Definitions [1]
Integration by substitution is a method in which we replace a part of the integral by a new variable to simplify the integration.
Formulae [8]
| f(x) | ∫ f(x) dx |
|---|---|
| xⁿ | \[\frac{x^{\mathrm{n+1}}}{\mathrm{n+1}}\], n ≠ -1 |
| \[\frac{1}{x}\] | log |
| eˣ | eˣ |
| aˣ | \[\frac{a^x}{\log a}\] (a ≠ 1, a > 0) |
| log x | x(log x − 1) |
| sin x | −cos x |
| cos x | sin x |
| sec² x | tan x |
| cosec² x | −cot x |
| sec x tan x | sec x |
| cosec x cot x | −cosec x |
| tan x | \[-\log|\cos x|\] or \[\log\left|\sec x\right|\] |
| cot x | \[\log|\sin x|\] or \[-\log|\operatorname{cosec}x|\] |
| sec x | \[\log|\sec x+\tan x|\] or \[\log\left|\tan\left(\frac{\pi}{4}+\frac{x}{2}\right)\right|\] |
| cosec x | \[\log|\operatorname{cosec}x-\cot x|\] or \[\log\left|\tan\frac{x}{2}\right|\] |
| \[\frac{1}{\sqrt{1-x^2}}\] | sin⁻¹ x or cos⁻¹ x |
| \[\frac{1}{1+x^2}\] | tan⁻¹ x or −cot⁻¹ x |
| \[\frac{1}{|x|\sqrt{x^2-1}}\] | \[sec^{-1}x\] or \[-cosec^{-1}x\] |
| \[\frac{1}{x^2+a^2}\] | \[\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right)\] |
| \[\frac{1}{x^2-a^2}\] | \[\frac{1}{2\mathrm{a}}\mathrm{log}\left|\frac{x-\mathrm{a}}{x+\mathrm{a}}\right|\] |
| \[\frac{1}{\mathbf{a}^2-x^2}\] | \[\frac{1}{2\mathrm{a}}\mathrm{log}\left|\frac{\mathrm{a}+x}{\mathrm{a}-x}\right|\] |
| \[\frac{1}{\sqrt{x^2+a^2}}\] | \[\log\left|x+\sqrt{x^{2}+\mathrm{a}^{2}}\right|\] |
| \[\frac{1}{\sqrt{x^2-a^2}}\] | \[\log\left|x+\sqrt{x^{2}-a^{2}}\right|\] |
| \[\frac{1}{\sqrt{a^2-x^2}}\] | \[\sin^{-1}\left(\frac{x}{a}\right)\] |
| \[\frac{1}{x\sqrt{x^2-a^2}}\] | \[\frac{1}{a}\sec^{-1}\left(\frac{x}{a}\right)\] |
| \[\sqrt{a^{2}-x^{2}}\] | \[\frac{x}{2}\sqrt{\mathrm{a}^{2}-x^{2}}+\frac{\mathrm{a}^{2}}{2}\sin^{-1}\left(\frac{x}{\mathrm{a}}\right)\] |
| \[\sqrt{x^2+a^2}\] | \[\frac{x}{2}\sqrt{x^{2}+\mathrm{a}^{2}}+\frac{\mathrm{a}^{2}}{2}\log\left|x+\sqrt{x^{2}+\mathrm{a}^{2}}\right|\] |
| \[\sqrt{x^{2}-a^{2}}\] | \[\frac{x}{2}\sqrt{x^{2}-\mathrm{a}^{2}}-\frac{\mathrm{a}^{2}}{2}\log\left|x+\sqrt{x^{2}-\mathrm{a}^{2}}\right|\] |
\[\int\left(\mathrm{u.v}\right)\mathrm{dx}=\mathrm{u}\int\mathrm{v}\mathrm{dx}-\int\left(\frac{\mathrm{du}}{\mathrm{dx}}\right).\left(\int\mathrm{v}\mathrm{dx}\right)\mathrm{dx}\]
Special Result:
∫ eˣ [f(x) + f′(x)] dx = eˣ f(x) + C
\[\int u\mathrm{~}dv=uv-\int v\mathrm{~}du\]
