Practice the most common integrals you need for Calculus 2.
Evaluate \(\displaystyle\int\cos x\,dx\)
1st: Recognize the basic form
The derivative of \(\sin x\) is \(\cos x\).
2nd: Integrate
\(\displaystyle\int\cos x\,dx = \sin x + C\)
Final answer
\(\sin x + C\)
Direct basic integrals
\(\displaystyle\int\sec^{2}x\,dx\)
1st: Recognize the form
The derivative of \(\tan x\) is \(\sec^{2}x\).
2nd: Integrate
\(\displaystyle\int\sec^{2}x\,dx = \tan x + C\)
Final answer
\(\tan x + C\)
\(\displaystyle\int e^{x}\,dx\)
1st: Recognize the form
The derivative of \(e^{x}\) is \(e^{x}\).
2nd: Integrate
\(\displaystyle\int e^{x}\,dx = e^{x} + C\)
Final answer
\(e^{x} + C\)
\(\displaystyle\int\dfrac{1}{x}\,dx\)
1st: Recognize the form
The derivative of \(\ln|x|\) is \(\dfrac{1}{x}\).
2nd: Integrate
\(\displaystyle\int\dfrac{1}{x}\,dx = \ln|x| + C\)
Why the absolute value?
\(\ln x\) is only defined for \(x > 0\), but \(\dfrac{1}{x}\) is defined for both positive and negative \(x\).
Using \(\ln|x|\) makes the antiderivative valid on both sides of zero:
• \(x > 0\) → \(\ln|x| = \ln x\)
• \(x < 0\) → \(\ln|x| = \ln(-x)\)
Final answer
\(\ln|x| + C\)
Requires an identity or simple rewrite
\(\displaystyle\int\tan^{2}x\,dx\)
1st: Use a Pythagorean identity
\(\tan^{2}x = \sec^{2}x – 1\)
2nd: Rewrite the integral
\(\displaystyle\int\tan^{2}x\,dx = \int(\sec^{2}x – 1)\,dx\)
3. Integrate term by term
\(\int\sec^{2}x\,dx – \int 1\,dx = \tan x – x + C\)
Final answer
\(\tan x – x + C\)
\(\displaystyle\int\sin^{2}x\,dx\)
1st: Use the power-reduction formula
\(\sin^{2}x = \dfrac{1 – \cos 2x}{2}\)
2nd: Rewrite the integral
\(\displaystyle\int\sin^{2}x\,dx = \int\dfrac{1 – \cos 2x}{2}\,dx = \dfrac{1}{2}\int(1 – \cos 2x)\,dx\)
3. Integrate
\(\dfrac{1}{2}\left(x – \dfrac{1}{2}\sin 2x\right) + C = \dfrac{x}{2} – \dfrac{1}{4}\sin 2x + C\)
Final answer
\(\dfrac{x}{2} – \dfrac{1}{4}\sin 2x + C\)
\(\displaystyle\int\sec x\tan x\,dx\)
1st: Recognize the form
The derivative of \(\sec x\) is \(\sec x\tan x\).
2nd: Integrate
\(\displaystyle\int\sec x\tan x\,dx = \sec x + C\)
Final answer
\(\sec x + C\)
\(\displaystyle\int\cos^{2}x\,dx\)
1st: Use the power-reduction formula
\(\cos^{2}x = \dfrac{1 + \cos 2x}{2}\)
2nd: Rewrite and integrate
\(\dfrac{1}{2}\int(1 + \cos 2x)\,dx = \dfrac{1}{2}\left(x + \dfrac{1}{2}\sin 2x\right) + C\)
Final answer
\(\dfrac{x}{2} + \dfrac{1}{4}\sin 2x + C\)
Ready to master basic integration?