Formulae [6]
Formula: Integration by Parts
\[\int u\mathrm{~}dv=uv-\int v\mathrm{~}du\]
Formula: Integration by Substitution
If x = φ(t) is a differentiable function of t, then \[\int f\left(x\right)\quad dx=\int f\left[\phi\left(t\right)\right]\phi^{\prime}\left(t\right)dt.\]
| \[\int f\left(ax+b\right)\quad dx=g\left(ax+b\right)\frac{1}{a}+c\] |
| \[\int[f(x)]^n\cdot f^{\prime}(x)dx=\frac{[f(x)]^{n+1}}{n+1}+c\] |
| \[\int\frac{f^{\prime}(x)}{f(x)}dx=\log|f(x)|+C\] |
Formula: Integration by Partial Fractions
| Rational Form | Partial Fraction Form |
|---|---|
| \[\frac{P(x)}{(x-a)(x-b)(x-c)}\] | \[\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}\] |
| \[\frac{P(x)}{(x-a)^2(x-b)}\] | \[\frac{A}{x-a}+\frac{B}{(x-a)^2}+\frac{C}{x-b}\] |
| \[\frac{P(x)}{(x-a)(x^2+bx+c)}\] | \[\frac{A}{x-a}+\frac{Bx+C}{x^2+bx+c}\] |
| \[\frac{P(x)}{(x^2+bx+c)^2}\] | \[\frac{Ax+B}{x^2+bx+c}+\frac{Cx+D}{(x^2+bx+c)^2}\] |
Formula: Elementary Integration Formulae
| Function | Integral |
|---|---|
| \[x^{n}\] | \[\frac{x^{n+1}}{n+1}+C\] |
| $$\int (ax + b)^n dx$$ | $$\int (ax + b)^n dx$$\[\frac{x^{n+1}}{n+1}+C\] |
| \[\int a^xdx\] | \[\frac{a^x}{\log a}+C\] |
| \[\int A^{ax+b}dx\] | \[\frac{A^{ax+b}}{a\log A}+C\] |
| \[\int e^xdx\] | \[e^{x}+C\] |
| \[\int e^{ax+b}dx\] | \[\frac{e^{ax+b}}{a}+C\] |
| \[\int\cos xdx\] | \[\sin x+C\] |
| \[\int\cos(ax+b)dx\] | \[\frac{\sin(ax+b)}{a}+C\] |
| $$\int \sin x dx$$ | \[-\cos x+C\] |
| \[\int\sin(ax+b)dx\] | \[-\frac{\cos(ax+b)}{a}+C\] |
| \[\int\sec^2xdx\] | \[\tan x+C\] |
| \[\int\sec^2(ax+b)dx\] | \[\frac{\tan(ax+b)}{a}+C\] |
| $$\int \sec x \tan x dx$$ | \[\sec x+C\] |
| \[\int\sec(ax+b)\tan(ax+b)dx\] | \[\frac{\sec(ax+b)}{a}+C\] |
| \[\int\operatorname{cosec}x\cot xdx\] | \[-\operatorname{cosec}x+C\] |
| \[\int\operatorname{cosec}(ax+b)\cot(ax+b)dx\] | \[-\frac{\mathrm{cosec}(ax+b)}{a}+C\] |
| \[\int\mathrm{cosec}^2xdx\] | \[-\cot x+C\] |
| \[\int\mathrm{cosec}^2(ax+b)dx\] | \[-\frac{\cot(ax+b)}{a}+C\] |
| \[\int\frac{1}{x}dx\] | \[\log x+C\] |
| \[\int\frac{1}{ax+b}dx\] | \[\frac{1}{a}\log(ax+b)+C\] |
Formula: Standard Substitutions
| Form in integral | Best substitution |
|---|---|
| \[\sqrt{a^2-x^2}\] | (or a cosθ) |
| \[\sqrt{a^2+x^2}\] | x = a tanθ |
| \[\sqrt{x^2-a^2}\] | x = a secθ |
| \[\sqrt{\frac{a-x}{a+x}}\] | x = a cos2θ |
Formula: Some Special Integrals
| No. | Standard Integral | Result |
|---|---|---|
| 1 | \[\int\frac{1}{x^2+a^2}dx\] | \[\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)+C\] |
| 2 | \[\int\frac{1}{x^2-a^2}dx\] | \[\frac{1}{2a}\log\left(\frac{x-a}{x+a}\right)+c\] |
| 3 | \[\int\frac{1}{a^2-x^2}\quad dx\] | \[\frac{1}{2a}\log\left(\frac{a+x}{a-x}\right)+C\] |
| 4 | \[\int\frac{1}{\sqrt{a^2-x^2}}\quad dx\] | \[\sin^{-1}\left(\frac{X}{a}\right)+c\] |
