\(\int \sin(x^2 + 3) 2x \mathrm{d}x\)

We will use substitution

\(u = x^2 + 3\)

\(\mathrm{d}u = 2x \mathrm{d}x\)

This leads to the solution:

\(\int \sin u \mathrm{d}u\)

\(= -\cos u + C\)

\(= -\cos (x^2 + 3) + c\)

\(\int x^2 \sqrt[4]{x^3 - 2} \mathrm{d}x\)

We will use substitution

\(u = x^3 - 2\)

\(\mathrm{d}u = 3x^2 \mathrm{d}x\), \(\dfrac{\mathrm{d}u}{3} = x^2 \mathrm{d}x\)

This leads to the solution:

\(\int \sqrt[4]{u} \dfrac{1}{2} \mathrm{d}u\)

(1/4) + 1

5/4

\(= \dfrac{u^{5/4}}{5/2} + C\)

\(= \dfrac{2*(u)^{5/4}}{5} + C\)

\(= \dfrac{2*(x^3 - 2)^{5/4}}{5} + C\)

\(\int x^2 (\cos x^3) \mathrm{d}x\)

We will use substitution

\(u = x^3\)

\(\mathrm{d}u = 3x^2 \mathrm{d}x\)

\(\dfrac{\mathrm{d}u}{3} = x^2 \mathrm{d}x\)

This leads to the solution:

\(\int \dfrac{\cos u}{3} \mathrm{d}u\)

\(= \dfrac{\sin u}{3} + C\)

\(= \dfrac{\sin (x^3)}{3} + C\)

\(\int \sin x \cos^3 x \mathrm{d}x\)

We will use substitution

\(u = \cos x\)

\(\mathrm{d}u = -\sin x \mathrm{d}x\)

\(-\mathrm{d}u = \sin x \mathrm{d}x\)

This leads to the solution:

\(\int -u^3 \mathrm{d}u\)

\(= \dfrac{-u^4}{4} + C\)

\(= \dfrac{-(\cos x)^4}{4}\)

\(\int x^2 \sin (x^3) \cos^3 (x^3) \mathrm{d}x\)

We will use substitution

\(u = x^3\)

\(\mathrm{d}u = 3x^2 \mathrm{d}x\)

\(\dfrac{\mathrm{d}u}{3} = x^2 \mathrm{d}x\)

This leads to the solution:

\(\int \sin u \cos^3 u \dfrac{1}{3} \mathrm{d}u\)

We know from solution 4, that the solution is

\(= -\dfrac{\cos^4 u}{12} + C\)

\(= -\dfrac{\cos^4 (x^3)}{12} + C\)

\(\int \dfrac{x \mathrm{d}x}{(x+1)^3}\)

We will use substitution

Let \(u = x + 1\)

\(\mathrm{d}u = \mathrm{d}x\)

This leads to the solution:

\(\int \dfrac{u-1}{u^3} \mathrm{d}u\)

\(= \int 1 - \dfrac{1}{u} \mathrm{d}u\)

\(= u - \dfrac{u^{-3+1}}{-2} + C\)

\(= u - \dfrac{1}{-2u^{2}} + C\)

\(= x + 1 - \dfrac{1}{-2(x+1)^{2}} + C\)

\(= x + 1 - \dfrac{1}{-2(x+1)^2} + C\)

\(\int \dfrac{\mathrm{d}x}{\sqrt{x}(\sqrt{x} + 1)^4}\)

We will use substitution

Let \(u = \sqrt{x} + 1\)

\(\mathrm{d}u = \dfrac{-1}{2*\sqrt{x}} dx\)

\(\dfrac{-2*du}{1} = \dfrac{1}{\sqrt{x}} dx\)

This leads to the solution:

\(\int \dfrac{-2*du}{(u)^4}\)

\(= \dfrac{-2*u^{-3}}{-3} + C\)

\(= \dfrac{-2}{3(\sqrt{x} + 1)^3} + C\)

\(\int \dfrac{\sqrt{x} dx}{(\sqrt{x} + 1)^4}\)

We will use substitution

Let \(u = \sqrt{x}\)

diff(sqrt(x), x).expand()

1/2/sqrt(x)

\(du = \dfrac{1}{2*\sqrt{x}} dx\)

\(du = \dfrac{\sqrt{x}}{2*x} dx\)

\(\dfrac{du}{2x} = \sqrt{x} dx\)

This leads to the solution:

\(\int \dfrac{2x du}{(u + 1)^4}\)

\(= \int \dfrac{2u^2 du}{(u+1)^4}\)

\(= \int \dfrac{2u^2 du}{u^4(1 + 1/u)^4}\)

\(= \int \dfrac{2 du}{u^2(1+1/u)^4}\)

Let \(n = \dfrac{1}{u}\)

u = var('u')
diff(1/u, u).expand()

-1/u^2

\(dn = -\dfrac{1}{u^2} du\)

\(-dn = dfrac{1}{u^2} du\)

\(= \int \dfrac{-dn}{(1 + n)^4}\)

\(= \int \dfrac{-1}{(1+n)^4} dn\)

Let \(k = 1 + n\)

