Derivative of \(\ln(x)\)
By Definition of the Derivative, we can write
\[f'(x) = \lim_{h \to 0} \frac{\ln(x+h) - \ln(x)}{h}\] Applying Logarithmic Properties, we get \[f'(x) = \lim_{h \to 0} \frac{1}{h} \ln\left(\frac{x+h}{x}\right) = \lim_{h \to 0} \frac{1}{h} \ln\left(1 + \frac{h}{x}\right)\] Using the Power Rule for Logs, we get \[f'(x) = \lim_{h \to 0} \ln\left(1 + \frac{h}{x}\right)^{\frac{1}{h}}\] Manipulation for the Constant \(e\), we get \[f'(x) = \lim_{h \to 0} \ln\left[\left(1 + \frac{h}{x}\right)^{\frac{x}{h}}\right]^{\frac{1}{x}}\] Evaluating the Limit, we get \[f'(x) = \frac{1}{x} \ln \left[ \lim_{u \to 0} (1+u)^{\frac{1}{u}} \right]\] So we have \[f'(x) = \frac{1}{x} \cdot 1 = \frac{1}{x}\]
\[f'(x) = \lim_{h \to 0} \frac{\ln(x+h) - \ln(x)}{h}\] Applying Logarithmic Properties, we get \[f'(x) = \lim_{h \to 0} \frac{1}{h} \ln\left(\frac{x+h}{x}\right) = \lim_{h \to 0} \frac{1}{h} \ln\left(1 + \frac{h}{x}\right)\] Using the Power Rule for Logs, we get \[f'(x) = \lim_{h \to 0} \ln\left(1 + \frac{h}{x}\right)^{\frac{1}{h}}\] Manipulation for the Constant \(e\), we get \[f'(x) = \lim_{h \to 0} \ln\left[\left(1 + \frac{h}{x}\right)^{\frac{x}{h}}\right]^{\frac{1}{x}}\] Evaluating the Limit, we get \[f'(x) = \frac{1}{x} \ln \left[ \lim_{u \to 0} (1+u)^{\frac{1}{u}} \right]\] So we have \[f'(x) = \frac{1}{x} \cdot 1 = \frac{1}{x}\]
Derivative of \(\sin(x)\)
By Definition of the Derivative, we get
\[f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h}\]
Now we Apply Sum Identity, then
\[f'(x) = \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h}\]
We Rearrange and Factor, then
\[f'(x) = \lim_{h \to 0} \left[ \frac{\sin x (\cos h - 1)}{h} + \frac{\cos x \sin h}{h} \right]\]
\[f'(x) = \sin x \cdot \lim_{h \to 0} \left( \frac{\cos h - 1}{h} \right) + \cos x \cdot \lim_{h \to 0} \left( \frac{\sin h}{h} \right)\]
We Evaluate Special Limits, then
\[\lim_{h \to 0} \frac{\sin h}{h} = 1 \quad \text{and} \quad \lim_{h \to 0} \frac{\cos h - 1}{h} = 0\]
Now, we Substitute and Simplify, then
\[f'(x) = \sin x \cdot (0) + \cos x \cdot (1)\]
\[f'(x) = \cos x\]
Derivative of \(e^x\)
By Definition of the Derivative, we get
\[f'(x) = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h}\]
Using the exponent rule \(e^{x+h} = e^x \cdot e^h\), we get
\[f'(x) = \lim_{h \to 0} \frac{e^x e^h - e^x}{h}\]
\[f'(x) = \lim_{h \to 0} \frac{e^x(e^h - 1)}{h}\]
By Extracting Constants from the Limit, we get
\[f'(x) = e^x \cdot \lim_{h \to 0} \frac{e^h - 1}{h}\]
