### Math Notes

Subjects

#### Integral Calculus Solutions

##### Topics || Problems

I. Integration Using Derivatives

Note that: $$\frac{d}{{dx}}f\left( x \right) = f'\left( x \right)$$. Thus,

$$\int {\frac{d}{{dx}}f\left( x \right)} = \int {f'\left( x \right)}$$

$$f\left( x \right) = \int {f'\left( x \right)}$$

Example: Find the integral value of $$y' = {\sec ^2}x$$.

From derivatives

$$\frac{d}{{dx}}\tan x = {\sec ^2}x$$

Integrating both sides

$$\int {d\left( {\tan x} \right)} = \int {{{\sec }^2}xdx}$$

$$\tan x = \int {{{\sec }^2}xdx}$$

Thus:

$$y' = {\sec ^2}x$$

$$\int {d\left( y \right)} = \int {{{\sec }^2}x} dx$$

$$\ {y} = \tan x + c$$

2. Integration by trigonometric substitution

This method is applicable to problems involving the following forms:

1. $$\sqrt {{a^2} - {u^2}}$$

Assume or let u = a sin $$\theta$$

2. $$\sqrt {{a^2} + {u^2}}$$

Let u = a tan $$\theta$$

3. $$\sqrt {{u^2} - {a^2}}$$

Let u = a sec $$\theta$$

Example:

1. Integrate $$\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}}$$

$$\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}}$$

Let

$$x = 3\tan \theta$$

$$dx = 3{\sec ^2}\theta$$

$$x = 9{\tan ^2}x$$

$$\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \int {\frac{{3{{\sec }^2}\theta d\theta }}{{\sqrt {9 + 9{{\tan }^2}\theta } }}}$$

$$\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \frac{3}{3}\int {\frac{{{{\sec }^2}\theta d\theta }}{{\sqrt {1 + {{\tan }^2}\theta } }}}$$

$$\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \int {\frac{{{{\sec }^2}\theta d\theta }}{{\sqrt {{{\sec }^2}\theta } }}}$$

$$\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \int {\frac{{{{\sec }^2}\theta d\theta }}{{\sec \theta }}}$$

$$\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \int {\sec \theta d\theta }$$

$$\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \int {\sec \theta \left( {\frac{{\sec \theta + \tan \theta }}{{\sec \theta + \tan \theta }}} \right)} d\theta$$

$$\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \int {\frac{{{{\sec }^2}\theta + \sec \theta \tan \theta }}{{\sec \theta + \tan \theta }}} d\theta$$

$$\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \ln \left( {\sec \theta + \tan \theta } \right) + C$$

$$\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }}} = \ln \left( {\frac{{1 + \sin \theta }}{{\cos \theta }}} \right) + C$$

But from the assumption $$x = 3\tan \theta$$

$$\sin \theta = \frac{x}{{\sqrt {{x^2} + 9} }}$$

$$\cos \theta = \frac{3}{{\sqrt {{x^2} + 9} }}$$

$$\frac{{1 + \sin \theta }}{{\cos \theta }} = \frac{{1 + \frac{x}{{\sqrt {{x^2} + 9} }}}}{{\frac{3}{{\sqrt {{x^2} + 9} }}}}$$

$$\frac{{1 + \sin \theta }}{{\cos \theta }} = \frac{{\frac{{\sqrt {{x^2} + 9} + x}}{{\sqrt {{x^2} + 9} }}}}{{\frac{3}{{\sqrt {{x^2} + 9} }}}}$$

$$\frac{{1 + \sin \theta }}{{\cos \theta }} = \frac{{\sqrt {{x^2} + 9} + x}}{3}$$

Thus

$$\int {\frac{{dx}}{{\sqrt {9 + {x^2}} }} = \ln \left( {\frac{{\sqrt {{x^2} + 9} + x}}{3}} \right) + C}$$

2. Integrate $$\int {y\sqrt {16 - {y^2}} dy}$$

$$\int\ y\sqrt {16 - {y^2}} dy$$

Let y =4 sin$$\theta$$

$$dy = 4\cos \theta d\theta$$

$${y^2} = 16{\sin ^2}\theta$$

$$\int\ y\sqrt {16 - {y^2}} dy = \int {\left( {4\sin \theta } \right)} \sqrt {16 - 16{{\sin }^2}\theta } \left( {4\cos \theta d\theta } \right)$$

$$\int\ y\sqrt {16 - {y^2}} dy = 64\int {\sin \theta \cos \theta \sqrt {1 - {{\sin }^2}\theta } d\theta }$$

$$\int\ y\sqrt {16 - {y^2}} dy = 64\int {\sin \theta {{\cos }^2}\theta d\theta }$$

$$\int\ y\sqrt {16 - {y^2}} dy = 64\left( {\frac{{{{\cos }^3}\theta }}{3}} \right) + C$$

