Using the most appropriate method, evaluate the following integrals: (a) ∫x(3x2 + 2)2 dx (b) ∫ (3x2 - 1)/(x3…

Question

Using the most appropriate method, evaluate the following integrals: (a) ∫x(3x2 + 2)2 dx (b) ∫ (3x2 - 1)/(x3 - x + 4)3 dx (c) ∫_-11 (x−1)(x+3)dx

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to the integration problems:

Question 1 (a)

a) To evaluate $\int x(3x^2 + 2)^2 dx$ , we use u-substitution.

Let $u = 3x^2 + 2$ .
Differentiate $u$ with respect to $x$ to find $du$
$du = \frac{d}{dx}(3x^2 + 2) dx = 6x \, dx$
Rearrange to find $x \, dx$
$x \, dx = \frac{1}{6} du$
Substitute $u$ and $x \, dx$ into the integral
$\int u^2 \left(\frac{1}{6}\right) du = \frac{1}{6} \int u^2 du$
Integrate with respect to $u$
$\frac{1}{6} \left(\frac{u^{2+1}}{2+1}\right) + C = \frac{1}{6} \left(\frac{u^3}{3}\right) + C = \frac{u^3}{18} + C$
Substitute back $u = 3x^2 + 2$
$\boxed{\frac{(3x^2 + 2)^3}{18} + C}$
b) To evaluate $\int \frac{3x^2 - 1}{(x^3 - x + 4)^3} dx$ , we use u-substitution.
Let $u = x^3 - x + 4$ .
Differentiate $u$ with respect to $x$ to find $du$
$du = \frac{d}{dx}(x^3 - x + 4) dx = (3x^2 - 1) \, dx$
Substitute $u$ and $du$ into the integral
$\int \frac{1}{u^3} du = \int u^{-3} du$
Integrate with respect to $u$
$\frac{u^{-3+1}}{-3+1} + C = \frac{u^{-2}}{-2} + C = -\frac{1}{2u^2} + C$
Substitute back $u = x^3 - x + 4$
$\boxed{-\frac{1}{2(x^3 - x + 4)^2} + C}$
c) To evaluate $\int_{-1}^{1} (x - 1)(x + 3) dx$ , we first expand the integrand.
Expand the product $(x - 1)(x + 3)$
$(x - 1)(x + 3) = x^2 + 3x - x - 3 = x^2 + 2x - 3$
Integrate the polynomial
$\int (x^2 + 2x - 3) dx = \frac{x^3}{3} + \frac{2x^2}{2} - 3x = \frac{x^3}{3} + x^2 - 3x$
Evaluate the definite integral using the limits from -1 to 1
$\left[ \frac{x^3}{3} + x^2 - 3x \right]_{-1}^{1} = \left( \frac{(1)^3}{3} + (1)^2 - 3(1) \right) - \left( \frac{(-1)^3}{3} + (-1)^2 - 3(-1) \right)$
Calculate the values at the upper and lower limits
$= \left( \frac{1}{3} + 1 - 3 \right) - \left( -\frac{1}{3} + 1 + 3 \right)$ $= \left( \frac{1}{3} - 2 \right) - \left( -\frac{1}{3} + 4 \right)$ $= \left( \frac{1}{3} - \frac{6}{3} \right) - \left( -\frac{1}{3} + \frac{12}{3} \right)$ $= -\frac{5}{3} - \frac{11}{3}$
Subtract the values
$= -\frac{16}{3}$ $\boxed{-\frac{16}{3}}$
Question 1 (b)

a) To integrate $\sin^5 x$ using Pythagoras' theorem of odd powers:
Rewrite $\sin^5 x$ as $\sin^4 x \sin x$
$\int \sin^5 x \, dx = \int \sin^4 x \sin x \, dx$
Use the identity $\sin^2 x = 1 - \cos^2 x$
$= \int (\sin^2 x)^2 \sin x \, dx = \int (1 - \cos^2 x)^2 \sin x \, dx$
Let $u = \cos x$ .
Then $du = -\sin x \, dx$ , so $\sin x \, dx = -du$ .
$= \int (1 - u^2)^2 (-du) = -\int (1 - 2u^2 + u^4) du$
Integrate the polynomial with respect to $u$
$= -\left( u - \frac{2u^3}{3} + \frac{u^5}{5} \right) + C$
Substitute back $u = \cos x$
$\boxed{-\cos x + \frac{2}{3}\cos^3 x - \frac{1}{5}\cos^5 x + C}$
b) To integrate $\cos^3 x$ using Pythagoras' theorem of odd powers:
Rewrite $\cos^3 x$ as $\cos^2 x \cos x$
$\int \cos^3 x \, dx = \int \cos^2 x \cos x \, dx$
Use the identity $\cos^2 x = 1 - \sin^2 x$
$= \int (1 - \sin^2 x) \cos x \, dx$
Let $u = \sin x$ .
Then $du = \cos x \, dx$ .
$= \int (1 - u^2) du$
Integrate the polynomial with respect to $u$
$= u - \frac{u^3}{3} + C$
Substitute back $u = \sin x$
$\boxed{\sin x - \frac{1}{3}\sin^3 x + C}$

Using the most appropriate method, evaluate the following integrals: (a) ∫x(3x2 + 2)2 dx (b) ∫ (3x2 - 1)/(x3 - x + 4)3 dx (c) ∫_-11 (x-1)(x+3)dx

Step-by-step solution

Let $u = 3x^2 + 2$ .