Now we Evaluate the Limit, then
\[\lim_{h \to 0} \frac{e^h - 1}{h} = 1\]
Hence
\[f'(x) = e^x \cdot 1 = e^x\]
Derivative of \(\sin^{-1}(x)\)
Initially we set up the inverse relationship, thus we suppose \(y = \sin^{-1}(x)\). This gives
\[\sin(y) = x\]
Now, we Differentiate implicitly with respect to \(x\), then
\[\frac{d}{dx}[\sin(y)] = \frac{d}{dx}[x]\]
Using the chain rule on the left side, we get
\[\cos(y) \cdot \frac{dy}{dx} = 1\]
Now, we solve for \(\frac{dy}{dx}\), then
\[\frac{dy}{dx} = \frac{1}{\cos(y)}\]
\[\frac{dy}{dx} = \frac{1}{\sqrt{1-sin ^2y}}\]
Hence, we have
\[\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}\]
Integration of \( \int \log (x) dx\)
We choose Integration by Parts, in which
\(u=\log (x)\) and \(v=1\)
Now , the integration is
Hence, the integral is
\( \int \log (x) dx = x \log (x) - \int dx = x \log (x) - x + c\)
\(u=\log (x)\) and \(v=1\)
Now , the integration is
| \(+/-\) | \(u\) (Derivative) | \(dv\) (Integral) | Product |
| + | \(\log (x)\)↘ | \(1\) | |
| - | \(\frac{1}{x}\)→ | ↘\(x\) | \(x \ln(x)\) |
| \(+\) | \(-\int \frac{1}{x} \cdot x \, dx\) |
\( \int \log (x) dx = x \log (x) - \int dx = x \log (x) - x + c\)
Integration of \( \int x^2 \sin x dx\)
We choose Integration by Parts, in which
\(u=x^2\) and \(dv=\sin(x)dx\)
Now, the integration is
Hence, the integral is
\( \int x^2 \sin (x) dx = -x^2 \cos (x) + 2x \sin (x) + 2 \cos (x) + c \)
\(u=x^2\) and \(dv=\sin(x)dx\)
Now, the integration is
| \(+/-\) | \(u\) (Derivative) | \(dv\) (Integral) | Product |
| + | \(x^2\)↘ | \(\sin(x)\) | |
| - | \(2x\)↘ | \(-\cos(x)\) | \(-x^2 \cos(x)\) |
| + | \(2\)↘ | \(-\sin(x)\) | \(+2x \sin(x)\) |
| - | \(0\) | \(\cos(x)\) | \(+2 \cos(x)\) |
Hence, the integral is
\( \int x^2 \sin (x) dx = -x^2 \cos (x) + 2x \sin (x) + 2 \cos (x) + c \)
Integration of \( \int x e^x dx\)
We choose Integration by Parts, in which
\(u=x\) and \(dv=e^x dx\)
Now, the integration is
Hence, the integral is
\( \int x e^x dx = x e^x - e^x + c = e^x(x - 1) + c \)
\(u=x\) and \(dv=e^x dx\)
Now, the integration is
| \(+/-\) | \(u\) (Derivative) | \(dv\) (Integral) | Product |
| \(x\)↘ | \(e^x\) | ||
| + | \(1\)↘ | \(e^x\) | \(x e^x\) |
| - | \(0\) | \(e^x\) | \(-e^x\) |
Hence, the integral is
\( \int x e^x dx = x e^x - e^x + c = e^x(x - 1) + c \)
Integration of \( \int e^x \sin x dx\)
We choose Integration by Parts, in which
\(u=\sin(x)\) and \(dv=e^x\)
Now, the integration is
Hence, the integral is
\( \int e^x \sin (x) dx =e^x \sin(x) - e^x \cos(x) + \int (- \sin x) e^x dx \)
