Integration is known as the inverse process of derivatives, also called anti-derivative. There are many techniques to find integrations. In this article, we will provide a few examples of integrations with detailed solutions.
Table of Contents
Integration Formulas
| Power rule of integration: ∫xn dx = $\dfrac{x^{n+1}}{n+1}$+C Integrations of constants: ∫k dx = kx+C ∫sin x dx = -cosx+C ∫cos x dx = sin x+C ∫sec2 x dx = tan x+C ∫cosec2 x dx = -cot x+C ∫sec x tan x dx = sec x+C ∫cosec x cot x dx = -cosec x+C ∫ex dx = ex+C ∫ax dx = $\dfrac{a^x}{\ln a}$+C; a(≠1)>0. ∫$\dfrac{dx}{x}$ = ln|x| +C |
Integration: Definition, Formulas, Properties, Examples
Integration Examples and Solutions
We will solve a few integrals here depending upon the type of the integrands. They will be categorized as follows:
Power Rule of Integration Examples
| Question 1: Evaluate ∫6x2 dx |
Answer:
By the power rule of integration,
∫6x2 dx = 6 x2+1/(2+1) + C
= 6 x3/3 + C
= 2x3 + C, where C is an integration constant.
| Question 2: Evaluate ∫(x2+x+1) dx |
Answer:
∫(x2+x+1) dx = ∫x2 dx + ∫x dx + ∫dx
= x3/3 + x2/2 + x + C
= 2x3 + C, where C is an integration constant.
| Question 2: Evaluate ∫1/x2 dx |
Answer:
Note that 1/x2 can be written as x-2 by the rule of indices. We have:
∫1/x2 dx = ∫x-2 dx
= x-2+1/(-2+1) + C
= x-1/-1 + C
= -1/x + C, where C is an integration constant.
Trigonometric Integration Examples
| Question 4: Evaluate ∫sin 2x dx |
Answer:
We will use the substitution method to integrate. Let us put
z=2x ⇒ dz=2 dx ⇒dx=dz/2.
Thus, ∫sin 2x dx = ∫sin z dz/2
= (1/2)∫sin z dz = (1/2) (-cos z) + C
= $-\dfrac{\cos 2x}{2}$ + C as z=2x.
Also Read:
| Question 5: Evaluate ∫cos2x dx |
Answer:
We will use the following trigonometric formula: cos 2x = 2cos2x -1. Thus, we have
cos2x = $\dfrac{1+\cos 2x}{2}$
∴ ∫cos2x dx = $\dfrac{1}{2} \int (1+\cos 2x) dx$
= $\dfrac{1}{2} (x+\dfrac{\sin 2x}{2})$ as we know that ∫cos mx dx = $\dfrac{\sin mx}{m}$ +C
= $\dfrac{x}{2} +\dfrac{\sin 2x}{4}$
Exponential Integration Examples
| Question 6: Evaluate ∫e2x dx |
Answer:
Put 2x=z. so we have dx=dz/2.
∴ ∫e2x dx = $\int e^z \dfrac{dz}{2}$
= $\dfrac{1}{2} \int e^z dz$
= $\dfrac{1}{2} e^z$ +C
= $\dfrac{1}{2} e^{2x}$ +C as z=2x.
| Question 7: Evaluate ∫(ex+e-x) dx |
Answer:
∫(ex+e-x) dx
= ∫ex dx + ∫e-x dx
= ex – e-x +C as we know that ∫emx dx = $\dfrac{e^{mx}}{m}$ +C.
| Question 8: Evaluate ∫2x dx |
Answer:
We know that ∫ax dx = $\dfrac{a^x}{\ln a}$ +C; a(≠1)>0. Here ln denotes the natural logarithm, that is, the logarithm with base e.
So applying the above formula for a=2, we obtain that
∫2x dx = $\dfrac{2^x}{\ln 2}$+C.
Logarithmic Integration Examples
| Question 9: Evaluate ∫logx dx |
Answer:
To find the integration of logx, we will use the integration by parts formula:
∫uv dx = u∫v dx $-\int [\frac{du}{dx}( \int v \, dx )]\, dx$, where u, v are functions of x.
Put u=logx, v=1
∴ ∫logx dx
= ∫(logx × 1) dx
= log x ∫dx – ∫x [d/dx(logx)] dx +C
= xlog x – ∫x (1/x) dx +C
= xlog x – ∫dx +C
= xlog x – x +C
So the integration of logx is equal to x(logx-1)+C.
| Question 10: Evaluate ∫$\dfrac{\log x}{x}$ dx |
Answer:
Put z=logx
So dz = dx/x.
∴ ∫$\dfrac{\log x}{x}$ dx
= ∫z dz
= z2/2 +C
= (log x)2/2 +C, where C is an integral constant.