But from the assumption y =4 sin$$\theta$$

$$\sin \theta = \frac{y}{4}$$ and using the a right triangle,$$\sin \theta = \frac{{opp}}{{hyp}}$$

$$\cos \theta = \frac{{\sqrt {16 - {y^2}} }}{4}$$

$${\cos ^3}\theta = {\left( {\frac{{\sqrt {16 - {y^2}} }}{4}} \right)^3}$$

$${\cos ^3}\theta = \frac{{{{\left( {16 - {y^2}} \right)}^{\frac{3}{2}}}}}{{64}}$$

thus,

$$\int {y\sqrt {16 - {y^2}} dy = 64\left( {\frac{{{{\cos }^3}\theta }}{3}} \right) + C}$$

$$\int {y\sqrt {16 - {y^2}} dy} = 64\left( {\frac{{{{\left( {16 - {y^2}} \right)}^{\frac{3}{2}}}}}{{64(3)}}} \right) + C$$

$$\int {y\sqrt {16 - {y^2}} } dy = \frac{{{{\left( {16 - {y^2}} \right)}^{\frac{3}{2}}}}}{3} + C$$

3. Integrate $$\int {\frac{{\sqrt {9{x^2} - 4} }}{x}} dx$$

$$\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx}$$

Let

$$3x = 2\sec \theta$$

$$9{x^2} = 4{\sec ^2}\theta$$

$$x = \frac{2}{3}\sec \theta$$

$$dx = \frac{2}{3}\sec \theta \tan \theta d\theta$$

$$\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = \int {\frac{{\sqrt {4{{\sec }^2}\theta - 4} }}{{\frac{2}{3}\sec \theta }}\left( {\frac{2}{3}\sec \theta \tan \theta d\theta } \right)}$$

$$\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = 2\int {\sqrt {{{\sec }^2}\theta - 1} \left( {\tan \theta d\theta } \right)}$$

$$\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = 2\int {\left( {{{\tan }^2}\theta d\theta } \right)}$$

$$\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = 2\int {\left( {{{\sec }^2}\theta - 1} \right)} d\theta$$

$$\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = 2\tan \theta - 2\theta + C$$

$$\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = 2\frac{{\sqrt {9{x^2} - 4} }}{2} - 2\arccos \left( {\frac{2}{{3x}}} \right) + C$$

$$\int {\frac{{\sqrt {9{x^2} - 4} }}{x}dx} = \sqrt {9{x^2} - 4} - 2\arccos \left( {\frac{2}{{3x}}} \right) + C$$

3. Integration by Parts

Integration by parts is from derivatives of a product.

$$\frac{d}{{dx}}\left( {uv} \right) = vdu + udv$$

$$\int {\frac{d}{{dx}}\left( {uv} \right)} = \int {vdu} + \int {udv}$$

$$uv = \int {vdu} + \int {udv}$$

$$\int {udv = uv - \int {vdu} }$$

How to integrate using integration by parts.

1. Assume for the values of u and dv, Sometimes this part is trial and error

2. Solve du and v

3. Substitute

Examples:

1. $$\int {x{e^x}dx}$$

Let u = x and $$dv = e^x dx$$

Thus, du = dx and $$v = e^x$$

$$\int {x{e^x}dx}$$ = $$\int {udv}$$

$$\int {x{e^x}dx}$$ = $$uv -\int{vdu}$$

$$\int {x{e^x}dx}$$ = $$xe^x -\int{e^xdx}$$

$$\int {x{e^x}dx}$$ = $$xe^x -e^x$$ +c

2. $$\int {{e^x}\cos xdx}$$

Let $$u = cos x$$ and $$e^xdx = dv$$

Thus$$du = -sin x dx$$ and $$e^x = v$$

$$\int {{e^x}\cos xdx}$$ = $$\int {udv}$$

$$\int {{e^x}\cos xdx}$$ = $$uv -\int{vdu}$$

$$\int {{e^x}\cos xdx}$$ = $$e^x cosx -\int{e^x(-sinx)dx}$$

$$\int {{e^x}\cos xdx}$$ = $$e^x cosx +\int{e^x(sinx)dx}$$

Repeate the process of by parts on $$\int{e^x(sinx)dx}$$

Let u = sin x and $$e^xdx = dv$$

Thus du = cos x dx and $$e^x = v$$

$$\int{e^x(sinx)dx}$$=$$\int{udv}$$

$$\int{e^x(sinx)dx}$$=$$uv-\int{vdu}$$

$$\int{e^x(sinx)dx}$$=$$e^x sin x -\int{e^x cos xdx}$$

$$\int {{e^x}\cos xdx}$$ = $$e^x cosx +\int{e^x(sinx)dx}$$

$$\int {{e^x}\cos xdx}$$ = $$e^x cosx + e^x sin x -\int{e^x cos xdx}$$

$$2\int {{e^x}\cos xdx}$$ = $$e^x cosx e^x sin x + C$$

$$\int {{e^x}\cos xdx}$$ = $$\frac{e^x cosx e^x sin x + C}{2}$$