Evaluate the integrals: 1. \ \int \frac{x^2}{2} - 2x + 6 \ dx \\ 2. \ \int x^2\cos \ 3x \ dx \\...

Question:

Evaluate the integrals:

{eq}1. \ \int \frac{x^2}{2} - 2x + 6 \ dx \\ 2. \ \int x^2\cos \ 3x \ dx \\ 3. \ \int \sin^3x \ dx {/eq}

Integration:

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.

Integration By Parts:

If given function are {eq}u,v {/eq} then

{eq}\displaystyle \int (u dv) = uv - \int (v du) {/eq}

Sum Rule:

If functions given are {eq}f(x) , g(x) {/eq} then

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

Formulas used:

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

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

{eq}\displaystyle \int \sin x dx = -\cos x {/eq}

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

Answer and Explanation:


1.

Given

{eq}\displaystyle \int (\frac{x^2}{2} - 2x +6 )dx {/eq}

Applying sum rule above function can be written as

{eq}\displaystyle =\int \frac{x^2}{2}dx -\int 2x dx +\int 6dx {/eq}

{eq}\displaystyle =\frac{x^3}{6} -x^2 +6x {/eq}

The answer is

{eq}\displaystyle \int (\frac{x^2}{2} - 2x +6 )dx=\frac{x^3}{6} -x^2 +6x +c {/eq}

where c is any constant.


2.

Given

{eq}\displaystyle \int x^2 \cos (3x) dx................(1) {/eq}

Assume

{eq}\displaystyle u=3x...................(2) {/eq}

On differentiating we get

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

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

{eq}\displaystyle =\int \frac{u^2\cos u}{27}du {/eq}

{eq}\displaystyle =\frac{1}{27}\int u^2 \cos u du {/eq}

By using Integration By Parts we assume {eq}u=u^2,v'=\cos u {/eq}

Now applying Integration By Parts we get

{eq}\displaystyle =\frac{1}{27}(u^2\sin u - \int 2u\sin u du) {/eq}

By using Integration By Parts for second term in above equation we assume {eq}u=u,v'=\sin u {/eq}

Now applying Integration By Parts we get

{eq}\displaystyle =\frac{1}{27}(u^2\sin u -2(-u\cos u - \int -\cos u du)) {/eq}

{eq}\displaystyle =\frac{1}{27}(u^2\sin u +2u\cos u - 2\sin u ) {/eq}

Substituting equation (2) in above equation we get

{eq}\displaystyle =\frac{1}{27}((3x)^2\sin u +2(3x)\cos (3x) - 2\sin (3x) ) {/eq}

{eq}\displaystyle =\frac{1}{27}(9x^2\sin u +6x\cos (3x) - 2\sin (3x) ) {/eq}

The answer is

{eq}\displaystyle \int x^2 \cos (3x) dx=\frac{1}{27}(9x^2\sin u +6x\cos (3x) - 2\sin (3x) )+c {/eq}

where c is any constant.


3.

Given

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

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

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

Assume

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

On differentiating we get

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

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

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

By using sum rule above equation can be written as

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

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

Substituting equation (2) in above equation we get

{eq}\displaystyle =-\cos x+\frac{(\cos^3 x)}{3} {/eq}

The answer is

{eq}\displaystyle \int \sin^3 x dx=-\cos x+\frac{(\cos^3 x)}{3}+c {/eq}

where c is any constant.


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Integration Problems in Calculus: Solutions & Examples

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13
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