = 1 - \cos^2 \theta\) to solve for \(\sin\,\theta\).
By completing the square, quadratic expressions in \(x\) can be put in one of the forms \(a^2 \pm u^2\) or \(u^2 - a^2\), enabling the use of the corresponding trigonometric substitution.
Example \(\PageIndex{1}\): trigsub3
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Solution
Evaluate \(~\displaystyle\int \frac{\dx}{(4x^2 + 8x - 5)^{3/2}}~\).
Solution: This integral cannot be evaluated by using the Power Formula, so try a trigonometric substitution. Complete the square on the expression \(4x^2 + 8x - 5\):
\[4x^2 ~+~ 8x ~-~ 5 ~=~ 4\,(x^2 + 2x) ~-~ 5 ~=~ 4\,(x^2 + 2x + 1) ~-~ 5 ~-~ 4 ~=~ 4\,(x + 1)^2 ~-~ 9 \nonumber \]
This expression is now of the form \(u^2 - a^2\) for \(u=2\,(x+1)\) and \(a=3\). Use the substitution \(u=a\,\sec\,\theta\), which means \(2\,(x+1) = 3\,\sec\,\theta\). Then \(2\,\dx = 3\,\sec\,\theta\;\tan\,\theta\;\dtheta\) and so:
\[\begin{aligned} \int \frac{\dx}{(4x^2 + 8x - 5)^{3/2}} ~&=~ \int \frac{\dx}{(4\,(x + 1)^2 ~-~ 9)^{3/2}}\
\[6pt] &=~ \frac{3}{2}\,\int \frac{\sec\,\theta\;\tan\,\theta~\dtheta}{(9\,\sec^2 \theta ~-~ 9)^{3/2}} ~=~ \frac{3}{2}\,\int \frac{\sec\,\theta\;\tan\,\theta~\dtheta}{(9\,(\sec^2 \theta ~-~ 1))^{3/2}}\
\[6pt] &=~ \frac{1}{18}\,\int \frac{\sec\,\theta\;\tan\,\theta~\dtheta}{\tan^3 \theta} ~=~ \frac{1}{18}\,\int \frac{\sec\,\theta~\dtheta}{\tan^2 \theta} ~=~ \frac{1}{18}\,\int \frac{\cos\,\theta~\dtheta}{\sin^2 \theta}\
\[6pt] &=~ \frac{1}{18}\,\int \csc\,\theta\;\cot\,\theta~\dtheta ~=~ -\frac{1}{18}\,\csc\,\theta ~+~ C\end{aligned} \nonumber \]
To get an expression for \(\csc\,\theta\) in terms of \(x\), draw a right triangle with an angle \(\theta\) such that \(\sec\,\theta = \frac{2(x+1)}{3}\), as in the drawing on the right. The side opposite \(\theta\) must then be \(\sqrt{4x^2 + 8x - 5}\) (by the Pythagorean Theorem), and hence:
\[\csc\,\theta ~=~ \frac{2(x+1)}{\sqrt{4x^2 + 8x - 5}} \nonumber \]
Putting the integral back in terms of \(x\) yields:
\[\begin{aligned} \int \frac{\dx}{(4x^2 + 8x - 5)^{3/2}} ~&=~ -\frac{1}{18}\,\frac{2(x+1)}{\sqrt{4x^2 + 8x - 5}} ~+~ C\
\[6pt] &=~ -\frac{x+1}{9\,\sqrt{4x^2 + 8x - 5}} ~+~ C\end{aligned} \nonumber \]
Note: Trigonometric identities could have been used to obtain \(\csc\,\theta\) by knowing \(\sec\,\theta\).
The following integrals from Section 5.4 might be helpful for the exercises:
[sec6dot3]
For Exercises 1-16, evaluate the given integral.
