Indefinite Integrals
1.0 Introduction
1.0 Introduction
If $f(x)$ and $g(x)$ are two different functions of $x$ in such a way that $$g'(x) = f(x) \Rightarrow \frac{{d\left\{ {g(x)} \right\}}}{{dx}} = f(x)$$ then symbolically, we define the indefinite integration of $f(x)$ with respect to $x$ as $$\int {f(x)dx = g(x) + C} $$ where $C$ is the integration constant.
- $\int {cf(x)dx = c\int {f(x)dx\quad (c{\text{ is constant)}}} } $
- $\int {\left[ {f(x) \pm g(x)} \right]} dx = \int {f(x)dx \pm \int {g(x)} dx} $
- $\int {f(x)dx = g(x) + C} \Rightarrow \int {f(px + q)dx} = \frac{{g(px + q)}}{p} + C'$
The following standard integration results of various functions are obtained using the standard differentiation.
| S.No | Differentiation | Integration |
| 1. | $$\frac{d}{{dx}}\left( {\frac{{{x^{n + 1}}}}{{n + 1}}} \right) = {x^n}$$ | $$\int {{x^n}dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C} $$ |
| 2. | $$\frac{d}{{dx}}\left\{ {\frac{{{{\left( {ax + b} \right)}^{n + 1}}}}{{a\left( {n + 1} \right)}}} \right\} = {\left( {ax + b} \right)^n}$$ | $$\int {{{\left( {ax + b} \right)}^n}} dx = \frac{{{{\left( {ax + b} \right)}^{n + 1}}}}{{a\left( {n + 1} \right)}} + C$$ |
| 3. | $$\frac{d}{{dx}}\left( {\log \left| x \right|} \right) = \frac{1}{x}$$ | $$\int {\frac{1}{x}} dx = \log \left| x \right| + C$$ |
| 4. | $$\frac{d}{{dx}}\left( {\frac{1}{a}\ln \left| {ax + b} \right|} \right) = \frac{1}{{ax + b}}$$ | $$\int {\frac{{dx}}{{ax + b}}} = \frac{1}{a}\ln \left| {ax + b} \right| + C$$ |
| 5. | $$\frac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$$ | $$\int {{e^x}dx} = {e^x} + C$$ |
| 6. | $$\frac{d}{{dx}}\left( {\frac{1}{a}{e^{ax + b}}} \right) = {e^{ax + b}}$$ | $$\int {{e^{ax + b}}dx} = \frac{1}{a}{e^{ax + b}} + C$$ |
| 7. | $$\frac{d}{{dx}}\left( {\frac{{{a^x}}}{{{{\log }_e}a}}} \right) = {a^x}$$ | $$\int {{a^x}} dx = \frac{{{a^x}}}{{{{\log }_e}a}} + C$$ |
| 8. | $$\frac{d}{{dx}}\left( {\frac{1}{p}\frac{{{a^{px + q}}}}{{\ln a}}} \right) = {a^{px + q}}$$ | $$\int {{a^{px + q}}dx} = \frac{1}{p}\frac{{{a^{px + q}}}}{{\ln a}} + C$$ |
| 9. | $$\frac{d}{{dx}}\left( { - \cos x} \right) = \sin x$$ | $$\int {\sin xdx} = - \cos x + C$$ |
| 10. | $$\frac{d}{{dx}}\left( { - \frac{1}{a}\cos (ax + b)} \right) = \sin (ax + b)$$ | $$\int {\sin (ax + b)dx} = - \frac{1}{a}\cos (ax + b) + C$$ |
| 11. | $$\frac{d}{{dx}}\left( {\sin x} \right) = \cos x$$ | $$\int {\cos xdx = \sin x + C} $$ |
| 12. | $$\frac{d}{{dx}}\left( {\frac{1}{a}\sin (ax + b)} \right) = \cos (ax + b)$$ | $$\int {\cos (ax + b)dx = \frac{1}{a}\sin (ax + b) + C} $$ |
| 13. | $$\frac{d}{{dx}}(\tan x) = {\sec ^2}x$$ | $$\int {{{\sec }^2}x} dx = \tan x + C$$ |
| 14. | $$\frac{d}{{dx}}\left( {\frac{1}{a}tan(ax + b)} \right) = {\sec ^2}(ax + b)$$ | $$\int {{{\sec }^2}(ax + b)dx = \frac{1}{a}tan(ax + b) + C} $$ |
