Question

1 Find ∫ e dx
2 Find ∫ e2xdx
4 Find the ∫ tan3xsec3xdx
5 Find ∫ sec3xtanxdx
6 Find the integral with respect to x : ∫ cosxsinxdx
7 Find the integral with respect to x : ∫ (ex x)(ex 1)dx
8 Integrate with respect to x : ∫ dx
2
−1
x2
(x3 + 4)2

Answer 1

Answer on Question #52101 – Math – Integral Calculus

1 Find $\int e \, dx$

2 Find $\int e^2 x \, dx$

4 Find the $\int \tan^3 x \sec^3 x \, dx$

5 Find $\int \sec^3 x \tan x \, dx$

6 Find the integral with respect to $x : \int \cos x \sin x \, dx$

7 Find the integral with respect to $x : \int (ex \, x)(ex \, 1) \, dx$

8 Integrate with respect to $x : \int dx$

2

-1

x2

$(x3 + 4)2$

Solution

Let $C$ be an arbitrary real constant.

1. $\int e \, dx = e \int dx = ex + C;$

2. $\int e^{2x} \, dx = \frac{1}{2} \int e^{2x} d(2x) = | \text{substitution } t = 2x | = \frac{1}{2} \int e^t dt = \frac{1}{2} e^t + C = \frac{1}{2} e^{2x} + C;$

4.

Method 1.

$$\begin{array}{l} \int \tan (3x) \sec (3x) \, dx = \int \frac{\sin (3x) \, dx}{\cos^2 (3x)} = - \frac{1}{3} \int \frac{d(\cos (3x))}{\cos^2 (3x)} = \left| \frac{\text{substitution}}{t = \cos (3x)} \right| = \\ = - \frac{1}{3} \int t^{-2} dt = - \frac{1}{3} \cdot \left(\frac{t^{-1}}{-1}\right) + C = \frac{1}{3t} + C = \frac{1}{3 \cos (3x)} + C = \frac{1}{3} \sec (3x) + C; \end{array}$$

Method 2.

$$\begin{aligned} \int \tan(3x) \sec(3x) dx &= \int \frac{\sin(3x) dx}{\cos^2(3x)} = \frac{1}{3} \int \sin(3x) d(\tan(3x)) \\ &= \left| \text{integration by parts } u = \sin(3x), dv = d(\tan(3x)) \right| \\ &= \frac{1}{3} \sin(3x) \tan(3x) - \frac{1}{3} \int 3 \tan(3x) \cos(3x) dx = \frac{1}{3} \cdot \frac{\sin^2(3x)}{\cos(3x)} \\ &\quad - \frac{1}{3} \int \sin(3x) d(3x) = \frac{1}{3} \cdot \frac{\sin^2(3x)}{\cos(3x)} + \frac{1}{3} \cos(3x) + C = \\ &= \frac{1}{3 \cos(3x)} \left( \sin^2(3x) + \cos^2(3x) \right) + C = \\ &= \left| \sin^2(3x) + \cos^2(3x) = 1 \right| = \frac{1}{3 \cos(3x)} + C = \frac{1}{3} \sec(3x) + C \end{aligned}$$

5. $\int \sec^3 x \tan x dx = \int \frac{\sin(x) dx}{\cos^4(x)} = -\int \frac{d(\cos 3x)}{\cos^4(3x)} = \left| \frac{\text{substitution}}{t = \cos(3x)} \right| =$

$$= -\int \frac{dt}{t^4} = -\int t^{-4} dt = -\frac{t^{-3}}{-3} + C = \frac{1}{3t^3} + C = \frac{1}{3 \cos^3(3x)} +$$ $$+C = \frac{1}{3} \sec^3(3x) + C.$$

6.

