Solution. Let \(\hspace {0.2cm} u = \cos x \hspace {0.2cm}\) so that \(\hspace {0.2cm} du = -\sin x dx\)

\begin {align*} \int \sin ^3x\cos ^{3/4}x\hspace {0.2cm}dx & = \int \sin ^2x\cos ^{3/4}x\sin x dx\\\\ & = \int \big (u^2 - 1\big )\hspace {0.2cm} u^{3/4}\hspace {0.2cm} du = \int \big (u^{11/4} - u^{3/4}\big )du\\\\ & = \frac {4}{15}u^{15/4} - \dfrac {4}{7}u^{7/4} + c\\\\ & = \frac {4}{15}\cos ^{15/4} x - \dfrac {4}{7}\cos ^{7/4}x + c \end {align*}

\[\therefore \hspace {0.5cm}\int \sin ^3x\cos ^{3/4}x\hspace {0.2cm}dx = \frac {4}{15}\cos ^{15/4} x - \dfrac {4}{7}\cos ^{7/4}x + c\]

If both \(m\) and \(n\) are even integers the integral can be evaluated by first altering the powers using the identities:

\(\sin ^2\theta = \dfrac {1}{2}\big (1 - \cos 2\theta \big )\)

\(\cos ^2\theta = \dfrac {1}{2}\big (1 + \cos 2\theta \big )\)

Solution. \[\sin ^2\theta = \dfrac {1}{2}\big (1 - \cos 2\theta \big )\]

\begin {align*} \cos ^4x & = \big (\cos ^2x\big )^2 = \Big [\dfrac {1}{2}\big (1 + \cos 2x\big )\Big ]^2\\ & = \dfrac {1}{4}\Big [1 + 2\cos 2x + \cos ^22x\Big ] = \dfrac {1}{4}\Big [1 + 2\cos 2x + \dfrac {1}{2}\big (1 + \cos 4x\big )\Big ]\\\\ & = \dfrac {1}{4}\Big [1 + 2\cos 2x + \dfrac {1}{2} + \dfrac {1}{2}\cos 4x\Big ]\\\\ & = \dfrac {1}{4}\Big [\dfrac {3}{2} + 2\cos 2x + \dfrac {1}{2}\cos 4x\Big ]\\ \end {align*}

\begin {align*} \sin ^2x\cos ^4x & = \dfrac {1}{2}\Big [1 - \cos 2x\Big ]\cdot \dfrac {1}{4}\Big [\dfrac {3}{2} + 2\cos 2x + \dfrac {1}{2}\cos 4x\Big ]\\\\ & = \dfrac {1}{8}\Big [\dfrac {1}{2}- \dfrac {1}{2}\cos 4x - \dfrac {1}{2}\cos 6x\Big ] \end {align*}

\begin {align*} \therefore \hspace {0.5cm} \int \sin ^2x\cos ^4x\hspace {0.2cm}dx & = \dfrac {1}{16}\int \big (1 - \cos 4x - \cos 6x\big )\hspace {0.2cm}dx\\\\ & = \dfrac {1}{16}\Big [x - \dfrac {1}{4}\sin 4x - \dfrac {1}{6}\sin 6x\Big ] + c\\ \end {align*}

Integrals of the form \(\int \tan ^m x\sec ^n x\,dx\)

If \(n\) is even substitute \(\hspace {0.2cm} u = \tan x\)

If \(m\) is odd substitute \(\hspace {0.2cm} u = \sec x\)

Solution. Let \(\hspace {0.2cm} u = \tan x\hspace {0.3cm}\) then \(\hspace {0.3cm} du = \sec ^2xdx\hspace {0.3cm}\) Thus

\begin {align*} \int \tan ^6x\sec ^4x\hspace {0.2cm} dx & = \int \tan ^6x\underbrace {\sec ^2x}_{1 + \tan ^2x}\sec ^2x\hspace {0.1cm}dx\\\\ & = \int u^6\big (1 + u^2\big )du = \int \big (u^6 + u^8\big )du\\\\ & = \dfrac {1}{7}u^7 + \dfrac {1}{9}u^9 + c\\\\ & = \dfrac {1}{7}\tan ^7x + \dfrac {1}{9}\tan ^9x + c\\ \end {align*}

\[\int \tan ^6x\sec ^4x\hspace {0.2cm} dx = \dfrac {1}{7}\tan ^7x + \dfrac {1}{9}\tan ^9x + c\]

Solution.

