Tag: Indefinite Integrals

Find indefinite integrals of powers of sin

by RoRi

October 21, 2015

From the previous two exercises (here and here) we know

$\int \sin^2 x \, dx = \frac{1}{2}x - \frac{1}{4} \sin (2x),$

and

$\int \sin^n x \, dx = - \frac{\sin^{n-1} x \cos x}{n} + \frac{n-1}{n} \int \sin^{n-2} x \, dx.$

Use these results to establish the following formulas:

$\displaystyle{ \int \sin^3 x \, dx = -\frac{3}{4} \cos x + \frac{1}{12} \cos (3x)}$ .
$\displaystyle{ \int \sin^4 x \, dx = \frac{3}{8} x - \frac{1}{4} \sin (2x) + \frac{1}{32} \sin (4x)}$ .
$\displaystyle{ \int \sin^5 x \, dx = -\frac{5}{8} \cos x + \frac{5}{48} \cos (3x) - \frac{1}{80} \cos (5x)}$ .

( Note: There is an error in my edition of the book for part (c). It requests we prove $\int \sin^5 x \, dx = -\frac{5}{8} x + \frac{5}{48} \cos (3x) - \frac{1}{80} \cos(5x)$ , omitting the $\cos x$ in the first term of the result. This is just an error in the book since the formula as stated is false.)

Using the second formula, and the triple angle identity for cosine ( $\cos (3x) = 4 \cos^3 x - 3 \cos x$ which implies $\cos^3 x = \frac{1}{4} \cos (3x) + \frac{3}{4} \cos x$ , derived in this exercise), we have
$\begin{align*} \int \sin^3 x \, dx &= - \frac{\sin^2 x \cos x}{3} + \frac{2}{3} \int \sin x \, dx \\[9pt] &= -\frac{1}{3} \left( \sin^2 x \cos x + 2 \cos x ) \\[9pt] &= -\frac{1}{3} \left( (1 - \cos^2 x) \cos x + 2 \cos x ) \\[9pt] &= -\frac{1}{3} \cos x (3 - \cos^2 x) \\[9pt] &= - \cos x + \frac{1}{3} \cos^3 x \\[9pt] &= - \cos x + \frac{1}{3} \left(\frac{1}{4} \cos (3x) + \frac{3}{4} \cos x \right) \\[9pt] &= -\frac{3}{4} \cos x + \frac{1}{12} \cos (3x). \qquad \blacksquare \end{align*}$
Using the second formula above and then the first we have (and also recalling the trig identity $\cos (2x) = 1 - 2 \sin^2 x$ which implies $\sin^2 x = \frac{1}{2} - \frac{1}{2} \cos (2x)$ ),
$\begin{align*} \int \sin^4 x \, dx &= - \frac{\sin^3 x \cos x}{4} + \frac{3}{4} \int \sin^2 x \, dx \\[9pt] &= -\frac{1}{4} (\sin^3 x \cos x) + \frac{3}{4} \left( \frac{1}{2} x - \frac{1}{4} \sin (2x) \right) \\[9pt] &= -\frac{1}{4} (\sin^2 x \sin x \cos x) + \frac{3}{8} x - \frac{3}{16} \sin (2x) \\[9pt] &= -\frac{1}{8} (\sin^2 x \sin (2x)) + \frac{3}{8} x - \frac{3}{16} \sin (2x) \\[9pt] &= -\frac{1}{16} (\sin (2x) - \sin (2x) \cos (2x)) + \frac{3}{8} x - \frac{3}{16} \sin (2x) \\[9pt] &= \frac{3}{8} - \frac{1}{4} \sin (2x) + \frac{1}{32} \sin (4x). \qquad \blacksquare \end{align*}$
Using the second formula above and then part (a) we have
$\begin{align*} \int \sin^5 x \, dx &= -\frac{\sin^4 x \cos x}{5} + \frac{4}{5} \int \sin^3 x \, dx \\[9pt] &= -\frac{1}{5} \left( (1-\cos^2 x)(1-\cos^2 x) \cos x) + \frac{4}{5} \left(-\frac{3}{4} \cos x + \frac{1}{12} \cos (3x) \right) \\[9pt] &= -\frac{1}{5} (\cos x - 2 \cos^3 x + \cos^5 x) + \frac{1}{16} (10 \cos x + 5 \cos (3x) + \cos (5x)) \\[9pt] & \qquad \qquad - \frac{3}{5} \cos x + \frac{1}{15} \cos (3x) \\[9pt] &= -\frac{1}{5} \left( \frac{1}{8} \cos x - \frac{3}{16} \cos (3x) + \frac{1}{16} \cos (5x) \right) - \frac{3}{5}\cos x + \frac{1}{15} \cos (3x) \\[9pt] &= -\frac{5}{8} \cos x + \frac{5}{48} \cos (3x) - \frac{1}{80} \cos (5x). \qquad \blacksquare \end{align*}$