| 15. | $$\frac{d}{{dx}}( - \cot x) = {\operatorname{cosec} ^2}x$$ | $$\int {{{\operatorname{cosec} }^2}x} dx = - \cot x + C$$ |
| 16. | $$\frac{d}{{dx}}\left( { - \frac{1}{a}cot(ax + b)} \right) = {\operatorname{cosec} ^2}(ax + b)$$ | $$\int {{{\operatorname{cosec} }^2}(ax + b)dx = - \frac{1}{a}cot(ax + b) + C} $$ |
| 17. | $$\frac{d}{{dx}}(secx) = \sec x\tan x$$ | $$\int {\sec x\tan x} dx = \sec x + C$$ |
| 18. | $$\frac{d}{{dx}}\left( {\frac{1}{a}sec(ax + b)} \right) = \sec (ax + b)tan(ax + b)$$ | $$\int {\sec (ax + b)tan(ax + b)dx = \frac{1}{a}sec(ax + b) + C} $$ |
| 19. | $$\frac{d}{{dx}}( - cosecx) = \operatorname{cosec} x\cot x$$ | $$\int {\operatorname{cosec} x\cot x} dx = - cosecx + C$$ |
| 20. | $$\frac{d}{{dx}}\left( { - \frac{1}{a}\operatorname{cosec} (ax + b)} \right) = cosec(ax + b)cot(ax + b)$$ | $$\int {cosec(ax + b)cot(ax + b)dx = - \frac{1}{a}\operatorname{cosec} (ax + b) + C} $$ |
| 21. | $$\frac{d}{{dx}}(\log \left| {\sin x} \right|) = \cot x$$ | $$\int {\cot x} dx = \log \left| {\sin x} \right| + C$$ |
| 22. | $$\frac{d}{{dx}}\left( {\frac{1}{a}\ln \left| {\sin (ax + b)} \right|} \right) = cot(ax + b)$$ | $$\int {cot(ax + b)dx = \frac{1}{a}\ln \left| {\sin (ax + b)} \right| + C} $$ |
| 23. | $$\frac{d}{{dx}}( - \log \left| {\cos x} \right|) = \tan x$$ | $$\int {\tan x} dx = - \log \left| {\cos x} \right| + C = \log \left| {\sec x} \right| + C$$ |
| 24. | $$\frac{d}{{dx}}\left( { - \frac{1}{a}\ln \left| {cos(ax + b)} \right|} \right) = \tan (ax + b)$$ | $$\int {tan(ax + b)dx = - \frac{1}{a}\ln \left| {cos(ax + b)} \right| + C = \frac{1}{a}\ln \left| {sec(ax + b)} \right| + C} $$ |
| 25. | $$\frac{d}{{dx}}(\log \left| {\sec x + \tan x} \right|) = \sec x$$ | $$\int {\sec x} dx = \log \left| {\sec x + \tan x} \right| + C$$ |
| 26. | $$\frac{d}{{dx}}(\log \left| {cosecx - \cot x} \right|) = cosecx$$ | $$\int {cosecx} dx = \log \left| {cosecx - \cot x} \right| + C$$ |
| 27. | $$\frac{d}{{dx}}\left( {{{\sin }^{ - 1}}\frac{x}{a}} \right) = \frac{1}{{\sqrt {{a^2} - {x^2}} }}$$ | $$\int {\frac{{dx}}{{\sqrt {{a^2} - {x^2}} }} = {{\sin }^{ - 1}}\frac{x}{a} + C} $$ |
| 28. | $$\frac{d}{{dx}}\left( {{{\cos }^{ - 1}}\frac{x}{a}} \right) = \frac{{ - 1}}{{\sqrt {{a^2} - {x^2}} }}$$ | $$\int {\frac{{ - 1}}{{\sqrt {{a^2} - {x^2}} }}dx = {{\cos }^{ - 1}}\frac{x}{a} + C} $$ |
| 29. | $$\frac{d}{{dx}}\left( {\frac{1}{a}{{\tan }^{ - 1}}\frac{x}{a}} \right) = \frac{1}{{{a^2} + {x^2}}}$$ | $$\int {\frac{1}{{{a^2} + {x^2}}}dx = \frac{1}{a}{{\tan }^{ - 1}}\frac{x}{a} + C} $$ |
| 30. | $$\frac{d}{{dx}}\left( {\frac{1}{a}{{\cot }^{ - 1}}\frac{x}{a}} \right) = \frac{{ - 1}}{{{a^2} + {x^2}}}$$ | $$\int {\frac{{ - 1}}{{{a^2} + {x^2}}}dx = \frac{1}{a}{{\cot }^{ - 1}}\frac{x}{a} + C} $$ |