Let \(\hspace {0.2cm} x = a\tan \theta \hspace {0.2cm}\) so that \(\hspace {0.2cm} dx = a\sec ^2\theta \hspace {0.2cm}d\theta \hspace {0.3cm}\) Thus

\begin {align*} \int \dfrac {dx}{\sqrt {a^2 + x^2}} & = \int \dfrac {a\sec ^2\theta \hspace {0.2cm}d\theta }{\sqrt {a^2 + a^2\tan ^2\theta }}\\\\ & = \int \dfrac {a\sec ^2\theta \hspace {0.2cm} d\theta }{\sqrt {a^2\big (1 + \tan ^2\theta \big )}} = \int \dfrac {\sec ^2\theta }{\sec \theta }d\theta \\\\ & = \int \sec \theta \hspace {0.2cm}d\theta \\\\ & = \int \dfrac {\sec \theta \big (\tan \theta + \sec \theta \big )\hspace {0.2cm}d\theta }{\tan \theta + \sec \theta } = \int \dfrac {\tan \theta \sec \theta + \sec ^2\theta }{\tan \theta + \sec \theta }\hspace {0.2cm}d\theta \\\\ & = \ln \left |\tan \theta + \sec \theta \right | + c\\\\ & = \ln \left |\dfrac {x}{a} + \dfrac {\sqrt {x^2 + a^2}}{a}\right | + c\\ \end {align*}

\[\text {Note:}\hspace {0.5cm} \int \sec \theta \hspace {0.2cm}d\theta = \ln \left |\tan \theta + \sec \theta \right | + c\]

Solution.

Let \(\hspace {0.2cm} x = 3\sin \theta \hspace {0.2cm}\). Then \(\hspace {0.2cm} dx = 3\cos \theta \hspace {0.2cm}d\theta \hspace {0.2cm}\). Thus

\begin {align*} \int \dfrac {dx}{x^2\sqrt {9 - x^2}} & = \int \dfrac {3\cos \theta \hspace {0.2cm}d\theta }{\big (3\sin \theta \big )^2\sqrt {9 - 9\sin ^2\theta }} = \int \dfrac {\cos \theta }{3\sin ^2\theta \cdot 3 \sqrt {1 - \sin ^2\theta }}\hspace {0.2cm}d\theta \\\\ & = \dfrac {1}{9}\int \dfrac {1}{\sin ^2\theta }\hspace {0.2cm}d\theta \\\\ & = \dfrac {1}{9}\int \csc ^2\theta \hspace {0.2cm}d\theta \\\\ & = \dfrac {-1}{9}\cot \theta + c\\\\ & = \dfrac {-1}{9}\dfrac {\sqrt {9 - x^2}}{x} + c\\\\ \end {align*}

Solution.

Let \(\hspace {0.2cm} x = 2\sec \theta \hspace {0.2cm}\) so that \(\hspace {0.2cm} dx = 2\tan \theta \sec \theta \hspace {0.2cm}d\theta \)

\begin {align*} \int \dfrac {\sqrt {x^2 - 4}}{x}\hspace {0.2cm}dx & = \int \dfrac {\sqrt {4\sec ^2\theta - 4}}{2\sec \theta }\cdot 2\tan \theta \sec \theta \hspace {0.2cm} d\theta \\\\ & = 2 \int \tan ^2\theta \hspace {0.2cm} d\theta = 2\int \big (\sec ^2\theta - 1\big )d\theta \\\\ & = 2\Bigg [\int \sec ^2\theta d\theta - \int d\theta \Bigg ]\\\\ & = 2\tan \theta - 2\theta + c\\\\ & = \dfrac {2\sqrt {x^2 - 4}}{2} - 2 \sec ^{-1}\Big (\dfrac {x}{2}\Big ) + c=\sqrt {x^2 - 4} - 2\sec ^{-1}\Big (\dfrac {x}{2}\Big ) + c \end {align*}