User:Aman maurya

In Calculus, the two important processes are differentiation and integration. We know that differentiation is finding the derivative of a function, whereas integration is the inverse process of differentiation. Here, we are going to tell some formulas of indefinite integrals.

=Integration formulae=


 * $$\int a\,dx = ax + C$$


 * $$\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \qquad\text{(for } n\neq -1\text{)}$$ (Cavalieri's quadrature formula)


 * $$\int a^x\,dx=\frac{a^x}{log_ea}+C$$


 * $$\int\frac{dx}{x}=ln\left|x\right|+C$$


 * $$\int\frac{1}{a^2+x^2}\,dx=\frac{1}{a}tan^{-1}\frac{x}{a}+C$$


 * $$\int\frac{1}{x^2-a^2}\,dx=\frac{1}{2a}ln \left|\frac{x-a}{x+a}\right|+C$$


 * $$\int\frac{1}{\sqrt{x^2+a^2}}\,dx=ln\left|x+\sqrt{x^2+a^2}\right|+C$$


 * $$\int\frac{1}{\sqrt{x^2-a^2}}\,dx=ln\left|x+\sqrt{x^2-a^2}\right|+C$$


 * $$\int\frac{1}{\sqrt{a^2-x^2}}\,dx=sin^{-1}\frac{x}{a}+C$$


 * $$\int\frac{x}{\sqrt{x^2+a^2}}\,dx=\sqrt{x^2+a^2}+C$$


 * $$\int\frac{x}{\sqrt{x^2-a^2}}\,dx=\sqrt{x^2-a^2}+C$$


 * $$\int\frac{x}{\sqrt{a^2-x^2}}\,dx=-\sqrt{a^2-x^2}+C$$


 * $$\int \sqrt{x^2+a^2}\,dx = \frac{1}{2}[x\sqrt{x^2+a^2}+a^2 ln\left|x+\sqrt{x^2+a^2}\right|]+C$$


 * $$\int \sqrt{x^2-a^2}\,dx = \frac{1}{2}[x\sqrt{x^2-a^2}-a^2 ln\left|x+\sqrt{x^2-a^2}\right|]+C$$


 * $$\int \sqrt{a^2-x^2}\,dx = \frac{1}{2}[x\sqrt{a^2-x^2}+a^2 sin^{-1}\frac{x}{a}]+C$$


 * $$\int ln\left|x\right|\,dx=xln\left|x\right|-x+C$$


 * $$\int x^n lnx\,dx=\frac{x^{n+1}}{(n+1)^2}[lnx^{n+1} -1] + C$$

=Indefinite integration of trigonometric functions=
 * $$\int sinx\,dx= -cosx +C$$
 * $$\int cosx\,dx= sinx +C$$
 * $$\int tanx\,dx= ln\left|secx\right|+C=-ln\left|cosx\right|+C$$
 * $$\int cotx\,dx= ln\left|sinx\right|+C=-ln\left|cosecx\right|+C $$


 * $$\int secx\,dx= ln\left|secx+tanx\right|+C$$


 * $$\int cosecx\,dx= ln\left|cosecx-cotx\right|+C=-ln\left|cosecx+cotx\right|+C=ln\left|tan(\frac{x}{2})\right|+C $$


 * $$\int secxtanx\,dx= secx +C$$


 * $$\int cosecxcotx\,dx= -cosecx +C$$


 * $$\int sec^2x\,dx= tanx +C$$


 * $$\int cosec^2x\,dx= -cotx +C$$

=To solve the integral of forms=

\, or\,\,\int\frac{1}{a+bsinx}\,dx\,\,or\,\int\frac{1}{asinx+bcosx}\,dx$$
 * $$\int\frac{1}{a+bcosx}\,dx \,

$$Put\, sinx=\frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}} \,and \,cosx=\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}\, and\, proceed...$$


 * $$\int\frac{1}{a+bsin^2x}\,dx\,\,or\,\,\int\frac{1}{a+bcos^2x}\,dx\,\,or\,\,\int\frac{1}{asin^2x+bcos^2x}\,dx$$

Then divide $$N^r$$ and $$ D^r$$ by $$sin^2x$$ or $$cos^2x$$ to make it in the form of $$tan^2x$$ and $$sec^2x $$ then suppose $$tanx$$ as t and proceed...


 * $$\int sinax\,cosbx\,dx$$

$$Then\, \,apply\,\,2sinAcosB=sin(A+B)+sin(A-B)$$


 * $$\int sin^nx\,dx\,or\,\int cos^nx\,dx$$

When n is even then reduce the trigonometric power in linear.


 * $$\int sin^mx\,cos^nx\,dx$$

When $$m+n=-ve\, even\, no.$$

Then convert trigonometric ratio in tan or cot and proceed...


 * $$\int\frac{pcosx+qsinx+r}{acosx+bsinx+c}\,dx$$

Suppose $$N^r=\lambda D^r+\mu\frac{d}{dx}D^r+\delta$$ and find $$\lambda,\mu,\delta$$ Then replace $$N^r=\lambda D^r+\mu\frac{d}{dx}D^r+\delta$$ and solve.