Differentiate $u$ with respect to $x$ to find $du$

Rearrange to find $x \, dx$

Substitute $u$ and $x \, dx$ into the integral

Integrate with respect to $u$

Substitute back $u = 3x^2 + 2$

Let $u = x^3 - x + 4$ .

Differentiate $u$ with respect to $x$ to find $du$

Substitute $u$ and $du$ into the integral

Integrate with respect to $u$

Substitute back $u = x^3 - x + 4$

Expand the product $(x - 1)(x + 3)$

Integrate the polynomial

Evaluate the definite integral using the limits from -1 to 1

Calculate the values at the upper and lower limits

Subtract the values

Rewrite $\sin^5 x$ as $\sin^4 x \sin x$

Use the identity $\sin^2 x = 1 - \cos^2 x$

Let $u = \cos x$ .

Integrate the polynomial with respect to $u$

Substitute back $u = \cos x$

Rewrite $\cos^3 x$ as $\cos^2 x \cos x$

Use the identity $\cos^2 x = 1 - \sin^2 x$

Let $u = \sin x$ .

Integrate the polynomial with respect to $u$

Substitute back $u = \sin x$

Still stuck on this one?

More Mathematics Questions

Using the most appropriate method, evaluate the following integrals: (a) ∫x(3x2 + 2)2 dx (b) ∫ (3x2 - 1)/(x3 - x + 4)3 dx (c) ∫_-11 (x-1)(x+3)dx

Step-by-step solution

Let u=3x2+2u = 3x^2 + 2u=3x2+2.

Differentiate uuu with respect to xxx to find dududu

Rearrange to find x dxx \, dxxdx

Substitute uuu and x dxx \, dxxdx into the integral

Integrate with respect to uuu

Substitute back u=3x2+2u = 3x^2 + 2u=3x2+2

Let u=x3−x+4u = x^3 - x + 4u=x3−x+4.

Differentiate uuu with respect to xxx to find dududu

Substitute uuu and dududu into the integral

Integrate with respect to uuu

Substitute back u=x3−x+4u = x^3 - x + 4u=x3−x+4

Expand the product (x−1)(x+3)(x - 1)(x + 3)(x−1)(x+3)

Integrate the polynomial

Evaluate the definite integral using the limits from -1 to 1

Calculate the values at the upper and lower limits

Subtract the values

Rewrite sin⁡5x\sin^5 xsin5x as sin⁡4xsin⁡x\sin^4 x \sin xsin4xsinx

Use the identity sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 xsin2x=1−cos2x

Let u=cos⁡xu = \cos xu=cosx.

Integrate the polynomial with respect to uuu

Substitute back u=cos⁡xu = \cos xu=cosx

Rewrite cos⁡3x\cos^3 xcos3x as cos⁡2xcos⁡x\cos^2 x \cos xcos2xcosx

Use the identity cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 xcos2x=1−sin2x

Let u=sin⁡xu = \sin xu=sinx.

Integrate the polynomial with respect to uuu

Substitute back u=sin⁡xu = \sin xu=sinx

Still stuck on this one?

More Mathematics Questions

Let $u = 3x^2 + 2$ .

Differentiate $u$ with respect to $x$ to find $du$

Rearrange to find $x \, dx$

Substitute $u$ and $x \, dx$ into the integral

Integrate with respect to $u$

Substitute back $u = 3x^2 + 2$

Let $u = x^3 - x + 4$ .

Differentiate $u$ with respect to $x$ to find $du$

Substitute $u$ and $du$ into the integral

Integrate with respect to $u$

Substitute back $u = x^3 - x + 4$

Expand the product $(x - 1)(x + 3)$

Rewrite $\sin^5 x$ as $\sin^4 x \sin x$

Use the identity $\sin^2 x = 1 - \cos^2 x$

Let $u = \cos x$ .

Integrate the polynomial with respect to $u$

Substitute back $u = \cos x$

Rewrite $\cos^3 x$ as $\cos^2 x \cos x$

Use the identity $\cos^2 x = 1 - \sin^2 x$

Let $u = \sin x$ .

Integrate the polynomial with respect to $u$

Substitute back $u = \sin x$