\(dk = dn\)

\(= \int \dfrac{-1}{k^4} dk\)

\(= \dfrac{-k^{-3}}{-3}\)

\(= \dfrac{1}{3k^3}\)

\(= \dfrac{1}{3(1+n)^3}\)

\(= \dfrac{1}{3(1+(1/u))^3}\)

\(= \dfrac{1}{3(1+(1/\sqrt{x}))^3}\)

\(\int \sec^2 x \tan^2 x dx\)

We will use substitution

We know that \(\tan^2 x = \sec^2 x - 1\)

\(\int (\tan^2 x + 1)\tan^2 x dx\)

We know that

x = var('x')
diff(tan(x), x)

tan(x)^2 + 1

Let \(u = \tan x\)

\(du = \tan^2 x + 1 dx\)

\(\int u^2 du\)

\(\dfrac{u^3}{3} + C\)

\(\dfrac{\tan^3 x}{3} + C\)

\(\int \sqrt{1 + \sin x} \cos x dx\)

We will use substitution

Let \(u = 1 + \sin x\)

\(du = \cos x dx\)

\(\int \sqrt{u} du\)

\(\dfrac{u^{3/2}}{3/2} + C\)

\(2 \dfrac{(1+\sin x)^(3/2)}{3} + C\)

\(\int_0^4 \dfrac{x dx}{\sqrt{x^2 + 9}}\)

Let's evaluate the indefinite integral.

We will use substitution

Let \(u = x^2 + 9\)

\(du = 2x dx\)

\(\dfrac{du}{2} = x dx\)

\(\int \dfrac{1}{2*\sqrt{u}} du\)

\(= \sqrt{u} + C\)

\(= \sqrt{x^2 + 9} + C\)

From theorem 5.4.3,

\(\int_0^4 \dfrac{x dx}{\sqrt{x^2 + 9}} = \sqrt{4^2 + 9} - \sqrt{4^2 + 9}\)

x = var('x')
f = sqrt(x^2 + 9)
f(4) - f(0)

2

\(\int_0^1 (x+1)(x^2 + 2x - 1)^4 dx\)

Let's evaluate the indefinite integral.

We will use substitution

Let \(u = x^2 + 2x - 1\)

\(du = 2x + 2 dx\)

\(\dfrac{du}{2} = (x+1) dx\)

\(\int \dfrac{u du}{2}\)

\(= \dfrac{u^2}{4} + C\)

\(= \dfrac{(x^2 + 2x - 1)^2}{4} + C\)

From theorem 5.4.3,

\(\int_0^1 (x+1)(x^2 + 2x - 1)^4 dx = \dfrac{(1^2 + 2 - 1)^2}{4} - \dfrac{(0^2 + 0 - 1)^2}{4}\)

x = var('x')
f = ((x^2 + 2*x - 1)^2)/4
f(1) - f(0)

3/4

\(\int_0^{\pi/3} \sec^3 x \tan x dx\)

We know that:

x = var('x')
diff(sec(x))

sec(x)*tan(x)

Let's evaluate the indefinite integral.

We will use substitution

Let \(u = \sec x\)

\(du = \sec x \tan x dx\)

\(\int u^2 du\)

\(= \dfrac{u^3}{3} + C\)

\(= \dfrac{(\sec x)^3}{3} + C\)

From theorem 5.4.3,

\(\int_0^{\pi/3} \sec^3 x \tan x dx = \dfrac{(\sec (\pi/3))^3}{3} - \dfrac{(\sec 0)^3}{3}\)

x = var('x')
f = ((sec(x))^3)/3
f(pi/3) - f(0)

7/3

\(\int_{\pi^2}{4\pi^2} \dfrac{\cos (\sqrt{x}/4)}{\sqrt{x}} dx\)

We know that:

x = var('x')
diff((sqrt(x)), x)

1/2/sqrt(x)

Let's evaluate the indefinite integral.

We will use substitution

Let \(u = \dfrac{\sqrt{x}}{4}\)

\(8 du = \dfrac{1}{\sqrt{x}} dx\)

\(\int 8 \cos (u) du\)

\(= -8\sin u + C\)

\(= -8\sin (\dfrac{\sqrt{x}}{4}) + C\)

From theorem 5.4.3,

\(\int_{\pi^2}{4\pi^2} \dfrac{\cos (\sqrt{x}/4)}{\sqrt{x}} dx = -8\sin (\dfrac{\sqrt{4\pi^2}}{4}) + -8\sin (\dfrac{\sqrt{\pi^2}}{4})\)

x = var('x')
f = -8*sin(sqrt(x)/4)
f(4*(pi^2)) - f(pi^2)

4*sqrt(2) - 8

\(\int \dfrac{\sin^2 x}{\cos^2 x} dx\)

We know that \(\sin^2 x = 1 - \cos^2 x\)

\(= \int \dfrac{1 - \cos^2 x}{\cos^2 x} dx\)

\(= \int \dfrac{1}{\cos^2 x} dx - \int 1 dx\)

\(= \int \sec^2 x dx - \int 1 dx\)

\(= \tan x - x + C\)