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Integrating standard functions

Integrating standard functions

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Tutor: Bilal

Summary

Integrating standard functions

​​In a nutshell

It is possible to integrate functions other than those of the form axnax^n. Knowing some standard integrals will help solve problems quicker.



Standard integrals

Here is a list of integrals you need to know. In it, nn and CC are constants.​


function

integral

xnx^n​​

xn dx=1n+1xn+1+C,n1\int x^n \, dx = \dfrac{1}{n+1}x^{n+1}+C, n\ne 1​​

1x\dfrac1x​​

1x dx=lnx+C\int \dfrac1x \, dx = \ln \vert x\vert +C​​

exe^x​​

ex dx=ex+C\int e^x \, dx = e^x+C​​

sin(x)\sin(x)​​

sin(x) dx=cos(x)+C\int \sin(x) \, dx = -\cos(x)+C​​

cos(x)\cos(x)​​

cos(x) dx=sin(x)+C\int \cos(x) \, dx = \sin(x)+C​​

sec2(x)\sec^2 (x)​​

sec2(x) dx=tan(x)+C\int \sec^2(x) \, dx = \tan(x)+C​​

cosec2(x)\cosec^2(x)​​

cosec2(x) dx=cot(x)+C\int \cosec^2 (x) \, dx = -\cot(x)+C​​

sec(x)tan(x)\sec(x) \tan(x)​​

sec(x)tan(x) dx=sec(x)+C\int \sec(x) \tan(x) \, dx = \sec(x)+C​​

cosec(x)cot(x)\cosec(x)\cot(x)​​

cosec(x)cot(x) dx=cosec(x)+C\int \cosec(x)\cot(x) \, dx = -\cosec(x)+C​​


Note: The trigonometric integrals assume that xx is in radians, not degrees.


Example 1

Compute the following integrals:

i) 14x dx\int \dfrac{1}{4x} \, dx

ii) π23π2(4cot(x)sin(x)cos(x)) dx\int_\frac{\pi}{2}^\frac{3\pi}{2} \left(\dfrac{-4\cot(x)}{\sin(x)}-\cos(x) \right) \, dx


Part i):

Use the fact that kf(x) dx=kf(x) dx\int kf(x)\, dx = k\int f(x)\, dx to turn the integral into one of the standard integrals.

14x dx=14×1x dx=141x dx=14lnx+C\begin{aligned} \int\dfrac{1}{4x} \, dx &= \int \dfrac{1}{4} \times \dfrac1x \, dx\\&=\dfrac14 \int \dfrac1x \, dx\\&=\dfrac14 \ln \vert x\vert +C \end{aligned}​​


14x dx=14lnx+C\int \dfrac{1}{4x}\, dx= \underline{\dfrac14 \ln \vert x\vert +C}​​


Part ii):

Use the fact that (f(x)+g(x)) dx=f(x) dx+g(x) dx\int (f(x)+g(x))\, dx = \int f(x)\, dx + \int g(x)\, dx together with the procedure for definite integration.


π23π2(4cot(x)sin(x)cos(x)) dx=π23π2(4cot(x)×1sin(x)cos(x)) dx=π23π2(4cot(x)cosec(x)cos(x)) dx=4π23π2cot(x)cosec(x) dxπ23π2cos(x) dx=[4(cosec(x))(sin(x))]π23π2=[4cosec(x)sin(x)]π23π2=(4sin(3π2)sin(3π2))(4sin(π2)sin(π2))=(41(1))(41(1))=(3)(3)=6\begin{aligned} \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \left(\dfrac{-4\cot(x)}{\sin(x)}-\cos(x) \right) \, dx &= \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \left(-4\cot(x) \times\dfrac{1}{\sin(x)} - \cos(x)\right)\, dx\\&=\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \left(-4\cot(x)\cosec(x) -\cos(x)\right) \, dx\\&=-4\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \cot(x) \cosec(x) \, dx - \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \cos(x)\, dx\\&=\left[-4(-\cosec(x)) - (\sin(x))\right]_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\\&=\left[4\cosec(x) - \sin(x)\right]_{\frac{\pi}{2}}^{\frac{3\pi}{2}}\\&=\left(\dfrac{4}{\sin \left(\frac{3\pi}{2} \right)} - \sin\left(\frac{3 \pi}{2}\right)\right) - \left(\dfrac{4}{\sin \left(\frac{\pi}{2}\right)} - \sin\left(\frac{ \pi}{2}\right)\right)\\&=\left(\dfrac{4}{-1} - (-1)\right) - \left(\dfrac{4}{1} - (1)\right)\\&=(-3)-(3)\\&=-6 \end{aligned}​​


π23π2(4cot(x)sin(x)cos(x)) dx=6\int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \left(\dfrac{-4\cot(x)}{\sin(x)}-\cos(x) \right) \, dx= \underline{-6}​​



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