Question

9 Integrate with respect to x : ∫ dx
3
−1
x
7 + x2
10 Integrate with respect to x : ∫ dx
4
1
x + 1
√ x
1 Integrate with respect to x : ∫ secxtanxdx
2 Integrate with respect to x : ∫ csc2xdx
3 Integrate with respect to x : ∫ sinxdx
4 Integrate with respect to Q: ∫ 2sinQ − 5cosQdQ
π3
0
5 Integrate with respect to v : ∫ 4v − 2 dv
1
63
5 Integrate completely with respect to x : ∫ (7 − 4x dx
1
0

Answer 1

Answer on Question #52103 - Math - Integral Calculus

Integrate with respect to $x$: $\int_{-1}^{3} \frac{x}{7 + x^2} dx$

Solution

$$\begin{aligned} \int_{-1}^{3} \frac{x}{7 + x^2} dx &= \frac{1}{2} \int_{-1}^{3} \frac{1}{7 + x^2} d(x^2 + 7) = \left| \text{substitution } t = x^2 + 7, \quad \frac{dt}{t} = d(\ln(t)) \right| \\ &= \frac{1}{2} \ln(x^2 + 7) \Big|_{-1}^{3} = \frac{1}{2} \big( \ln(3^2 + 7) - \ln((-1)^2 + 7) \big) = \frac{1}{2} \ln \left( \frac{3^2 + 7}{(-1)^2 + 7} \right) = \frac{\ln 2}{2}. \end{aligned}$$

Answer: $\frac{\ln 2}{2}$.

$$\int_{1}^{4} \left( x + \frac{1}{\sqrt{x}} \right) dx$$

Solution

$$\begin{aligned} \int_{1}^{4} \left( x + \frac{1}{\sqrt{x}} \right) dx &= \int_{1}^{4} x \, dx + \int_{1}^{4} \frac{1}{\sqrt{x}} dx = \frac{x^2}{2} \Big|_{1}^{4} + 2 \int_{1}^{4} \frac{1}{2\sqrt{x}} dx = \frac{4^2}{2} - \frac{1^2}{2} + 2 \int_{1}^{4} d\sqrt{x} = \frac{15}{2} + 2\sqrt{x} \Big|_{1}^{4} = \\ &= \frac{15}{2} + 2(\sqrt{4} - \sqrt{1}) = \frac{15}{2} + 2(2 - 1) = \frac{15}{2} + 2 = \frac{19}{2} = 9.5. \end{aligned}$$

Answer: 9.5.

Integrate with respect to $x$: $\int \sec x \cdot \tan x \, dx$.

Solution

Method 1 (substitution).

$$\int \sec x \cdot \tan x \, dx = \int \frac{\sin x}{\cos^2 x} dx = \left| u = \sec x, \, du = d\left(\frac{1}{\cos x}\right) = - \frac{-\sin x}{\cos^2 x} dx = \sec x \cdot \tan x \, dx \right| = \int du = u + C = \sec x + C, \text{ where } C \text{ is an arbitrary real constant}.$$

Method 2 (integration by parts).

$$\begin{aligned} \int \sec(x) \cdot \tan(x) \, dx &= \int \frac{\sin(x)}{\cos^2(x)} dx = \int \sin(x) d(\tan(x)) = |u = \sin(x), \, dv = d(\tan(x)), \, du = \cos(x) \, dx, \, v = \tan(x)| = \sin(x) \tan(x) - \int \tan(x) \cos(x) \, dx = \frac{\sin^2(x)}{\cos(x)} - \int \sin(x) \, dx = \\ &= \frac{\sin^2(x)}{\cos(x)} + \cos(x) + C = \frac{\sin^2(x) + \cos^2(x)}{\cos(x)} + C = \frac{1}{\cos(x)} + C = \sec(x) + C, \end{aligned}$$

where $C$ is an arbitrary real constant.

Answer: $\sec x + C$.

Integrate with respect to $x$: $\int \csc 2x \, dx$.

Solution

First we use Trigonometric Identity, namely Reciprocal Identity: $\csc x = \frac{1}{\sin x}$. So, $\csc 2x = \frac{1}{\sin 2x}$.

$$\int \csc 2x \, dx = \int \frac{1}{\sin 2x} \, dx.$$

Method 1.

