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Calculus

Basic Trigonometric Integrals

The trigonometric integrals of the six functions are

\begin{align} \int\sin u \,du&=-\cos u+C\\ \int\cos u \,du&=\sin u+C \\ \int\sec^2 u \,du&=\tan u+C \\ \int\csc^2 u \,du&=-\cot u+C \\ \int\sec u\tan u \,du&=\sec u+C \\ \int\csc u\cot u \,du&=-\csc u+C\end{align}

Example 1: Integrate \[\int 0.3 \sec^2 3\theta \,d\theta\]

Solution: From the antiderivatives above, we know the integral of \(\sec^2 u\) turns into \(\tan u\).

Let \(u=3\theta\), then \(du=3\,d\theta\)

\begin{align} &0.3\int \sec^2 u \left(\frac{du}{3}\right) \\ =& \frac{0.3}{3} \tan u+C \\ =& \frac{1}{10} \tan 3\theta+C  \end{align}


Example 2: Integrate \[\int \left(\cos^2 4x-\sin^2 4x\right)dx\]

Solution: We want to simplify the expression inside the integral by using the double angle identity \(\cos 2u = \cos^2 u-\sin^2 u\)

\begin{align} \int \left(\cos^2 4x-\sin^2 4x\right)dx &= \int \cos 2(4x) dx \\ &= \int \cos 8x dx \\ &= \frac{\sin 8x}{8} +C \end{align}

Integrals of Tangent and Reciprocal Trigonometric Functions

We know that \[\tan u = \frac{\sin u}{\cos u}\]

Thus,

\[\int \tan u \,du = \int \frac{\sin u}{\cos u}du\]

We can integrate using the logarithmic form by letting \(x=\cos u\). Thus, \(dx=-\sin u \,du\)

\begin{align} &\int \frac{\sin u}{\cos u}du \\ =& -\int \frac{1}{x}dx \\ =& -\ln|x|+C \\ =& -\ln|\cos u|+C \end{align}

By a similar method we can show that \[\int \cot u \,du=\ln|\sin u|+C\]


For the integral of \(\sec u\,du\), we want to multiply the top and bottom of the function by \(\sec u + \tan u\).

Not that the derivative of \(\sec u + \tan u\) is \(\left(\sec u\tan u + \sec^2 u\right)\)

Thus,

\begin{align} \int \sec u\,du &= \int \frac{\sec u(\sec u + \tan u)du}{\sec u + \tan u} \\&= \int \frac{\sec^2 u + \sec u \tan u}{\sec u + \tan u}du \\ &= \ln|\sec u + \tan u|+C \end{align}

By multiplying the top and bottom by \(\csc u - \cot u\), we can prove that 

\[\csc u \,du=\ln|\csc u - \cot u|+C\]

Example 1: Solve

\[\int_{0.5}^{1} x^2\cot x^3\,dx\]

Solution: Let \(u=x^3\), then \(du=3x^2\,dx\)

\begin{align} \int_{0.5}^{1} x^2\cot x^3\,dx &= \int_{0.5}^{1} \cot u\left(\frac{du}{3}\right) \\ &= \left(\ln|\sin u|\right)_{0.5}^{1} \\ \left(\ln|\sin x^3|\right)_{0.5}^{1} \\ &= \ln|\sin (1)^3 - \ln|\sin (0.5)^3 = 2.079\end{align}


Example 2: Find the volume generated by revolving the region bounded by \(y=\sec x\), \(x=0\), \(x=\frac{\pi}{3}\), and \(y=0\) about the x-axis.

Solution: The volume is given by the formula \[V=\int_{a}^{b} \pi[f(x)]^2dx\]

\begin{align} V&=\int_{0}^{\frac{\pi}{3}} \pi[\sec x]^2dx \\ &=\pi\left(\ln|\sec x+\tan x|\right)_{0}^{\frac{\pi}{3}} \\&=4.137 \end{align}

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Designed by Matthew Cheung. This work is licensed under a Creative Commons Attribution 4.0 International License.

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