or\( 2\int e^x \sin (x) dx =e^x \sin(x) - e^x \cos(x) \)
or\( \int e^x \sin (x) dx =\frac{1}{2} [e^x \sin(x) - e^x \cos(x)] +c\)
\(u=\sin(x)\) and \(dv=e^x\)
Now, the integration is
| \(+/-\) | \(u\) (Derivative) | \(dv\) (Integral) | Product |
| + | \(\sin(x)\)↘ | \(e^x\) | |
| - | \(\cos(x)\)↘ | \(e^x\) | \(e^x \sin(x)\) |
| + | \(-\sin(x)\) \(\rightarrow\) | \(e^x\) | \(-e^x \cos(x)\) |
Hence, the integral is
\( \int e^x \sin (x) dx =e^x \sin(x) - e^x \cos(x) + \int (- \sin x) e^x dx \)
or\( 2\int e^x \sin (x) dx =e^x \sin(x) - e^x \cos(x) \)
or\( \int e^x \sin (x) dx =\frac{1}{2} [e^x \sin(x) - e^x \cos(x)] +c\)
Integration of \( \int e^{2x} \sin x dx\)
We choose Integration by Parts, in which
\(u=\sin(x)\) and \(dv=e^{2x}\)
Now, the integration is
Hence, the integral is
\( \int e^{2x} \sin (x) dx =\frac{1}{2} e^{2x} \sin(x) - \frac{1}{4} e^{2x} \cos(x) - \int \frac{1}{4} e^{2x} \sin(x) dx \)
or\( \frac{5}{4} \int e^{2x}\sin (x) dx =\frac{1}{2} e^{2x} \sin(x) - \frac{1}{4} e^{2x} \cos(x) +c\)
or\( \int e^{2x}\sin (x) dx =\frac{1}{5} [2 e^{2x} \sin(x) - e^{2x} \cos(x)] +c\)
\(u=\sin(x)\) and \(dv=e^{2x}\)
Now, the integration is
| \(+/-\) | \(u\) (Derivative) | \(dv\) (Integral) | Product |
| \(\sin(x)\)↘ | \(e^{2x}\) | ||
| + | \(\cos(x)\)↘ | \(\frac{1}{2}e^{2x}\) | \(\frac{1}{2}e^{2x} \sin(x)\) |
| - | \(-\sin(x)\) \(\rightarrow\) | \(\frac{1}{4}e^{2x}\) | \(-\frac{1}{4}e^{2x} \cos(x)\) |
Hence, the integral is
\( \int e^{2x} \sin (x) dx =\frac{1}{2} e^{2x} \sin(x) - \frac{1}{4} e^{2x} \cos(x) - \int \frac{1}{4} e^{2x} \sin(x) dx \)
or\( \frac{5}{4} \int e^{2x}\sin (x) dx =\frac{1}{2} e^{2x} \sin(x) - \frac{1}{4} e^{2x} \cos(x) +c\)
or\( \int e^{2x}\sin (x) dx =\frac{1}{5} [2 e^{2x} \sin(x) - e^{2x} \cos(x)] +c\)
Integration of \( \int \sin^{-1} (x) dx\)
We choose Integration by Parts, in which
\(u=\sin^{-1} (x)\) and \(v=1\)
Now, the integration is
Hence, the integral is
\( \int \sin^{-1} (x) dx = x \sin^{-1} (x) + \sqrt{1-x^2} + c \)
\(u=\sin^{-1} (x)\) and \(v=1\)
Now, the integration is
| \(+/-\) | \(u\) (Derivative) | \(dv\) (Integral) | Product |
| \(\sin^{-1} (x)\)↘ | \(1\) | ||
| + | \(\frac{1}{\sqrt{1-x^2}}\)↘ | \(x\) | \(x \sin^{-1}(x)\) |
| \(-\) | ↘ | \(-\int \frac{x}{\sqrt{1-x^2}} dx\) |
Hence, the integral is
\( \int \sin^{-1} (x) dx = x \sin^{-1} (x) + \sqrt{1-x^2} + c \)
Integration of \( \int \log x \sin^{-1} x \, dx\)
We choose Integration by Parts, in which
\(u=\sin^{-1}(x)\) and \(dv=\ln(x)dx\)
