Wize University Calculus 1 Textbook > Integrals

Antiderivatives of Trig Functions

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Antiderivatives (Indefinite Integral) of Trig Functions

Let kk be a nonzero constant. Then we have the following indefinite integral rules:
∫sin⁡(kx)dx=−1kcos⁡(kx)+C∫cos⁡(kx)dx=1ksin⁡(kx)+C∫sec⁡2(kx)dx=1ktan⁡(kx)+C∫sec⁡(kx)tan⁡(kx)dx=1ksec⁡(kx)+C∫csc⁡2(kx)dx=−1kcot⁡(kx)+C∫csc⁡(kx)cot⁡(kx)dx=−1kcsc⁡(kx)+C\begin{array}{|l|l|} \hline\\ \displaystyle\int\sin{(kx)}dx=-\dfrac{1}{k}\cos{( kx)}+C & \displaystyle\int\cos{ (kx)}dx=\dfrac{1}{k}\sin{( kx)}+C \\ \\ \hline\\ \displaystyle\int\sec^2{(kx)}dx=\dfrac{1}{k}\tan (kx)+C &\displaystyle\int\sec (kx)\tan (kx)dx=\dfrac{1}{k}\sec (kx)+C \\ \\ \hline\\ \displaystyle\int\csc^2(kx)dx=-\dfrac{1}{k}\cot (kx)+C & \displaystyle\int\csc (kx)\cot (kx)dx=-\dfrac{1}{k}\csc( kx)+C \\ \\ \hline \end{array}
Additionally
∫tan⁡xdx=−ln⁡∣cos⁡x∣+C∫cot⁡xdx=ln⁡∣sin⁡x∣∫sec⁡xdx=ln⁡∣sec⁡x+tan⁡x∣+C∫csc⁡xdx=−ln⁡∣csc⁡x+cot⁡x∣+C\begin{array}{|l|l|} \hline\\ \displaystyle \int\tan xdx=-\ln|\cos x|+C & \displaystyle \int \cot xdx=\ln|\sin x| \\ \\ \hline\\ \displaystyle\int \sec x dx=\ln|\sec x+\tan x| + C &\displaystyle \int \csc x dx=-\ln|\csc x+\cot x|+C \\ \\ \hline \end{array}

Wize Tip
These last four indefinite integrals can be found using future techniques of integration and are not always required to be memorized.

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Example: Trig Indefinite Integrals

Evaluate the following indefinite integral

∫1+tan⁡2θ dθ\displaystyle \int\sqrt{1+\tan^2\theta} \ d\theta

∫1+tan⁡2θ dθ\displaystyle \int\sqrt{1+\tan^2\theta} \ d\theta

=∫sec⁡2θ dθ=\displaystyle \int\sqrt{\sec^2\theta} \ d\theta

=∫sec⁡θ dθ=\displaystyle \int\sec \theta \ d\theta

=ln⁡∣sec⁡θ+tan⁡θ∣+C\displaystyle=\ln|\sec \theta+\tan \theta| + C
Evaluate the following indefinite integral

∫(x2+2sin⁡x−3cos⁡x)dx\displaystyle \int( x^2+2\sin x-3\cos x )dx
Evaluate the following indefinite integral

∫(sec⁡2x)(cos⁡22x)dx\displaystyle \int (\sec 2x)( \cos^2 2x)dx

Extra Practice