Find the integrals. A) integral of x/(sqrt(x + 1)) dx B) integral of sqrt(sin x) cos^3 x dx C)...

Question:

Find the integrals.

A) {eq}\int \frac{x}{\sqrt{x + 1}} \, \mathrm{d}x {/eq}

B) {eq}\int \sqrt{\sin x} \cos^3 x \, \mathrm{d}x {/eq}

C) {eq}\int 3x(5x^2 - 4)^7 \, \mathrm{d}x {/eq}

D) {eq}\int \frac{1}{x^2} \sqrt{2 - \frac{1}{x}} \, \mathrm{d}x {/eq}

Integration By Substitution:

If a function contains large terms then those large terms are considered as variables and then the problem is solved.

Let us consider a function f(t) where t is time, then the integral to that function is found out as follows:

{eq}\displaystyle \int f(t) \, dt = F(t) + C {/eq}.

From the above equation, {eq}F(t) {/eq} is the integral and C is an arbitrary constant.

Sum Rule:

Let us consider the functions {eq}f(x) , g(x) {/eq}

{eq}\displaystyle \int (f(x)+g(x))dx = \int f(x)dx - \int g(x)dx {/eq}

Formulas used are:

{eq}\displaystyle \int x^n dx = \frac{x^{n+1}}{n+1} {/eq}

{eq}\displaystyle \frac{d\sin x}{dx}=\cos x {/eq}

{eq}\displaystyle \cos^2 x = 1- \sin^2 x {/eq}

Answer and Explanation:


a.

Given

{eq}\displaystyle \int \frac{x}{\sqrt {x+1}}.........................(1) {/eq}

Assume

{eq}\displaystyle u=\sqrt {x+1}...................(2) {/eq}

On differentiating we get

{eq}\displaystyle du=\frac{dx}{\sqrt {x+1}}...................(3) {/eq}

On substituting equation (2) and equation (3) in equation (1) we get

{eq}\displaystyle \int 2(u^2-1)du {/eq}

By sum rule above equation can be written as

{eq}\displaystyle =2(\int u^2 du - \int 1 du) {/eq}

{eq}\displaystyle =2(\frac{u^3}{3} - u) {/eq}

Now substituting equation (2) in above equation we get

{eq}\displaystyle =2(\frac{(\sqrt {x+1})^3}{3} - \sqrt {x+1}) {/eq}

The answer is

{eq}\displaystyle \int \frac{x}{\sqrt {x+1}}dx=2(\frac{1}{3}(x+1)^\frac{3}{2} - \sqrt {x+1}) +c {/eq}

where c is any constant.


b.

{eq}\displaystyle \int \sqrt {\sin x}\cos^3 x dx {/eq}

{eq}\displaystyle =\int \sqrt {\sin x}\cos^2 x \cos x dx {/eq}

{eq}\displaystyle =\int \sqrt {\sin x}(1-\sin^2 x)\cos x dx.....................(1) {/eq}

Assume

{eq}\displaystyle u=\sin x...................(2) {/eq}

On differentiating we get

{eq}\displaystyle du=\cos x dx...................(3) {/eq}

On substituting equation (2) and equation (3) in equation (1) we get

{eq}\displaystyle \int \sqrt u (1-u^2)du {/eq}

{eq}\displaystyle =\int (\sqrt u - u^\frac{5}{2})du {/eq}

By Applying sum rule above equation can be written as

{eq}\displaystyle =\int \sqrt {u}du - \int u^\frac{5}{2}du {/eq}

{eq}\displaystyle =\frac{2u^\frac{3}{2}}{3} - \frac{2u^\frac{7}{2}}{7} {/eq}

Now substituting equation (2) in above equation we get

{eq}\displaystyle =\frac{2\sin^\frac{3}{2} x}{3} - \frac{2\sin^\frac{7}{2} x}{7} {/eq}

The answer is

{eq}\displaystyle \int \sqrt {\sin x}\cos^3 x dx=\frac{2\sin^\frac{3}{2} x}{3} - \frac{2\sin^\frac{7}{2} x}{7} +c {/eq}

where c is any constant.


c.

{eq}\displaystyle \int 3x(5x^2 - 4)^7 dx.....................(1) {/eq}

Assume

{eq}\displaystyle u=5x^2 - 4...................(2) {/eq}

On differentiating we get

{eq}\displaystyle du=10x dx...................(3) {/eq}

On substituting equation (2) and equation (3) in equation (1) we get

{eq}\displaystyle \int \frac{3}{10}u^7 du {/eq}

{eq}\displaystyle =\frac{3}{10}\frac{u^8}{8} {/eq}

Now substituting equation (2) in above equation we get

{eq}\displaystyle =\frac{3}{80}(5x^2 - 4)^8 {/eq}

The answer is

{eq}\displaystyle \int 3x(5x^2 - 4)^7 dx=\frac{3}{80}(5x^2 - 4)^8+c {/eq}

where c is any constant.


d.

{eq}\displaystyle \int \frac{1}{x^2} \sqrt{2 - \frac{1}{x}} dx...............(1) {/eq}

Assume

{eq}\displaystyle u= \frac{1}{x}...................(2) {/eq}

On differentiating we get

{eq}\displaystyle du=- \frac{1}{x^2}dx...................(3) {/eq}

On substituting equation (2) and equation (3) in equation (1) we get

{eq}\displaystyle \int -\sqrt {2-u}du......................(4) {/eq}

Assume

{eq}\displaystyle v=2-u...................(5) {/eq}

On differentiating we get

{eq}\displaystyle dv = -du...................(6) {/eq}

On substituting equation (2) and equation (3) in equation (1) we get

{eq}\displaystyle \int \sqrt v dv {/eq}

{eq}\displaystyle =\frac{2v^\frac{3}{2}}{3} {/eq}

Substituting equation (5) in above equation we get

{eq}\displaystyle =\frac{2(2-u)^\frac{3}{2}}{3} {/eq}

Substituting equation (2) in above equation we get

{eq}\displaystyle =\frac{2(2-\frac{1}{x})^\frac{3}{2}}{3} {/eq}

The answer is

{eq}\displaystyle \int \frac{1}{x^2} \sqrt{2 - \frac{1}{x}} dx=\frac{2(2-\frac{1}{x})^\frac{3}{2}}{3} + c {/eq}

where c is any constant.


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How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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