Now, the integration is
Hence, the integral is
\( \int \ln(x) \sin^{-1}(x) dx = (x\ln x - x)\sin^{-1}(x) - \int \frac{x\ln x - x}{\sqrt{1-x^2}} dx + c \)
\(u=\sin^{-1}(x)\) and \(dv=\ln(x)dx\)
Now, the integration is
| \(+/-\) | \(u\) (Derivative) | \(dv\) (Integral) | Product |
| \(\sin^{-1}(x)\)↘ | \(\ln(x)\) | ||
| + | \(\frac{1}{\sqrt{1-x^2}}\)↘ | \(x\ln(x)-x\) | \((x\ln x - x)\sin^{-1}(x)\) |
| - | ↘ | \(-\int \frac{x\ln(x)-x}{\sqrt{1-x^2}} dx\) |
Hence, the integral is
\( \int \ln(x) \sin^{-1}(x) dx = (x\ln x - x)\sin^{-1}(x) - \int \frac{x\ln x - x}{\sqrt{1-x^2}} dx + c \)
Integration of \( \int \tan x \, dx\)
We choose substitution, in which
\(u=\cos(x)\) then \(du=-\sin(x)dx\)
Now, the integration is
\( \int \tan(x) dx = \int \dfrac{\sin x}{\cos x} dx = -\int \dfrac{1}{u}du = -\log|\cos(x)| + c = \log|\sec(x)| + c \)
\(u=\cos(x)\) then \(du=-\sin(x)dx\)
Now, the integration is
\( \int \tan(x) dx = \int \dfrac{\sin x}{\cos x} dx = -\int \dfrac{1}{u}du = -\log|\cos(x)| + c = \log|\sec(x)| + c \)
Integration of \( \int \cot x \, dx\)
We choose substitution, in which
\(u=\sin(x)\) then \(du=\cos(x)dx\)
Now, the integration is
\( \int \cot(x) dx =\int \dfrac{\cos x}{\sin x} dx =\int \dfrac{1}{u}du= \log|\sin(x)| + c \)
\(u=\sin(x)\) then \(du=\cos(x)dx\)
Now, the integration is
\( \int \cot(x) dx =\int \dfrac{\cos x}{\sin x} dx =\int \dfrac{1}{u}du= \log|\sin(x)| + c \)
Integration of \( \int \sec x \, dx\)
We choose substitution, in which
\(u=\sec(x) + \tan(x)\) then \(du=(\sec x \tan x + \sec^2 x)dx\)
Now, the integration is
\( \int \sec(x) dx = \int \sec x \cdot \dfrac{\sec x + \tan x}{\sec x + \tan x} dx = \int \dfrac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} dx = \int \dfrac{1}{u}du = \log|\sec(x) + \tan(x)| + c \)
\(u=\sec(x) + \tan(x)\) then \(du=(\sec x \tan x + \sec^2 x)dx\)
Now, the integration is
\( \int \sec(x) dx = \int \sec x \cdot \dfrac{\sec x + \tan x}{\sec x + \tan x} dx = \int \dfrac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} dx = \int \dfrac{1}{u}du = \log|\sec(x) + \tan(x)| + c \)
Integration of \( \int \csc x \, dx\)
We choose substitution, in which
\(u=\csc(x) + \cot(x)\) then \(du=(-\csc x \cot x - \csc^2 x)dx\)
Now, the integration is
\( \int \csc(x) dx = \int \csc x \cdot \dfrac{\csc x + \cot x}{\csc x + \cot x} dx = \int \dfrac{\csc^2 x + \csc x \cot x}{\csc x + \cot x} dx = -\int \dfrac{1}{u}du = -\log|\csc(x) + \cot(x)| + c \)
\(u=\csc(x) + \cot(x)\) then \(du=(-\csc x \cot x - \csc^2 x)dx\)