If x = φ(t) is a differentiable function of t, then \[\int f\left(x\right)\quad dx=\int f\left[\phi\left(t\right)\right]\phi^{\prime}\left(t\right)dt.\]
| \[\int f\left(ax+b\right)\quad dx=g\left(ax+b\right)\frac{1}{a}+c\] |
| \[\int[f(x)]^n\cdot f^{\prime}(x)dx=\frac{[f(x)]^{n+1}}{n+1}+c\] |
| \[\int\frac{f^{\prime}(x)}{f(x)}dx=\log|f(x)|+C\] |
| Rational Form | Partial Fraction Form |
|---|---|
| \[\frac{P(x)}{(x-a)(x-b)(x-c)}\] | \[\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}\] |
| \[\frac{P(x)}{(x-a)^2(x-b)}\] | \[\frac{A}{x-a}+\frac{B}{(x-a)^2}+\frac{C}{x-b}\] |
| \[\frac{P(x)}{(x-a)(x^2+bx+c)}\] | \[\frac{A}{x-a}+\frac{Bx+C}{x^2+bx+c}\] |
| \[\frac{P(x)}{(x^2+bx+c)^2}\] | \[\frac{Ax+B}{x^2+bx+c}+\frac{Cx+D}{(x^2+bx+c)^2}\] |
| Function | Integral |
|---|---|
| \[x^{n}\] | \[\frac{x^{n+1}}{n+1}+C\] |
| $$\int (ax + b)^n dx$$ | $$\int (ax + b)^n dx$$\[\frac{x^{n+1}}{n+1}+C\] |
| \[\int a^xdx\] | \[\frac{a^x}{\log a}+C\] |
| \[\int A^{ax+b}dx\] | \[\frac{A^{ax+b}}{a\log A}+C\] |
| \[\int e^xdx\] | \[e^{x}+C\] |
| \[\int e^{ax+b}dx\] | \[\frac{e^{ax+b}}{a}+C\] |
| \[\int\cos xdx\] | \[\sin x+C\] |
| \[\int\cos(ax+b)dx\] | \[\frac{\sin(ax+b)}{a}+C\] |
| $$\int \sin x dx$$ | \[-\cos x+C\] |
| \[\int\sin(ax+b)dx\] | \[-\frac{\cos(ax+b)}{a}+C\] |
| \[\int\sec^2xdx\] | \[\tan x+C\] |
| \[\int\sec^2(ax+b)dx\] | \[\frac{\tan(ax+b)}{a}+C\] |
| $$\int \sec x \tan x dx$$ | \[\sec x+C\] |
| \[\int\sec(ax+b)\tan(ax+b)dx\] | \[\frac{\sec(ax+b)}{a}+C\] |
| \[\int\operatorname{cosec}x\cot xdx\] | \[-\operatorname{cosec}x+C\] |
| \[\int\operatorname{cosec}(ax+b)\cot(ax+b)dx\] | \[-\frac{\mathrm{cosec}(ax+b)}{a}+C\] |
| \[\int\mathrm{cosec}^2xdx\] | \[-\cot x+C\] |
| \[\int\mathrm{cosec}^2(ax+b)dx\] | \[-\frac{\cot(ax+b)}{a}+C\] |
| \[\int\frac{1}{x}dx\] | \[\log x+C\] |
| \[\int\frac{1}{ax+b}dx\] | \[\frac{1}{a}\log(ax+b)+C\] |
| 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 |
| 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
| Sr. No. | Integrand Form | Substitution |
|---|---|---|
| i | \[\sqrt{\mathrm{a}^2-x^2},\frac{1}{\sqrt{\mathrm{a}^2-x^2}},\mathrm{a}^2-x^2\] | x = a sinθ or x = a cosθ |
| ii | \[\sqrt{x^2+\mathrm{a}^2},\frac{1}{\sqrt{x^2+\mathrm{a}^2}},x^2+\mathrm{a}^2\] | x = a tanθ |
| iii | \[\sqrt{x^{2}-a^{2}},\frac{1}{\sqrt{x^{2}-a^{2}}},x^{2}-a^{2}\] | x = a secθ |
| iv | \[\sqrt{\frac{x}{a+x}},\sqrt{\frac{a+x}{x}},\]\[\sqrt{x(a+x)},\frac{1}{\sqrt{x(a+x)}}\] | x = a tan²θ |