Let's multiply and divine the integrand, i.e. the function that is to be integrated, $\frac{1}{\sin 2x}$ by $\sin 2x$:

$$\begin{aligned} \int \frac{1}{\sin 2x} \, dx &= \int \frac{\sin 2x}{\sin^2 2x} \, dx = \left| \frac{\text{Pythadorean identity}}{\sin^2 2x = 1 - \cos^2 2x} \right| = \int \frac{\sin 2x}{1 - \cos^2 2x} \, dx = -\frac{1}{2} \int \frac{d \cos 2x}{1 - \cos^2 2x} \\ &= |t = \cos 2x| = -\frac{1}{2} \int \frac{dt}{1 - t^2} = \left| \frac{1}{1 - t^2} = \frac{1}{(1 - t)(1 + t)} = \frac{1}{2} \left( \frac{1}{1 - t} + \frac{1}{1 + t} \right) \right| = \\ &= -\frac{1}{4} \int \frac{dt}{1 - t} - \frac{1}{4} \int \frac{dt}{1 + t} = \frac{1}{4} \int \frac{dt}{t - 1} - \frac{1}{4} \int \frac{dt}{t + 1} = -\frac{1}{4} \ln \left| \frac{t + 1}{t - 1} \right| + C = \frac{1}{4} \ln |1 - t| - \frac{1}{4} \ln |1 + t| + c \\ &= \frac{1}{4} \ln \left| \frac{1 - t}{1 + t} \right| + c. \end{aligned}$$

Thus,

$\int \csc 2x \, dx = \frac{1}{4} \ln \left| \frac{1 - \cos 2x}{1 + \cos 2x} \right| + c = \frac{1}{4} \ln \left| \frac{2 \sin^2(x)}{2 \cos^2(x)} \right| + c = \frac{1}{4} \ln |\tan^2(x)| + c = \frac{1}{2} \ln |\tan(x)| + c$, where $c$ is an arbitrary real constant.

Method 2.

$$\begin{aligned} \int \frac{1}{\sin 2x} \, dx &= \int \frac{dx}{2 \sin(x) \cos(x)} = \left| \text{divide the numerator and denominator by } \cos^2(x) \right| = \int \frac{\frac{dx}{\cos^2(x)}}{\frac{2 \sin(x) \cos(x)}{\cos^2(x)}} = \\ &\frac{1}{2} \int \frac{d(\tan(x))}{\tan(x)} = |y = \tan(x)| = \frac{1}{2} \int \frac{dy}{y} = \frac{1}{2} \ln |y| + c = \frac{1}{2} \ln |\tan(x)| + c, \text{ where } c \text{ is an arbitrary real constant}. \end{aligned}$$

Answer: $\frac{1}{2} \ln |\tan(x)| + c$

Integrate with respect to $x$: $\int \sin x \, dx$.

Solution

$$\int \sin x \, dx = -\cos x + C, C \in \mathbb{R}.$$

Answer: $-\cos x + C$.

Integrate with respect to $Q$: $\int_0^\pi (2 \sin Q - 5 \cos Q) \, dQ$.

Solution

$$\begin{aligned} \int_0^{\frac{\pi}{3}} (2 \sin Q - 5 \cos Q) \, dQ &= (-2 \cos Q - 5 \sin Q) \left| \begin{array}{c} \frac{\pi}{3} \\ 0 \end{array} \right| = -2 \cdot \frac{1}{2} - 5 \cdot \frac{\sqrt{3}}{2} + 2 - 0 = 1 - \frac{5\sqrt{3}}{2} = \frac{2 - 5\sqrt{3}}{2} \\ &\approx -3.33013. \end{aligned}$$

Answer: $\frac{2 - 5\sqrt{3}}{2} \approx -3.33013$.

Integrate with respect to $v$: $\int_{63}^{1}(4v - 2) \, dv$

Solution

Method 1

$$\int_{63}^{1} (4v - 2)dv = \left(4 \frac{v^2}{2} - 2v\right) \Big|_{63}^{1} = \left(4 \frac{1^2}{2} - 2 \cdot 1\right) - \left(4 \frac{63^2}{2} - 2 \cdot 63\right) = -7812.$$

Method 2

$$\begin{aligned} \int_{63}^{1} (4v - 2)dv &= \frac{1}{4} \int_{63}^{1} (4v - 2) d(4v - 2) = \left| \text{substitution } t = 4v - 2, t(63) = 250, t(1) = 2 \right| \\ &= \frac{1}{4} \int_{250}^{2} t dt = \left. \frac{1}{4} \frac{t^2}{2} \right|_{250}^{2} = \frac{1}{8} (2^2 - 250^2) = -7812 \end{aligned}$$

Answer: -7812.

Integrate with respect to $x$: $\int_0^1 (7 - 4x)^2 dx$.

Solution

Method 1.

$$\int_{0}^{1} (7 - 4x)^2 dx = \int_{0}^{1} (49 - 56x + 16x^2) dx = 49x - 28x^2 + \frac{16x^3}{3} \Big|_{0}^{1} = \frac{79}{3}.$$

Method 2.

$$\begin{aligned} \int_{0}^{1} (7 - 4x)^2 dx &= - \frac{1}{4} \int_{0}^{1} (7 - 4x)^2 d(7 - 4x) = \left| \text{substitution } t = 7 - 4x, t(0) = 7, t(1) = 3 \right| \\ &= - \frac{1}{4} \int_{7}^{3} t^2 dt = - \left. \frac{1}{4} \frac{t^3}{3} \right|_{7}^{3} = - \frac{1}{12} (3^3 - 7^3) = \frac{316}{12} = \frac{79}{3} \end{aligned}$$

Answer: $\frac{79}{3}$.

www.AssignmentExpert.com