Now, the integration is
\( \int \csc(x) dx = \int \csc x \cdot \dfrac{\csc x + \cot x}{\csc x + \cot x} dx = \int \dfrac{\csc^2 x + \csc x \cot x}{\csc x + \cot x} dx = -\int \dfrac{1}{u}du = -\log|\csc(x) + \cot(x)| + c \)
Integration of \( \int a^x \, dx\)
We choose substitution, in which
\(a^x = e^{x \ln a}\), let \(u = x \ln a\) then \(du = \ln a \, dx\)
Now, the integration is
\( \int a^x dx = \int e^{x \ln a} dx = \frac{1}{\ln a} \int e^u du = \frac{1}{\ln a} e^u + c = \dfrac{a^x}{\ln a} + c \)
\(a^x = e^{x \ln a}\), let \(u = x \ln a\) then \(du = \ln a \, dx\)
Now, the integration is
\( \int a^x dx = \int e^{x \ln a} dx = \frac{1}{\ln a} \int e^u du = \frac{1}{\ln a} e^u + c = \dfrac{a^x}{\ln a} + c \)
Introduction
In calculus, differentiation and integration are two fundamental operations that are inversely related to each other. If you differentiate a function, you find its rate of change or slope at any given point. Integration, on the other hand, finds the accumulation or total value of a function over a given interval.
When we say integration is the inverse operation of differentiation, we mean that integration "undoes" the effect of differentiation, and vice versa. More formally, if you integrate the derivative of a function, you'll get back the original function (up to a constant). Similarly, if you differentiate the integral of a function, you'll also get back the original function (up to a constant).
Mathematically, this is represented by the Fundamental Theorem of Calculus
Formula for Integral
Integral of algebraic functions_1
| SN | Derivative | Integral |
| 1 | \( \frac{d}{dx} \left ( x \right) =1 \) | \( \int \left ( 1 \right) dx=x+c \) |
| 2 | \( \frac{d}{dx} \left ( ax \right) =a \) | \( \int \left ( a \right) dx=ax+c \) |
| 3 | \( \frac{d}{dx} \left ( x^n \right) =n x^{n-1} \) | \( \int \left ( x^n \right) dx=\frac{x^{n+1}}{n+1}+c;n \ne -1 \) |
| 4 | \( \frac{d}{dx} (ax+b)^n=na (ax+b)^{n-1}\) | \( \int (ax+b)^n dx= \frac{(ax+b)^{n+1}}{a(n+1)}+c ;n \ne -1\) |
Integral of algebraic functions_2
| SN | Derivative | Integral |
| 1 | \( \frac{d}{dx} \log|x| =\frac{1}{x}\) | \( \int \frac{1}{x} dx=\log |x|+c \) |
| 2 | \( \frac{d}{dx} \log |ax+b| = \frac{a}{ax+b}\) | \( \int \frac{dx}{ax+b} dx= \frac{1}{a} \log |ax+b|+c\) |
Integral of trigonometric functions
| SN | Derivative | Integral |
| 1 | \( \frac{d}{dx} \left ( \cos x\right) =- \sin x\) | \( \int \left ( \sin x \right) dx=- \cos x+c \) |