| 31. | $$\frac{d}{{dx}}\left( {\frac{1}{a}{{\sec }^{ - 1}}\frac{x}{a}} \right) = \frac{1}{{x\sqrt {{x^2} - {a^2}} }}$$ | $$\int {\frac{1}{{x\sqrt {{x^2} - {a^2}} }}dx = \frac{1}{a}{{\sec }^{ - 1}}\frac{x}{a} + C} $$ |
| 32. | $$\frac{d}{{dx}}\left( {\frac{1}{a}cose{c^{ - 1}}\frac{x}{a}} \right) = \frac{{ - 1}}{{x\sqrt {{x^2} - {a^2}} }}$$ | $$\int {\frac{{ - 1}}{{x\sqrt {{x^2} - {a^2}} }}dx = \frac{1}{a}cose{c^{ - 1}}\frac{x}{a} + C} $$ |
| 33. | $$\frac{d}{{dx}}\left( {\ln \left| {x + \sqrt {{x^2} + {a^2}} } \right|} \right) = \frac{1}{{\sqrt {{x^2} + {a^2}} }}$$ | $$\int {\frac{1}{{\sqrt {{x^2} + {a^2}} }}dx = \ln \left| {x + \sqrt {{x^2} + {a^2}} } \right| + C} $$ |
| 34. | $$\frac{d}{{dx}}\left( {\ln \left| {x + \sqrt {{x^2} - {a^2}} } \right|} \right) = \frac{1}{{\sqrt {{x^2} - {a^2}} }}$$ | $$\int {\frac{1}{{\sqrt {{x^2} - {a^2}} }}dx = \ln \left| {x + \sqrt {{x^2} - {a^2}} } \right| + C} $$ |
| 35. | $$\frac{d}{{dx}}\left( {\frac{1}{{2a}}\ln \left| {\frac{{a + x}}{{a - x}}} \right|} \right) = \frac{1}{{{a^2} - {x^2}}}$$ | $$\int {\frac{1}{{{a^2} - {x^2}}}dx = \frac{1}{{2a}}\ln \left| {\frac{{a + x}}{{a - x}}} \right| + C} $$ |
| 36. | $$\frac{d}{{dx}}\left( {\frac{1}{{2a}}\ln \left| {\frac{{x + a}}{{x - a}}} \right|} \right) = \frac{1}{{{x^2} - {a^2}}}$$ | $$\int {\frac{1}{{{x^2} - {a^2}}}dx = \frac{1}{{2a}}\ln \left| {\frac{{x + a}}{{x - a}}} \right| + C} $$ |
| 37. | $$\frac{d}{{dx}}\left( {\frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\frac{x}{a}} \right) = \sqrt {{a^2} - {x^2}} $$ | $$\int {\sqrt {{a^2} - {x^2}} dx = \frac{x}{2}\sqrt {{a^2} - {x^2}} + \frac{{{a^2}}}{2}{{\sin }^{ - 1}}\frac{x}{a} + C} $$ |
| 38. | $$\frac{d}{{dx}}\left( {\frac{x}{2}\sqrt {{x^2} + {a^2}} + \frac{{{a^2}}}{2}\ln \left| {\frac{{x + \sqrt {{x^2} + {a^2}} }}{a}} \right|} \right) = \sqrt {{x^2} + {a^2}} $$ | $$\int {\sqrt {{x^2} + {a^2}} dx = \frac{x}{2}\sqrt {{x^2} + {a^2}} + \frac{{{a^2}}}{2}\ln \left| {\frac{{x + \sqrt {{x^2} + {a^2}} }}{a}} \right| + C} $$ |
| 39. | $$\frac{d}{{dx}}\left( {\frac{x}{2}\sqrt {{x^2} - {a^2}} - \frac{{{a^2}}}{2}\ln \left| {\frac{{x + \sqrt {{x^2} - {a^2}} }}{a}} \right|} \right) = \sqrt {{x^2} - {a^2}} $$ | $$\int {\sqrt {{x^2} - {a^2}} dx = \frac{x}{2}\sqrt {{x^2} - {a^2}} - \frac{{{a^2}}}{2}\ln \left| {\frac{{x + \sqrt {{x^2} - {a^2}} }}{a}} \right| + C} $$ |
Question 1. Evaluate
(i) $\int {\left( {{x^3} + 5{x^2} - 4 + \frac{7}{x} + \frac{2}{{\sqrt x }}} \right)} dx$
(ii) $\int {\frac{{{x^4}}}{{{x^2} + 1}}dx} $
(iii) $\int {\frac{{{{\sin }^6}x + {{\cos }^6}x}}{{{{\sin }^2}x{{\cos }^2}x}}} dx$
(iv) $\int {\cos 4x\cos 7x} dx$
Solution:
(i)$$ \int {\left( {{x^3} + 5{x^2} - 4 + \frac{7}{x} + \frac{2}{{\sqrt x }}} \right)} dx $$ $$ = \int {{x^3}} dx + \int {5{x^2}} dx - \int {4dx} + \int {\frac{7}{x}dx} + \int {\frac{2}{{\sqrt x }}dx} $$ $$ = \int {{x^3}} dx + 5\int {{x^2}} dx - 4\int {dx} + 7\int {\frac{{dx}}{x}} + 2\int {\frac{{dx}}{{\sqrt x }}} $$ $$ = \frac{{{x^4}}}{4} + 5.\frac{{{x^3}}}{3} - 4x + 7\ln \left| x \right| + 2\left( {\frac{{{x^{\frac{1}{2}}}}}{{1/2}}} \right) + C $$ $$ = \frac{{{x^4}}}{4} + 5.\frac{{{x^3}}}{3} - 4x + 7\ln \left| x \right| + 4\sqrt x + C $$
