Интеграли, съдържащи √x² - a²

Интеграли, съдържащи √a² - x²

Интеграли, съдържащи ax² + bx + c

Ако b² = 4ac, ax² + bx + c = a(x + b/2a)².

Интеграли, съдържащи √ax² + bx + c

В долните резултати, ако b² = 4ac, √ax² + bx + c = √a(x + b/2a)

Интеграли, съдържащи x³ + a³

Забележете, че формулата съдържаща x³ - a³ се заменя с - a.

Интеграли, съдържащи x⁴ ± a⁴

Интеграли, съдържащи xⁿ ± aⁿ

Интеграли, съдържащи sin ax

$\int_{}^{}sinaxdx=\frac{-cosax}{a}$

$\int_{}^{}x.sinaxdx=\frac{sinax}{a^2}-\frac{x.cosax}{a}$

$\int_{}^{}x^2.sinaxdx=\frac{2xsinax}{a^2}+\left(\frac{2}{a^3}-\frac{x^2}{a}\right)cosax$

$\int_{}^{}x^3sinaxdx=\left(\frac{3x^2}{a^2}-\frac{6}{a^4}\right)sinax+\left(\frac{6x}{a^3}-\frac{x^3}{ a}\right)cosax$

$\int_{}^{}\frac{sinaxdx}{x}=ax-\frac{ax^3}{3*3!}+\frac{ax^5}{5*5!}-...$

$\int_{}^{}\frac{sinaxdx}{x^2}=\frac{-sinax}{x}+a.\int_{}^{}\frac{cosaxdx}{x}$

$\int_{}^{}\frac{dx}{sinax}=\frac{ln(cscax-cotax)}{a}=\frac{ln(tan\left(\frac{ax}{2}\right))}{2}$

$\int_{}^{}sin^2axdx=\frac{x}{2}-\frac{sin2ax}{4a}$

$\int_{}^{}x.sin^2axdx=\frac{x^2}{4}-\frac{x.sin2ax}{4a}-\frac{cos2ax}{8a^2}$

$\int_{}^{}sin^3axdx=\frac{-cosax}{a}+\frac{cos^32ax}{3a}$

$\int_{}^{}sin^4axdx=\frac{3x}{8}-\frac{sin2ax}{4a}+\frac{sin4ax}{32a}$

$\int_{}^{}\frac{dx}{sin^2ax}=\frac{-cotax}{a}$

$\int_{}^{}\frac{dx}{sin^3ax}=\frac{-cosax}{2a.sin^2ax}+\frac{1}{2a}.ln(tan\left(\frac{ax}{ 2}\right))$

Интеграли, съдържащи cos ax

$\int_{}^{}cosaxdx=\frac{sinax}{a}$

$\int_{}^{}x.cosaxdx=\frac{cosax}{a^2}+\frac{x.sinax}{a}$

$\int_{}^{}x^2.cosaxdx=\frac{2x.cosax}{a^2}+\left(\frac{x^2}{a}-\frac{2}{a^3}\right)sinax$

$\int_{}^{}x^3.cosaxdx=\left(\frac{3x^2}{a^2}-\frac{6}{a^4}\right)cosax+\left(\frac{x^3}{a}-\frac{6x}{a^3}\right)sinax$

$\int_{}^{}\frac{cosaxdx}{x}=lnx-\frac{(ax)^2}{2*2!}+\frac{(ax)^4}{4*4!}-...$

$\int_{}^{}\frac{cosaxdx}{x^2}=\frac{-cosax}{x}-a.\int_{}^{}\frac{sinaxdx}{x}$

$\int_{}^{}\frac{dx}{cosax}=\frac{ln(secax+tanax)}{a}=\frac{ln(tan\left(\frac{\pi }{4}+\frac{ax}{2}\right))}{a}$

$\int_{}^{}cos^2axdx=\frac{x}{2}+\frac{sin2ax}{4a}$

$\int_{}^{}x.cos^2axdx=\frac{x^2}{4}+\frac{x.sin2ax}{4a}+\frac{cos2ax}{8a^2}$

$\int_{}^{}cos^3axdx=\frac{sinax}{a}-\frac{sin^32ax}{3a}$

$\int_{}^{}cos^4axdx=\frac{3x}{8}+\frac{sin2ax}{4a}+\frac{sin4ax}{32a}$

$\int_{}^{}\frac{dx}{cos^2ax}=\frac{tanax}{a}$

$\int_{}^{}\frac{dx}{cos^3ax}=\frac{sinax}{2a.cos^2ax}+\frac{1}{2a}.ln(tan\left(\frac{\pi}{4}+\frac{ax}{2}\right))$

Интеграли, съдържащи sin ax и cos ax

$\int_{}^{}sinax.cosaxdx=\frac{sin^2ax}{2a}$

$\int_{}^{}sin^2ax.cos^2axdx=\frac{x}{8}-\frac{sin4ax}{32a}$

$\int_{}^{}\frac{dx}{sinax.cosax}=\frac{ln(tanax)}{a}$

$\int_{}^{}\frac{dx}{sin^2ax.cosax}=\frac{ln(tan\left(\frac{\pi}{4}+\frac{ax}{2}\right))}{a}-\frac{1}{a}.sinax$

$\int_{}^{}\frac{dx}{sinax.cos^2ax}=\frac{ln(tan\left(\frac{ax}{2}\right)}{a}+\frac{1}{a}.cosax$

$\int_{}^{}\frac{dx}{sin^2ax.cos^2ax}=\frac{-2cot(2ax)}{a}$

$\int_{}^{}\frac{sin^2axdx}{cosax}=\frac{-sinax}{a}+\frac{1}{a}.ln(tan\left(\frac{ax}{2}+\frac{\pi}{4}\right))$

$\int_{}^{}\frac{cos^2axdx}{sinax}=\frac{cosax}{a}+\frac{1}{a}.ln(tan\left(\frac{ax}{2}\right))$

Интеграли, съдържащи tan ax

$\int_{}^{}tanaxdx=\frac{-1}{a}.ln(cosax)=\frac{1}{a}.ln(secax)$

$\int_{}^{}tan^2axdx=\frac{tanax}{a}-x$

$\int_{}^{}tan^3axdx=\frac{tan^2ax}{2a}+\frac{1}{a}.ln(cosax)$

$\int_{}^{}tan^nax.sec^2axdx=\frac{tan^{n+1}ax}{(n+1)a}$

$\int_{}^{}\frac{sec^2axdx}{tanax}=\frac{1}{a}.ln(tanax)$

$\int_{}^{}\frac{dx}{tanax}=\frac{1}{a}.ln(sinax)$

$\int_{}^{}\left(\frac{tanax}{x}\right)dx=ax+\frac{(ax)^3}{9}+\frac{2(ax)^5}{75}+....+\frac{2^{2n}(2^{2n}-1)B_n(ax)^{2n-1}}{(2n-1)(2n)!}+...$

$\int_{}^{}x.tan^2axdx=\frac{x.tanax}{a}+\left(\frac{1}{a^2}\right)ln(cosax)-\frac{x^2}{2}$