Find a primitive and apply the second fundamental theorem of calculus

by RoRi

October 14, 2015

Let

$f(x) = x^{\frac{4}{3}} - 5 \cos x.$

Find a function $P(x)$ such that $P'(x) = f(x)$ (i.e., a primitive of $f$ ). Use the second fundamental theorem of calculus to evaluate

$\int_a^b f(x) \, dx.$

The function

$P(x) = \frac{3}{7} x^{\frac{7}{3}} - 5 \sin x$

is a primitive of $f$ since

$P'(x) = x^{\frac{4}{3}} - 5 \cos x = f(x).$

Then, by the second fundamental theorem of calculus we have

$\begin{align*} \int_a^b f(x) \, dx &= P(b) - P(a) \\ &= \frac{3}{7} \left( b^{\frac{7}{3}} - a^{\frac{7}{3}} \right) - 5 (\sin b - \sin a). \end{align*}$

Find a primitive and apply the second fundamental theorem of calculus

by RoRi

October 14, 2015

Let

$f(x) = 3 \sin x + 2x^5.$

Find a function $P(x)$ such that $P'(x) = f(x)$ (i.e., a primitive of $f$ ). Use the second fundamental theorem of calculus to evaluate

$\int_a^b f(x) \, dx.$

The function

$P(x) = -3 \cos x + \frac{1}{3}x^6$

is a primitive of $f$ since

$P'(x) = 3 \sin x + 2 x^5 = f(x).$

Then, by the second fundamental theorem of calculus we have

$\begin{align*} \int_a^b f(x) \, dx &= P(b) - P(a) \\ &= -3 (\cos b - \cos a) + \frac{1}{3}(b^6 - a^6). \end{align*}$

Find a primitive and apply the second fundamental theorem of calculus

by RoRi

October 14, 2015

Let

$f(x) = 2x^{\frac{1}{3}} - x^{-\frac{1}{3}} \qquad x> 0.$

Find a function $P(x)$ such that $P'(x) = f(x)$ (i.e., a primitive of $f$ ). Use the second fundamental theorem of calculus to evaluate

$\int_a^b f(x) \, dx.$

The function

$P(x) = \frac{3}{2} x^{\frac{4}{3}} - \frac{3}{2} x^{\frac{2}{3}}$

is a primitive of $f$ since

$P'(x) = 2x^{\frac{1}{3}} - x^{-\frac{1}{3}} = f(x).$

Then, by the second fundamental theorem of calculus we have

$\begin{align*} \int_a^b f(x) \, dx &= P(b) - P(a) \\ &= \frac{3}{2} \left( b^{\frac{4}{3}} - a^{\frac{4}{3}} + b^{\frac{2}{3}} - a^{\frac{2}{3}} \right). \end{align*}$