(ii)$$ \int {\frac{{{x^4}}}{{{x^2} + 1}}dx} $$ $$ = \int {\frac{{{x^4} - 1 + 1}}{{{x^2} + 1}}dx} $$ $$ = \int {\left( {\frac{{{x^4} - 1}}{{{x^2} + 1}} + \frac{1}{{{x^2} + 1}}} \right)} dx $$ $$ = \int {\frac{{{{\left( {{x^2}} \right)}^2} - {{\left( 1 \right)}^2}}}{{{x^2} + 1}}dx} + \int {\frac{{dx}}{{{x^2} + 1}}} $$ $$ = \int {\frac{{({x^2} + 1)({x^2} - 1)}}{{{x^2} + 1}}} dx + \int {\frac{{dx}}{{{x^2} + 1}}} $$ $$ = \int {({x^2} - 1)dx} + \int {\frac{{dx}}{{{x^2} + 1}}} $$ $$ = \frac{{{x^3}}}{3} - x + {\tan ^{ - 1}}x + C $$
(iii)$$ \int {\frac{{{{\sin }^6}x + {{\cos }^6}x}}{{{{\sin }^2}x{{\cos }^2}x}}} dx $$ $$ = \int {\frac{{{{\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}^3} - 3{{\sin }^2}x{{\cos }^2}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)}}{{{{\sin }^2}x{{\cos }^2}x}}dx} $$ $$ = \int {\frac{{1 - 3{{\sin }^2}x{{\cos }^2}x}}{{{{\sin }^2}x{{\cos }^2}x}}} dx $$ $$ = \int {\frac{{dx}}{{{{\sin }^2}x{{\cos }^2}x}} - 3\int {dx} } $$ $$ = \int {\frac{{({{\sin }^2}x + {{\cos }^2}x)}}{{{{\sin }^2}x{{\cos }^2}x}}dx} - 3x + C $$ $$ = \int {\frac{{dx}}{{{{\cos }^2}x}}} + \int {\frac{{dx}}{{{{\sin }^2}x}}} - 3x + C $$ $$ = \int {{{\sec }^2}x} dx + \int {{\text{cose}}{{\text{c}}^2}x} dx - 3x + C $$ $$ = \tan x - \cot x - 3x + C $$
(iv)$$ \int {\cos 4x\cos 7x} dx $$ $$ = \frac{1}{2}\int {\left[ {\cos \left( {4 - 7} \right)x + \cos \left( {4 + 7} \right)x} \right]} $$ $$ = \frac{1}{2}\int {\left( {\cos 3x + \cos 11x} \right)} dx $$ $$ = \frac{1}{2}\int {\cos 3xdx} + \frac{1}{2}\int {\cos 11xdx} $$ $$ = \frac{{\sin 3x}}{{2 \times 3}} + \frac{{\sin 11x}}{{2 \times 11}} + C $$ $$ = \frac{{\sin 3x}}{6} + \frac{{\sin 11x}}{{22}} + C $$
Note: In this type of integral, we use the following trigonometric product formulae i.e.,
$$ \int {\cos 4x\cos 7x} dx $$ $$ = \frac{1}{2}\int {\left[ {\cos \left( {4 - 7} \right)x + \cos \left( {4 + 7} \right)x} \right]} $$ $$ = \frac{1}{2}\int {\left( {\cos 3x + \cos 11x} \right)} dx $$ $$ = \frac{1}{2}\int {\cos 3xdx} + \frac{1}{2}\int {\cos 11xdx} $$ $$ = \frac{{\sin 3x}}{{2 \times 3}} + \frac{{\sin 11x}}{{2 \times 11}} + C $$ $$ = \frac{{\sin 3x}}{6} + \frac{{\sin 11x}}{{22}} + C $$ $$ 1.{\text{ }}\sin mx\cos nx = \frac{1}{2}\left\{ {\sin (m - n)x + \sin (m + n)x} \right\} $$ $$ 2.{\text{ }}\cos mx\sin nx = \frac{1}{2}\left\{ {\sin (m + n)x - \sin (m - n)x} \right\} $$ $$ 3.{\text{ }}\sin mx\sin nx = \frac{1}{2}\left\{ {cos(m - n)x - \cos (m + n)x} \right\} $$ $$ 4.{\text{ }}\cos mx\cos nx = \frac{1}{2}\left\{ {cos(m - n)x + \cos (m + n)x} \right\} $$