Method 1.

$$\int \cos(x) \sin(x) dx = |\sin(2x) - 2 \sin(x) \cos(x)| = \frac{1}{2} \int \sin(2x) dx = \frac{1}{4} \int \sin(2x) d(2x) = |\text{substitution } t = 2x| = \frac{1}{4} \int \sin(t) dt = \\ = -\frac{1}{4} \cos(t) + C_1 = -\frac{1}{4} \cos(2x) + C_1.$$

Method 2.

$$\begin{aligned} \int \cos(x) \sin(x) dx &= \int \sin(x) d(\sin(x)) = |\text{substitution } t = \sin(x)| = \\ &= \int t dt = \frac{t^2}{2} + C_2 = \frac{\sin^2 x}{2} + C_2. \end{aligned}$$

Recall $\cos(2x) = \cos^2 x - \sin^2 x = 1 - 2 \sin^2(x)$, therefore

$$- \frac {1}{4} \cos (2 x) + C _ {1} = - \frac {1}{4} \left(1 - 2 \sin^ {2} (x)\right) + C _ {1} = - \frac {1}{4} + \frac {\sin^ {2} x}{2} + C _ {1} = \frac {\sin^ {2} x}{2} + C _ {2},$$

so we can assume $C_2 = C_1 - \frac{1}{4}$ and methods give the same answer.

7.

**Method 1.**

$$\int (e ^ {x} - x) (e ^ {x} - 1) d x = \int (e ^ {2 x} - x e ^ {x} - e ^ {x} + x) d x = \int e ^ {2 x} d x - \int x e ^ {x} d x -$$ $$- \int e ^ {x} d x + \int x d x = \frac {1}{2} \int e ^ {2 x} d (2 x) - (x - 1) e ^ {x} - e ^ {x} + \frac {1}{2} x ^ {2} =$$ $$= | \text{substitution } t = 2 x | = \frac {1}{2} \int e ^ {t} d t - (x - 1) e ^ {x} - e ^ {x} + \frac {1}{2} x ^ {2} =$$ $$= \frac {1}{2} e ^ {t} - (x - 1) e ^ {x} - e ^ {x} + \frac {1}{2} x ^ {2} + C =$$ $$= \frac {1}{2} e ^ {2 x} - (x - 1) e ^ {x} - e ^ {x} + \frac {1}{2} x ^ {2} + C = \frac {(e ^ {x} - x) ^ {2}}{2} + C.$$

We used the integrals $\int e^{x}dx = e^{x} + C$ and

$$\int x e ^ {x} d x = \int x d e ^ {x} = \left| \begin{array}{c} \text{integration by parts} \quad u = x, dv = d e ^ {x}, \\ d u = d x, v = e ^ {x} \end{array} \right| = x e ^ {x} -$$ $$- \int e ^ {x} d x = x e ^ {x} - e ^ {x} + C = (x - 1) e ^ {x} + C.$$

**Method 2.**

$$\int (e ^ {x} - x) (e ^ {x} - 1) d x = \int (e ^ {x} - x) (e ^ {x} - x) ^ {\prime} d x = \int (e ^ {x} - x) d (e ^ {x} - x) =$$ $$| \text{substitution } t = (e ^ {x} - x), \quad d t = (e ^ {x} - 1) d x | = \int t d t = \frac {t ^ {2}}{2} + C =$$ $$= \frac {(e ^ {x} - x) ^ {2}}{2} + C.$$

8.

**Method 1** (straightforward computation). Using the Newton-Leibniz axiom (or the second fundamental theorem of calculus) and table integral of powers, compute

$$\int_ {- 1} ^ {2} x ^ {2} (x ^ {3} + 4) ^ {2} d x = \int_ {- 1} ^ {2} (x ^ {8} + 8 x ^ {5} + 16 x ^ {2}) d x =$$ $$= \left(\frac {1}{9} x ^ {9} + \frac {8}{6} x ^ {6} + \frac {16}{3} x ^ {3}\right) | _ {x = - 1} ^ {x = 2} = \frac {512}{9} + \frac {256}{3} + \frac {128}{3} + \frac {1}{9} - \frac {4}{3} + \frac {16}{3} = 189.$$

**Method 2 (substitution).** Using the Newton-Leibniz axiom (or the second fundamental theorem of calculus) and table integral of powers, compute

$$\begin{array}{l} \int_{-1}^{2} x^{2} (x^{3} + 4)^{2} dx = \frac{1}{3} \int_{-1}^{2} (x^{3} + 4)^{2} d(x^{3} + 4) = \\ = \left| \frac{\text{substitution}}{t = x^{3} + 4, t(-1) = 3, t(2) = 12} \right| = \frac{1}{3} \int_{3}^{12} t^{2} dt = \frac{1}{3} \cdot \frac{t^{3}}{3} \Big|_{3}^{12} = \frac{1}{9} (12^{3} - 3^{3}) = \\ = 189. \end{array}$$

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