=Inverse Trigonometric functions=


 * $$\int\frac{1}{\sqrt{1-x^2}}\,dx=sin^{-1}x+C=-cos^{-1}x+C$$


 * $$\int\frac{1}{1+x^2}\,dx=tan^{-1}x+C=-cot^{-1}x+C$$


 * $$\int\frac{1}{\left|x\right|\sqrt{x^2-1}}\,dx=sec^{-1}x+C=-cosec^{-1}x+C$$


 * $$\int\frac{1}{\sqrt{a^2-x^2}}\,dx=sin^{-1}\frac{x}{a}+C$$


 * $$\int\frac{1}{a^2+x^2}\,dx=\frac{1}{a}tan^{-1}\frac{x}{a}+C$$


 * $$\int\frac{1}{\left|x\right|\sqrt{x^2-a^2}}\,dx=\frac{1}{a}sec^{-1}\frac{x}{a}+C=-cosec^{-1}\frac{x}{a}+C$$


 * $$\int sin^{-1}x\,dx=xsin^{-1}x-\sqrt{1-x^2}+C$$


 * $$\int tan^{-1}x\,dx=xtan^{-1}x-ln\left|\sqrt{1+x^2}\right|+C$$

=Some Algebraic Functions of the forms=

$$1.\,\int\frac{\lambda}{ax^2+bx+c}\,dx\,\,or\,\,\int\frac{dx}{\sqrt{ax^2+bx+c}}\,\,or\,\,\int\sqrt{ax^2+bx+c}\,dx\,\,$$

Make $$(ax^2+bx+c) \,\,as\,\,a\,\, perfect \,\, square.$$

$$2.\,\int\frac{px+q}{ax^2+bx+c}\,dx\,\,or\,\,\int\frac{px+q}{\sqrt{ax^2+bx+c}}\,dx\,\,or\,\,\int(px+q)\sqrt{ax^2+bx+c}\,dx$$

Suppose   $$px+q=\lambda\frac{d}{dx}(ax^2+bx+c)+\mu$$ $$Find\,\lambda,\mu\,\, and\,\, replace\,\, px+q\,\,by \,\lambda\frac{d}{dx}(ax^2+bx+c)+\mu\,\, and\,\, then \,\,solve...$$

$$3.\,\int\frac{1}{linear\sqrt{linear}}\,dx\,\,or\,\,\int\frac{1}{quadratic\sqrt{linear}}\,dx$$

Then suppose  $$\sqrt{linear}=t$$

$$4.\,\int \frac{1}{linear\sqrt{quadratic}}\,dx\,\,or\,\,\int \frac{1}{(linear)^n\sqrt{quadratic}}\,dx$$

Then suppose  $$linear=\frac{1}{t}$$

$$5.\,\int\frac{1}{quadratic\sqrt{quadratic}}\,dx$$

Then suppose  $$x=\frac{1}{t}\,\, and\,\, solve$$

$$6.\,\int\frac{ax^2+bx+c}{(mx+n)\sqrt{px^2+qx+r}}\,dx$$

Then suppose $$(ax^2+bx+c)=\lambda(mx+n)\frac{d}{dx}(px^2+qx+r)+\mu(mx+n)+\delta$$ $$Find \,\,\lambda,\mu,\delta\,\,\,\, and \,\,\,\,solve$$

=General functions=
 * $$\int\frac{f'(x)}{f(x)}\,dx=lnf(x)+C$$


 * $$\int e^x[f(x)+f'(x)]\,dx = e^xf(x)+C$$


 * $$\int f'(x)e^{f(x)}\,dx = e^{f(x)} + C$$


 * $$\int f(x)+xf'(x)\,dx = xf(x)+C$$

=Some Reductions=

1.$$\int tan^{2n+1}x\,dx= \frac{tan^{2n}x}{2n}-\frac{tan^{2n-2}x}{2n-2}+\frac{tan^{2n-4}x}{2n-4}-....(-1)^nln\left|secx\right|+ C$$ Where $$n=+ve\,\,integer$$

2.$$\int tan^{2n}x\,dx= \frac{tan^{2n-1}x}{2n-1}-\frac{tan^{2n-3}x}{2n-3}+\frac{tan^{2n-5}x}{2n-5}-....(-1)^nx+ C$$ Where $$n=+ve\,\,integer$$

3.$$\int sec^{2n+1}x\,dx=\frac{1}{2n}[sec^{2n-1}x\,tanx+\frac{2n-1}{2n-2}(sec^{2n-3}\,tanx+\frac{2n-3}{2n-4}(sec^{2n-5}\,tanx......)+\frac{3}{2}(secx\,tanx+ln\left|secx+tanx\right|))]+C$$ Where $$n=+ve\,\,integer$$

4.$$\int cot^{2n+1}x\,dx=-\frac{cot^{2n}x}{2n}+\frac{cot^{2n-2}x}{2n-2}- \frac{cot^{2n-4}x}{2n-4}+....(-1)^nln\left|sinx\right|+C$$ Where $$n=+ve\,\,integer$$

5.$$\int cot^{2n}x\,dx=-\frac{cot^{2n-1}x}{2n-1}+\frac{cot^{2n-3}x}{2n-3}- \frac{cot^{2n-5}x}{2n-5}+....(-1)^n[cotx+x]+C$$ Where $$n=+ve\,\,integer$$

= See also =