4
\(\displaystyle\int \sqrt{9 + 4x^2}~\dx\)
\(\displaystyle\int \sqrt{2 - 3x^2}~\dx\)
\(\displaystyle\int \sqrt{4x^2 - 9}~\dx\)
\(\displaystyle\int \sqrt{x^2 + 2x + 10}~\dx\)
4
\(\displaystyle\int \frac{\sqrt{1 - x^2}}{x^2}~\dx\)
\(\displaystyle\int \frac{x^2~\dx}{\sqrt{x^2 - 9}}\vphantom{\displaystyle\int \frac{\sqrt{1 - x^2}}{x^2}}\)
\(\displaystyle\int \frac{\dx}{x\,\sqrt{1 + x^2}}\vphantom{\displaystyle\int \frac{\sqrt{1 - x^2}}{x^2}}\)
\(\displaystyle\int \frac{\dx}{x^2\,\sqrt{a^2 + x^2}}~~~(a > 0)\vphantom{\displaystyle\int \frac{\sqrt{1 - x^2}}{x^2}}\)
4
\(\displaystyle\int \frac{x^3~\dx}{\sqrt{x^2 + 4}}\vphantom{\displaystyle\int \frac{x^2~\dx}{\sqrt{a^2 - x^2}}}\)
\(\displaystyle\int \frac{dx}{(4x^2 - 9)^{3/2}}\vphantom{\displaystyle\int \frac{x^2~\dx}{\sqrt{a^2 - x^2}}}\)
\(\displaystyle\int \frac{dx}{(9 + 4x^2)^2}\vphantom{\displaystyle\int \frac{x^2~\dx}{\sqrt{a^2 - x^2}}}\)
\(\displaystyle\int \frac{x^2~\dx}{\sqrt{a^2 - x^2}}~~~(a > 0)\)
4
\(\displaystyle\int \frac{x^3~\dx}{\sqrt{9 - x^2}}\vphantom{\displaystyle\int \frac{\sqrt{4 - x^2}}{x}}\)
\(\displaystyle\int \frac{\sqrt{4 - x^2}}{x}~\dx\)
\(\displaystyle\int \frac{(x - 4)~\dx}{\sqrt{-9x^2 + 36x - 32}}\vphantom{\displaystyle\int \frac{\sqrt{4 - x^2}}{x}}\)
\(\displaystyle\int \frac{dx}{(4x^2 + 16x + 15)^{3/2}}\vphantom{\displaystyle\int \frac{\sqrt{4 - x^2}}{x}}\)
Prove formula ([eqn:sqrta2u2sin]) directly by using the substitution \(u=a \sin\,\theta\).
2
Prove formula ([eqn:sqrta2u2tan]).
Prove formula ([eqn:sqrtu2a2sec]).
Show that using the substitution \(u=a \cot\,\theta\) to evaluate the integral \(\int \sqrt{a^2 + u^2}\,\dx~\) leads to an antiderivative equivalent to the one in formula ([eqn:sqrta2u2tan]).
Show that using the substitution \(u=a \csc\,\theta\) to evaluate the integral \(\int \sqrt{u^2 - a^2}\,\dx~\) leads to an antiderivative equivalent to the one in formula ([eqn:sqrtu2a2sec]). [[1.]]
The integrals \(~\displaystyle\int \frac{\dx}{\sqrt{x^2 \pm a^2}}~\) can be evaluated without the use of trigonometric substitutions, by using differentials:
- For \(u^2 = x^2 \pm a^2\), show that
\[\frac{\dx}{u} ~=~ \frac{d\,(x+u)}{x+u} ~. \nonumber \]
- Integrate both sides of the result from part (a).
Note: In general, many integrals involving \(\sqrt{x^2 \pm a^2}\) can be handled with a similar manipulation of differentials, with varying complexity.
According to Newtonian physics the path of a photon grazing the surface of the Sun should be deflected by the Sun’s gravitational field by an angle \(\theta\), given approximately by
\[\theta ~=~ \ABS{\frac{2\,GMR}{c^2}\,\int_{\infty}^0 \frac{\dy}{(R^2 + y^2)^{3/2}}} \nonumber \]
where \(c= 2.998 \times 10^8\) m/s is the speed of light, \(G = 6.67 \times 10^{-11}\) N/m2/kg2 is the gravitational constant, \(M = 1.99 \times 10^{30}\) kg is the mass of the Sun, and \(R = 6.96 \times 10^{8}\) m is the radius of the Sun. Show that
\[\theta ~=~ \frac{2\,GM}{c^2 R} ~=~ 4.24 \times 10^{-6}~\text{radians} ~=~ 2.43 \times 10^{-4}~\text{degrees} ~\approx~ 0.875~\text{seconds of arc,} \nonumber \]
where 1 second of arc \(= 1/3600\) of 1 degree.3