Find a primitive and apply the second fundamental theorem of calculus

by RoRi

October 14, 2015

Let

$f(x) = \frac{2x^2 - 6x + y}{2\sqrt{x}}, \qquad x > 0.$

Find a function $P(x)$ such that $P'(x) = f(x)$ (i.e., a primitive of $f$ ). Use the second fundamental theorem of calculus to evaluate

$\int_a^b f(x) \, dx.$

First, let’s rewrite $f(x)$ ,

$\begin{align*} f(x) &= \frac{2x^2 - 6x + y}{2 \sqrt{x}} \\ &= x^{\frac{3}{2}} - 3x^{\frac{1}{2}} + \frac{7}{2} x^{-\frac{1}{2}} \end{align*}$

Then,

$P(x) = \frac{2}{5} x^{\frac{5}{2}} - 2x^{\frac{3}{2}} + 7 x^{\frac{1}{2}}$

is a primitive of $f$ since

$P'(x) = x^{\frac{3}{2}} - 3x^{\frac{1}{2}} + \frac{7}{2} x^{-\frac{1}{2}} = f(x).$

Then, by the second fundamental theorem of calculus we have

$\begin{align*} \int_a^b f(x) \, dx &= P(b) - P(a) \\ &= \frac{2}{5} \left( b^{\frac{5}{2}} - a^{\frac{5}{2}} \right) - 2 \left( b^{\frac{3}{2}} - a^{\frac{3}{2}} \right) + 7 \left( b^{\frac{1}{2}} - a^{\frac{1}{2}} \right). \end{align*}$

Find a primitive and apply the second fundamental theorem of calculus

by RoRi

October 14, 2015

Let

$f(x) = \sqrt{2x} + \sqrt{\frac{1}{2} x}.$

Find a function $P(x)$ such that $P'(x) = f(x)$ (i.e., a primitive of $f$ ). Use the second fundamental theorem of calculus to evaluate

$\int_a^b f(x) \, dx.$

First, let’s rewrite $f(x)$ ,

$\begin{align*} f(x) &= \sqrt{2x} + \sqrt{\frac{1}{2} x} \\ &= \left( \sqrt{2} + \frac{1}{\sqrt{2}} \right) x^{\frac{1}{2}} \\ &= \frac{3\sqrt{2}}{2} \sqrt{x}. \end{align*}$

Then,

$P(x) = \sqrt{2} x^{\frac{3}{2}}$

is a primitive of $f$ since

$P'(x) = \frac{3\sqrt{2}}{2} x^{\frac{1}{2}} = f(x).$

Then, by the second fundamental theorem of calculus we have

$\begin{align*} \int_a^b f(x) \, dx &= P(b) - P(a) \\ &= \sqrt{2} (b^{\frac{3}{2}} - a^{\frac{3}{2}} ). \end{align*}$

Find a primitive and apply the second fundamental theorem of calculus

by RoRi

October 14, 2015

Let

$f(x) = \left( 1 + \sqrt{x} \right)^2 \qquad x> 0.$

Find a function $P(x)$ such that $P'(x) = f(x)$ (i.e., a primitive of $f$ ). Use the second fundamental theorem of calculus to evaluate

$\int_a^b f(x) \, dx.$

First, let’s expand $f(x)$ ,

$f(x) = \left( 1 + \sqrt{x} \right)^2 = x + 2 x^{\frac{1}{2}} + 1.$

Then,

$P(x) = \frac{1}{2}x^2 + \frac{4}{3} x^{\frac{3}{2}} + x$

is a primitive of $f$ since

$P'(x) = x + 2x^{\frac{1}{2}} + 1 = \left( 1 + \sqrt{x} \right)^2 = f(x).$

Then, by the second fundamental theorem of calculus we have

$\begin{align*} \int_a^b f(x) \, dx &= P(b) - P(a) \\ &= \frac{1}{2} (b^2-a^2) + \frac{4}{3} \left( b^{\frac{3}{2}} - a^{\frac{3}{2}} \right) + (b-a). \end{align*}$