| 5 | \[\int\frac{1}{\sqrt{x^2-a^2}}dx\] | \[\log\left(x+\sqrt{x^{2}-a^{2}}\right)+c\] |
| 6 | \[\int\frac{1}{\sqrt{a^2+x^2}}\quad dx\] | \[\log\left(x+\sqrt{a^{2}+x^{2}}\right)+c\] |
| 7 | \[\int\frac{1}{x\sqrt{x^2-a^2}}dx\] | \[\frac{1}{a}\sec^{-1}\left(\frac{x}{a}\right)+c\] |
Theorems and Laws [1]
Prove that: `int sqrt(a^2 - x^2) * dx = x/2 * sqrt(a^2 - x^2) + a^2/2 * sin^-1(x/a) + c`
Let I = `int sqrt(a^2 - x^2) dx`
= `int sqrt(a^2 - x^2)*1 dx`
= `sqrt(a^2 - x^2)* int 1 dx - int [d/dx (sqrt(a^2 - x^2))* int 1 dx]dx`
= `sqrt(a^2 - x^2)*x - int [1/(2sqrt(a^2 - x^2))*d/dx (a^2 - x^2)*x]dx`
= `sqrt(a^2 - x^2)*x - int 1/(2sqrt(a^2 - x^2))(0 - 2x)*x dx`
= `sqrt(a^2 - x^2)*x - int (-x)/sqrt(a^2 - x^2)*x dx`
= `xsqrt(a^2 - x^2) - int (a^2 - x^2 - a^2)/sqrt(a^2 - x^2)dx`
= `xsqrt(a^2 - x^2) - int sqrt(a^2 - x^2)dx + a^2 int dx/sqrt(a^2 - x^2)`
= `xsqrt(a^2 - x^2) - I + a^2sin^-1(x/a) + c_1`
∴ 2I = `xsqrt(a^2 - x^2) + a^2sin^-1(x/a) + c_1`
∴ I = `x/2 sqrt(a^2 - x^2) + a^2/2 sin^-1(x/a) + c_1/2`
∴ `int sqrt(a^2 - x^2)dx = x/2 sqrt(a^2 - x^2) + a^2/2sin^-1(x/a) + c`, where `c = c_1/2`.
Key Points
Key Points: LIATE Rule
| Order | Function Type |
|---|---|
| L | Logarithmic |
| I | Inverse Trigonometric |
| A | Algebraic |
| T | Trigonometric |
| E | Exponential |
Important Questions [39]
- Evaluate : ∫sinx/√(36−cos^2x)dx
- Evaluate : ∫π0 x/(a^2cos^2 x+b^2 sin^2 x)dx
- Show that: ∫1/(x^2sqrt(a^2+x^2))dx=-1/a^2(sqrt(a^2+x^2)/x)+c
- Evaluate :∫x logx dx
- Prove that int_a^bf(x)dx=f(a+b-x)dx
- Evaluate: ∫tanxsinxcosxdx
- Evaluate : ∫1/(3+2sinx+cosx)dx
- Evaluate Integral 1/(X(X-1))Dx
- Solve: dy/dx = cos(x + y)
- Evaluate `Integration 1/(3+ 2 Sinx + Cosx) Dx`
- Int"Dx"/9x2+1=
- Evaluate:
- Prove That:
- Show that
- Integrate the following function w.r.t. x: x9.sec2(x10)
- Evaluate the following: ∫125-9x2⋅dx
- Evaluate the following : ∫sinxsin3x.dx
- int sqrt(1 + sin2x) dx
- ∫cos3x dx = ______.
- Write ∫cotx dx.
- Evaluate the following: ∫x2sin3x dx
- Prove that: int sqrt(a^2 – x^2) * dx = x/2 * sqrt(a^2 – x^2) + a^2/2 * sin^–1(x/a) + c
- Integrate : sec^3x w. r. t. x.
- Prove that: ∫x2-a2dx=x2x2-a2-a22log|x+x2-a2|+c
- If ∫π/2 −π/2 sin^4 x/ (sin^4 x+cos^4 x)dx, then the value of I is:
- Evaluate: ∫ x tan^-1 x dx
- Integrate the following functions w.r.t. x : esin-1x.[x+1-x21-x2]
- Evaluate: ∫1x2+25dx
- ∫1/x logxdx=
- If U and V Are Two Functions of X Then Prove that ∫Uvdx=U∫Vdx−∫ Du/Dx∫Vdx Dx
- If u and v ore differentiable functions of x. then prove that: ∫uv dx=u∫v dx-∫[dud∫v dx]dx Hence evaluate ∫logx dx
- Prove that: ∫x2+a2dx=x2x2+a2+a22log|x+x2+a2|+c
- Evaluate: ∫5ex(ex+1)(e2x+9)dx
- Evaluate: ∫8/((x+2)(x^2+4))dx
- Evaluate : ∫(x+1)/((x+2)(x+3))dx
- Evaluate: ∫dx2+cosx-sinx
- Evaluate: int (2x^2 - 3)/((x^2 - 5)(x^2 + 4))dx
- Evaluate : ∫x^2/((x^2+2)(2x^2+1))dx
- Find: I=intdx/(sinx+sin2x)
