$\displaystyle \int_{a}^{b}f(x) \ dx$

In order to evaluate this integral, first find the antiderivative of $\displaystyle f(x).$

In this case, $\displaystyle f(x)= \sin(10x-3)$.

The antiderivative is  $\displaystyle F(x) = -\frac{\cos(10x-3)}{10}$.

Using the Fundamental Theorem of Calculus, evaluate the integral using the antiderivative: 

$\displaystyle \int_{1}^{2} \sin(10x-3) \ dx = \left ( -\frac{\cos(10x-3)}{10} \right )_{1}^{2}$

$\displaystyle = -\frac{\cos(10(2)-3)}{10} - \left ( -\frac{\cos(10-3)}{10} \right )$

$\displaystyle = 0.004$

$\displaystyle \int f(x) dx$

In order to evaluate this integral, first find the antiderivative of $\displaystyle f(x).$

If $\displaystyle f(x) = a$ then $\displaystyle F(x) = ax + C$

If $\displaystyle f(x) = x a$ then $\displaystyle F(x) = \frac{x^{a+1}}{a+1} + C$

If $\displaystyle f(x) = \frac{1}{x}$ then $\displaystyle F(x) = ln(x) + C$

If $\displaystyle f(x) = e ^x$ then $\displaystyle F(x) = e^x + C$

If $\displaystyle f(x) = \cos(x)$ then $\displaystyle F(x) = \sin(x) + C$

If $\displaystyle f(x) = \sin(x)$ then $\displaystyle F(x) = - \cos(x) + C$

If $\displaystyle f(x) = \sec^2 (x)$ then $\displaystyle F(x) = \tan(x) + C$

In this case, $\displaystyle f(x) = 3$.

The antiderivative is  $\displaystyle F(x) = 3x+ C$.

This integral can be evaluated using $\displaystyle u$-substitution.

$\displaystyle u = x^2$

$\displaystyle du=2x dx$

Then we can proceed as follows

$\displaystyle \int \frac{3x}{1+x^4}dx$ (Start)

$\displaystyle = \frac{3}{2}\int \frac{2x}{1+x^4}dx$ (Factor out a $\displaystyle 3/2$, leaving a $\displaystyle 2x$ in the numerator)

$\displaystyle =\frac{3}{2}\int \frac{1}{1+u^2}du$ (Substitute the equations for $\displaystyle u, du$)

$\displaystyle =\frac{3}{2}\tan^{-1}(u)+C$ (Integrate, recall that $\displaystyle \frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}$)

$\displaystyle =\frac{3}{2}\tan^{-1}(x^2)+C$ (Substitute $\displaystyle x^2$ back in)

The anti-derivative of $\displaystyle e^x$ is simply $\displaystyle e^x$.  This is the perfect function that no matter if you take the derivative or the anti-derivative, it always returns back the original function.  The integral of the constant returns the constant multiplied by the integration variable.  Then, since this is an indefinite integral, we must include the integration constant, $\displaystyle C.$

$\displaystyle \int(e^x+4)dx= e^x+4x+C.$

The indefinite integral is a reverse chain rule.  Remember, anti-derivatives are the exact opposite of the derivative of this function.  So, let's start there:

$\displaystyle cos(3x)'=-3sin(3x).$

Therefore, to undo this answer, we would have to get rid of the negative sign and divide by the chain rule part.  We also add our integration constant.  You would get the same result (but longer time) if you used u-substitution. 

$\displaystyle \int cos(3x)dx= \frac{sin(3x)}{3}+C$.

This integral requires use of the power rule for antiderivatives, which simplifies as follows:

$\displaystyle \int (x^4-2x^2+6)dx=\int x^4dx-2\int x^2 dx+6\int dx$

$\displaystyle =(\frac{1}{5}\cdot x^5)-2 (\frac{1}{3}\cdot x^3)+6(x)+C=\frac{x^5}{5}-\frac{2x^3}{3}+6x+C$

Make the substitution:

$\displaystyle u=3x^3$

where

$\displaystyle du=9x^2dx\rightarrow \frac{du}{9}=x^2dx$

Substituting this into the original expression:

$\displaystyle \int x^2\sec^2(3x^3)dx=\int\frac{\sec^2(u)}{9}du=\frac{\tan(u)}{9}+C=\frac{\tan(3x^3)}{9}+C$

To solve this with integration by parts, we rewrite the expression in the form

$\displaystyle \int u \: dv$

where

$\displaystyle u=\ln(x), \: \: \:dv=x\, dx$

and

$\displaystyle du=\frac{1}{x}dx,\: \: \:v=\frac{x^2}{2}$

To integrate, apply the formula for integration by parts:

$\displaystyle \int u \: dv=uv-\int vdu=\frac{1}{2}x^2\ln(x)-\int (\frac{x^2}{2}\cdot \frac{1}{x})dx$

$\displaystyle =\frac{x^2}{2}\ln(x)-\frac{1}{2}\int xdx=\frac{x^2\ln(x)}{2}-\frac{x^2}{4}+C=\frac{x^2}{4}(2\ln(x)-1)+C$

Make the trigonometric substitution:

$\displaystyle x=3\sin(a),\: \: \:dx=3\cos(a)da$

Applying this to the expression given:

$\displaystyle \int\sqrt{9-x^2}\, dx=\int3\cos(a)\sqrt{9(1-\sin^2a)}\, da=\int9\cos^2(a)da$

$\displaystyle =9[\frac{1}{2}\cos(a)\sin(a)+\frac{a}{2}=\frac{x\sqrt{9-x^2}}{2}+\frac{9}{2}\sin^{-1}(\frac{x}{3})+C$

Make the trigonometric substitution:

$\displaystyle x=5\sec(a),\,\,\,dx=5\tan(a)\sec(a)da$

Applying this substitution to the expression given, we can integrate:

$\displaystyle \int\frac{1}{x^2\sqrt{x^2-25}}\,dx=\int\frac{5\tan(a)\sec(a)}{25\sec^2(a) \sqrt{25\sec^2(a)-25}} da$

$\displaystyle =\int\frac{5\tan(a)}{125\sec(a)\tan(a)}da=\int\frac{da}{25\sec(a)}=\int\frac{\cos(a)}{25}da$

$\displaystyle =\frac{sin(a)}{25}+C=\frac{\sqrt{x^2-25}}{25x}+C$