| 2 | \( \frac{d}{dx} \left ( \sin x\right) = \cos x\) | \( \int \left ( \cos x \right) dx=- \sin x+c \) |
| 3 | \( \frac{d}{dx} \left ( \tan x\right) = \sec ^2x\) | \( \int \left ( \sec ^2x \right) dx= \tan x+c \) |
| \( \int \left ( \tan x \right) dx=- \log |\cos x|+c \) | ||
| 4 | \( \frac{d}{dx} \left ( \cot x\right) = -\csc ^2x\) | \(\int \left ( \csc ^2x \right) dx= -\cot x+c \) |
| \( \int \left ( \cot x \right) dx= \log |\sin x|+c \) | ||
| 5 | \( \frac{d}{dx} \left ( \sec x\right) = \sec x \tan x\) | \( \int \left ( \sec x \tan x \right) dx= \sec x+c \) |
| \( \int \left ( \sec x \right) dx= \log |\sec x+\tan x|+c \) | ||
| 6 | \( \frac{d}{dx} \left ( \csc x\right) = -\csc x \cot x\) | \( \int \left ( \csc x \cot x \right) dx= -\csc x+c \) |
| \( \int \left ( \csc x \right) dx= -\log |\csc x+\cot x|+c \) |
Integral of exponential/logarithm functions
| SN | Derivative | Integral |
| 1 | \( \frac{d}{dx} e^x = e^x\) | \( \int e^x dx=e^x+c\) |
| 2 | \( \frac{d}{dx} e^{ax} = ae^{ax}\) | \( \int e^{ax} dx=\frac{1}{a} e^{ax}+c\) |
| 3 | \( \frac{d}{dx} a^{x} = a^{x} \log a\) | \( \int a^{x} dx=\frac{a^{x}}{\log a} +c\) |
Integral of inverse functions
| SN | Derivative | Integral |
| 1 | \( \frac{d}{dx} \left ( \sin ^{-1} x\right) = \frac{1}{\sqrt{1-x^2}}\) | \(\int \left ( \frac{1}{\sqrt{1-x^2}} \right) dx=\sin ^{-1} x+c \) |
| 2 | \( \frac{d}{dx} \left ( \tan ^{-1} x\right) = \frac{1}{1+x^2}\) | \( \int \left ( \frac{1}{1+x^2} \right) dx=\tan ^{-1} x+c \) |
| 3 | \( \frac{d}{dx} \left ( \sec ^{-1} x\right) = \frac{1}{|x| \sqrt{x^2-1}} \) | \( \int \left ( \frac{1}{|x| \sqrt{x^2-1}} \right) dx=\sec ^{-1} x+c \) |
| 4 | \( \frac{d}{dx} \left ( \cos ^{-1} x\right) = \frac{-1}{\sqrt{1-x^2}}\) | \(\int \left ( \frac{-1}{\sqrt{1-x^2}} \right)dx=\cos ^{-1} x+c \) |
| 5 | \( \frac{d}{dx} \left ( \cot ^{-1} x\right) = \frac{-1}{1+x^2}\) | \( \int \left ( \frac{-1}{1+x^2} \right) dx=\cot ^{-1} x+c \) |
| 6 | \( \frac{d}{dx} \left ( \csc ^{-1} x\right) = \frac{-1}{|x| \sqrt{x^2-1}} \) | \( \int \left ( \frac{-1}{|x| \sqrt{x^2-1}} \right) dx=\csc ^{-1} x+c \) |
Basic Integral of algebraic functions: Activity 1
Compute the following integrals.- \( \int 60 dt\)
- \( \int 2x dx\)
- \( \int 5x^3 dx\)
- \( \int 7x^{5/2}dx\)
- \( \int 4x^{-5} dx\)
- \( \int 2x^{-\frac{7}{2}} dx \)
- \( \int (2x+4)dx \)
- \( \int (w^2-3)dw \)
- \( \int (x^2+2) dx\)
- \( \int (3x^2+2x+1) dx\)
- \( \int (2x+1)(3x+2) dx\)
- \( \int (4x^{1/3}+5x^{2/3}+\frac{1}{x^2}) dx \)
- \( \int (x^{3/4}+x^{1/2}+4x^{1/3} )dx\)
