QUESTION IMAGE
Question
quiz basic integration formulas
- $\int du =$ 2. $\int kdu =$ 3. $\int (du + dv) =$
- $\int u^n du =$ 5. $\int \frac{du}{u} =$ 6. $\int \sin udu =$
- $\int \cos udu =$ 8. $\int \sec^2 udu =$ 9. $\int \csc^2 udu =$
- $\int \sec u \tan udu =$ 11. $\int \csc u \cot udu =$ 12. $\int \tan udu =$
- $\int \cot udu =$ 14. $\int e^u du =$ 15. $\int a^u du =$
- $\int \frac{du}{\sqrt{a^2 - u^2}} =$ 17. $\int \frac{du}{a^2 + u^2} =$ 18. $\int \frac{du}{u\sqrt{u^2 - a^2}} =$
- $\int \sec x dx =$ 20. $\int \csc x dx =$
Step1: Recall basic integration formulas
- \(\int du=u + C\) (where \(C\) is the constant of integration. The integral of a differential \(du\) with respect to \(u\) is just \(u\) plus a constant)
- \(\int kdu=ku + C\) (by the constant - multiple rule of integration. If \(k\) is a constant, we can factor it out of the integral)
- \(\int(du + dv)=\int du+\int dv=u + v + C\) (by the sum rule of integration)
- \(\int u^{n}du=\frac{u^{n + 1}}{n+1}+C\), \(n
eq - 1\) (power rule for integration)
- \(\int\frac{du}{u}=\ln|u|+C\) (integral of the reciprocal function)
- \(\int\sin udu=-\cos u + C\) (derivative of \(-\cos u\) is \(\sin u\))
- \(\int\cos udu=\sin u + C\) (derivative of \(\sin u\) is \(\cos u\))
- \(\int\sec^{2}udu=\tan u + C\) (derivative of \(\tan u\) is \(\sec^{2}u\))
- \(\int\csc^{2}udu=-\cot u + C\) (derivative of \(-\cot u\) is \(\csc^{2}u\))
- \(\int\sec u\tan udu=\sec u + C\) (derivative of \(\sec u\) is \(\sec u\tan u\))
- \(\int\csc u\cot udu=-\csc u + C\) (derivative of \(-\csc u\) is \(\csc u\cot u\))
- \(\int\tan udu=\int\frac{\sin u}{\cos u}du=-\ln|\cos u|+C=\ln|\sec u|+C\) (using substitution \(t = \cos u\), \(dt=-\sin udu\))
- \(\int\cot udu=\int\frac{\cos u}{\sin u}du=\ln|\sin u|+C\) (using substitution \(t=\sin u\), \(dt = \cos udu\))
- \(\int e^{u}du=e^{u}+C\) (derivative of \(e^{u}\) is \(e^{u}\))
- \(\int a^{u}du=\frac{a^{u}}{\ln a}+C\), \(a>0,a
eq1\) (since \(\frac{d}{du}(\frac{a^{u}}{\ln a})=a^{u}\))
- \(\int\frac{du}{\sqrt{a^{2}-u^{2}}}=\sin^{- 1}(\frac{u}{a})+C\), \(|u|
- \(\int\frac{du}{a^{2}+u^{2}}=\frac{1}{a}\tan^{-1}(\frac{u}{a})+C\) (derivative of \(\tan^{-1}(x)\) is \(\frac{1}{1 + x^{2}}\), using substitution \(u=a\tan t\))
- \(\int\frac{du}{u\sqrt{u^{2}-a^{2}}}=\frac{1}{a}\sec^{-1}(\frac{|u|}{a})+C\), \(|u|>a>0\) (using substitution \(u = a\sec t\))
- \(\int\sec xdx=\ln|\sec x+\tan x|+C\) (multiply numerator and denominator by \(\sec x+\tan x\) and then use substitution)
- \(\int\csc xdx=\ln|\csc x-\cot x|+C\) (multiply numerator and denominator by \(\csc x-\cot x\) and then use substitution)
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- \(u + C\)
- \(ku + C\)
- \(u + v + C\)
- \(\frac{u^{n + 1}}{n + 1}+C\), \(n
eq-1\)
- \(\ln|u|+C\)
- \(-\cos u + C\)
- \(\sin u + C\)
- \(\tan u + C\)
- \(-\cot u + C\)
- \(\sec u + C\)
- \(-\csc u + C\)
- \(\ln|\sec u|+C\)
- \(\ln|\sin u|+C\)
- \(e^{u}+C\)
- \(\frac{a^{u}}{\ln a}+C\), \(a>0,a
eq1\)
- \(\sin^{-1}(\frac{u}{a})+C\), \(|u|
- \(\frac{1}{a}\tan^{-1}(\frac{u}{a})+C\)
- \(\frac{1}{a}\sec^{-1}(\frac{|u|}{a})+C\), \(|u|>a>0\)
- \(\ln|\sec x+\tan x|+C\)
- \(\ln|\csc x-\cot x|+C\)