| v | \[\sqrt{\frac{x}{a-x}},\sqrt{\frac{a-x}{x}},\]\[\sqrt{x(a-x)},\frac{1}{\sqrt{x(a-x)}}\] | x = a sin²θ |
| vi | \[\sqrt{\frac{x}{x-a}},\sqrt{\frac{x-a}{x}},\]\[\sqrt{x(x-\mathrm{a})},\frac{1}{\sqrt{x(x-\mathrm{a})}}\] | x = a sec²θ |
| vii | \[\sqrt{\frac{\mathrm{a}-x}{\mathrm{a}+x}},\sqrt{\frac{\mathrm{a}+x}{\mathrm{a}-x}}\] | x = a cos 2θ |
| viii | \[\sqrt{\frac{x-\alpha}{\beta-x}},\sqrt{(x-\alpha)(\beta-x)},\]\[(\beta>\alpha)\] | x = α cos²θ + β sin²θ |
First function should be chosen in the following order of preference:
L → Logarithmic function
I → Inverse trigonometric function
A → Algebraic function
T → Trigonometric function
E → Exponential function
Note:
For the integration of logarithmic or inverse trigonometric functions alone, take unity (1) as the second function.
Standard forms:
i) \[\int\sqrt{x^{2}+a^{2}}dx=\frac{1}{2}\left[ \begin{array} {c}{x\sqrt{x^{2}+a^{2}}} {+a^{2}\log|x+\sqrt{x^{2}+a^{2}|}} \end{array}\right]+C\]
ii) \[\int\sqrt{a^{2}-x^{2}}dx=\frac{1}{2}\left[x\sqrt{a^{2}-x^{2}}+a^{2}\sin^{-1}\left(\frac{x}{a}\right)\right]+C\]
iii) \[\int\sqrt{x^{2}-a^{2}}dx=\frac{1}{2}[x\sqrt{x^{2}-a^{2}}-a^{2}\log|x+\sqrt{x^{2}-a^{2}}|]\] + C
| Type | Rational Form | Partial Form |
|---|---|---|
| Type I (Non-repeated linear factors) | \[\frac{\mathrm{p}x+\mathrm{q}}{(x-\mathrm{a})(x-\mathrm{b})}\] | \[\frac{\mathrm{A}}{x-\mathrm{a}}+\frac{\mathrm{B}}{x-\mathrm{b}}\] |
| \[\frac{\mathrm{p}x^{2}+\mathrm{q}x+\mathrm{r}}{(x-\mathrm{a})(x-\mathrm{b})(x-\mathrm{c})}\] | \[\frac{\mathrm{A}}{x-\mathrm{a}}+\frac{\mathrm{B}}{x-\mathrm{b}}+\frac{\mathrm{C}}{x-\mathrm{c}}\] | |
| Type II (Repeated linear factors) | \[\frac{\mathrm{p}x+\mathrm{q}}{\left(x-\mathrm{a}\right)^2}\] | \[\frac{\mathrm{A}}{(x-\mathrm{a})}+\frac{\mathrm{B}}{(x-\mathrm{a})^{2}}\] |
| \[\frac{\mathrm{p}x^{2}+\mathrm{q}x+\mathrm{r}}{\left(x-\mathrm{a}\right)^{2}\left(x-\mathrm{b}\right)}\] | \[\frac{\mathrm{A}}{(x-\mathrm{a})}+\frac{\mathrm{B}}{(x-\mathrm{a})^{2}}+\frac{\mathrm{C}}{(x-\mathrm{b})}\] | |
| Type III (Linear × Quadratic) | \[\frac{\mathrm{p}x^{2}+\mathrm{q}x+\mathrm{r}}{(x-\mathrm{a})(x^{2}+\mathrm{b}x+\mathrm{c})}\] | \[\frac{\mathrm{A}}{(x-\mathrm{a})}+\frac{\mathrm{B}x+\mathrm{C}}{(x^{2}+\mathrm{b}x+\mathrm{c})}\] |
| 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: int sinx/(sin 3x) . 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)
