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 [5]
\[\int\left(\mathrm{u.v}\right)\mathrm{dx}=\mathrm{u}\int\mathrm{v}\mathrm{dx}-\int\left(\frac{\mathrm{du}}{\mathrm{dx}}\right).\left(\int\mathrm{v}\mathrm{dx}\right)\mathrm{dx}\]
Special Result:
∫ eˣ [f(x) + f′(x)] dx = eˣ f(x) + C
| Integral | Result |
|---|---|
| \[\int x^ndx\] | \[\frac{x^{n+1}}{n+1}+C\] |
| \[\int(ax+b)^ndx\] | \[\frac{(ax+b)^{n+1}}{a(n+1)}+C\] |
| \[\int\frac{1}{x}dx\] | \[log/x/+c\] |
| \[\int\frac{1}{ax+b}dx\] | \[\frac{\log\left|ax+b\right|}{a}+c\] |
| \[\int a^xdx\] | \[\frac{a^x}{\log a}+C\] |
| \[\int a^{px+q}dx\] | \[\frac{a^{px+q}}{p\log a}+C\] |
| \[\int e^xdx\] | \[e^{x}+C\] |
| \[\int e^{px+q}dx\] | \[\frac{e^{px+q}}{p}+C\] |
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\[\int\frac{1}{x^2-a^2}dx=\frac{1}{2a}log\left|\frac{x-a}{x+a}\right|+c\]
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\[\int\frac{1}{a^2-x^2}dx=\frac{1}{2a}log\left|\frac{a+x}{a-x}\right|+c\]
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\[\int\frac{1}{\sqrt{x^2+a^2}}dx=log\left|x+\sqrt{x^2+a^2}\right|+c\]
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\[\int\frac{1}{\sqrt{x^2-a^2}}dx=log\left|x+\sqrt{x^2-a^2}\right|+c\]
- \[\int
\begin{bmatrix}
f(x)
\end{bmatrix}^nf^{\prime}(x)dx=\frac{
\begin{bmatrix}
f(x)
\end{bmatrix}^{n+1}}{(n+1)}+c\] -
\[\int\left[\frac{f^{\prime}(x)}{f(x)}\right]dx=\log f(x)+c\]
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\[\int\left[\frac{f^{\prime}(x)}{\sqrt{f(x)}}\right]dx=2\sqrt{f(x)}+c\]
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\[\int\left[\frac{f^{\prime}(x)}{\sqrt[n]{f(x)}}\right]dx=\frac{n\sqrt[n]{\left[f(x)\right]^{n-1}}}{n-1}+c\]
| Type | Rational Form | Partial Fraction Form |
|---|---|---|
| 1 | \[\frac{px\pm q}{(x-a)(x-b)}\] | \[\frac{A}{x-a}+\frac{B}{x-b}\] |
| 2 | \[\frac{px^2\pm qx\pm r}{(x-a)(x-b)(x-c)}\] | \[\frac{A}{x-a}+\frac{B}{x-b}+\frac{C}{x-c}\] |
| 3 | \[\frac{px\pm q}{\left(x-a\right)^2}\] | \[\frac{A}{x-a}+\frac{B}{\left(x-a\right)^2}\] |
| 4 | \[\frac{px^2\pm qx\pm r}{(x-a)^2(x-b)}\] | \[\frac{A}{x-a}+\frac{B}{\left(x-a\right)^2}+\frac{C}{x-b}\] |
| 5 | \[\frac{px^2\pm qx\pm r}{(x-a)^3(x-b)}\] | \[\frac{A}{x-a}+\frac{B}{\left(x-a\right)^2}+\frac{C}{\left(x-a\right)^3}+\frac{D}{x-b}\] |
| 6 | \[\frac{px^2\pm qx\pm r}{(x-a)(ax^2\pmb x\pm c)}\] | \[\frac{A}{x-a}+\frac{Bx+C}{ax^2 \pm b\pmb x\pm c}\] |
Theorems and Laws [3]
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`.
If x = ϕ(t), then \[\int f(x)dx=\int f(\phi(t))\phi^{\prime}(t)dt\]
If u and v are two functions of x, then
\[\int u.vdx=u\int vdx-\int\left[\int vdx.\frac{du}{dx}\right]dx\]
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) | \[\frac{\mathrm{p}x^{2}+\mathrm{q}x+\mathrm{r}}{(x-\mathrm{a})(x^{2}+\mathrm{b}x+\mathrm{c})}\] | \[\frac{\mathrm{A}}{(x-\mathrm{a})}+\frac{\mathrm{B}x+\mathrm{C}}{(x^{2}+\mathrm{b}x+\mathrm{c})}\] |
Sum Rule:
\[\int[f(x)+g(x)]dx=\int f(x)dx+\int g(x)dx\]
Difference Rule:
\[\int[f(x)-g(x)]dx=\int f(x)dx-\int g(x)dx\]
Constant Multiple Rule:
\[\int kf(x)dx=k\int f(x)dx\]
Substitution Result:
If \[\int f(x)dx=F(x)+C\] Then \[\int f(ax+b)dx=\frac{1}{a}F(ax+b)+C\]
L → A → E
-
L = Logarithmic (log x)
-
A = Algebraic (x, x², polynomial)
-
E = Exponential (eˣ)
Important Questions [22]
- If f '(x) = xx1x+x and f(1) = 52, then f(x) = log x + xx22 + ______
- To find the value of ∫(1+logx)xdx the proper substitution is ______.
- Evaluate the following. ∫1x(x6+1) dx
- Evaluate the following. ∫1/(4x^2 - 20x + 17) dx
- Evaluate the following. ∫x4x4-20x2-3dx
- Evaluate the following. ∫17+6x-x2 dx
- If f'(x) = 4x3 − 3x2 + 2x + k, f(0) = 1 and f(1) = 4, find f(x).
- The value of d∫dx1-x is ______.
- ∫(1-x)-2 dx = (1-x)-1+c
- ∫(1-x)-2dx = ______.
- Complete the following activity: ∫02dx4+x-x2 = ∫02dx-x2+□+□ = ∫02dx-x2+x+14-□+4 = ∫02dx(x-12)2-(□)2 = 117log(20+41720-417)
- Evaluate the following. ∫𝑥2 ⋅𝑒3𝑥dx
- ∫1x2-9dx = ______.
- State whether the following statement is true or false. If ∫4ex-252ex-5 dx = Ax – 3 log |2ex – 5| + c, where c is the constant of integration, then A = 5.
- ∫x(x+2)(x+3)dx = ______ + ∫3x+3dx
- Evaluate the following : ∫x3.logx.dx
- Evaluate: ∫exe2x+4ex+13 dx
- ∫(x+1x)3dx = ______.
- For x - 1x + 1exdxex∫x - 1(x + 1)3 exdx=ex f(x) + c, f(x) = (x + 1)2.
- Evaluate: ∫2x+1x(x-1)(x-4)dx.
- Evaluate: ∫x(x-1)2(x+2)dx
- Dxxx∫dx(x-8)(x+7)=