- \( \int (x^2-\frac{1}{x^2}) dx\)
- \( \int \sqrt{x} (x^2-5) dx\)
- \( \int (x^2+3x+5)x^{-1/2}dx\)
- \( \int \left ( \sqrt{x}-\frac{1}{\sqrt{x}} \right ) dx\)
- \( \int (x-3)^2 dx\)
- \( \int (4x+5)^4dx \)
- \( \int (3x+5)^4 dx\)
- \( \int (a-bx)^5 dx\)
- \( \int (c+dx)^{-3/2}dx\)
- \( \int \frac{1}{\sqrt{2x+7} }dx\)
- \( \int \left ( x+\frac{1}{(x+3)^2} \right ) dx\)
- \( \int \left ( 4+\frac{1}{(5x+1)^2} \right ) dx\)
Basic Integral of algebraic functions: Activity 2
Compute the following integrals.- \( \int \frac{1}{\sqrt{x+1}-\sqrt{x}}dx\)
- \( \int \frac{1}{\sqrt{x+a}-\sqrt{x-b}} dx\)
- \( \int \frac{1}{\sqrt{2x+1}-\sqrt{2x-3}}dx\)
- \( \int \frac{1}{\sqrt{x+3}-\sqrt{x-1}} dx\)
- \( \int (2x+3)(4x+5)^4 dx \)
- \( \int (3x+2)(2x+5)^3 dx \)
- \( \int x \sqrt{x+1} dx\)
- \( \int 2x \sqrt{2x+3} dx\)
- \( \int x \sqrt{ax+b}dx\)
- \( \int 5x \sqrt{5x+2}dx\)
- \( \int (x+2) \sqrt {3x+2} dx\)
- \( \int (5x+3) \sqrt{4x+1} dx\)
- \( \int (2x+3) \sqrt{3x+1}dx\)
- \( \int \frac{3x+4}{\sqrt{x+1}} dx\)
- \( \int \frac{x+2}{\sqrt{x+1}} dx\)
- \( \int \frac{2x+1}{\sqrt{3x+2}}dx\)
Basic Integral of algebraic functions: Activity 3
Compute the following integrals.- \( \int \frac{(3x+2)}{(5x+1)^2} dx \)
- \( \int \frac{x+3}{x-3} dx\)
- \( \int \frac{3x^2-5x+2}{x}dx\)
- \( \int \frac{x^2+3x+3}{x+2} dx \)
- \( \int \frac{ax^2+bx+c}{x^2} dx\)
- \( \int \frac{3x-1}{x-2} dx\)
- \( \int \frac{x^2+5}{x+2} dx\)
Basic Integral of trigonometric functions: Activity 4
Compute the following integrals.- \( \int \sin 5x dx\)
- \( \int \sin^2 ax dx\)
- \( \int \tan ^2 a x dx\)
- \( \int \cos^4 n x dx\)
- \( \int \cos (a^2x+b)dx\)
- \( \int \frac{1}{\sec^2 xx \tan^2 x } dx\)
- \( \int \sqrt{1+\cos n x} dx\)
- \( \int \sqrt{1+\sin 2a x} dx\)
- \( \int \frac{1}{1-\cos nx } dx\)
- \( \int \sin (ax+b) dx\)
- \( \int \sec^2 (2x+3) dx\)
- \( \int \cos ^2 bx dx\)
- \( \int \sin ^4 x dx\)
- \( \int \frac{1}{\cos^2 x \sin ^2 x} dx\)
- \( \int \sqrt{1-\cos px } dx\)
- \( \int \frac{1}{1+\cos m x} dx\)
- \( \int \frac{1}{1-\sin a x} dx\)
- \( \int \sin 6 x \cos 8x dx\)
- \( \int \sin^7 x \sin ^5 x dx\)
- \( \int \cos px \cos q x dx\)
- \( \int \frac{\cos x- \cos 2x}{1-\cos x} dx\)
Basic Integral of Logarithm functions: Activity 5
Compute the following integrals.- \( \int (e^{px}+e^{-qx})dx\)
- \( \int (e^{px}+e^{-px})^2 dx\)
- \( \int \frac{e^{2x}+e^x+1}{e^x} dx\)
- \( \int e^x (e^{2x}+1) dx\)
- \( \int \frac{e^{ 6\log x}-e^{ 5\log x}}{e^{ 4\log x}-e^{ 3\log x}} dx\)
No comments:
Post a Comment