# 1. Find integral cos square root{x} d x. Hint: Let t = square root{x} so that t^2 = x....

## Question:

1. Find {eq}\int \cos \sqrt{x} dx {/eq}

Hint: Let {eq}t = \sqrt{x} \ so \ that \ t^2 = x.{/eq} Substitute and use integration by parts on the resulting integral.

2.Evaluate {eq}x^3 \sqrt{x^2 + 1} d x {/eq}.

3.Evaluate {eq}\int \frac{ 2 x + 1}{(x + 1)(x - 3)(x + 3)} dx {/eq}.

## Integration :

We will use the substitution method to solve the integrals presented in the questions. We will add a constant C after performing the required integration.

The required steps to evaluate.

{eq}\displaystyle{\int f(x)\quad dx}\\ {/eq}

by substitution are:

1. Put x=g(y), dx = g'(y) dy in the integrand

2. Evaluate the resulting integral in y.

3. Express the result obtained in terms of x.

Integration by parts states that if u and v are the two continuous functions of x and u has

continuous differential coefficient and v continuous integral w.r.t x, then

{eq}\displaystyle{\int uv\hspace{5pt}dx=u\int v\hspace{5pt}dx-\int \frac{d}{dx}(u)(\int v\hspace{5pt}dx)dx\\} {/eq}

In words it says Integral of product of two function= first function * integral of second

function- integral of differentiation coefficient of first * integral of second

If the integrand is in the form of an algebraic function and the integral cannot be evaluated by the simple methods, the fraction needs to be expressed in partial

fraction before integration takes place.

steps to do partial fraction:

1. Always start with proper rational fraction {eq}\frac{P(x)}{Q(x)}{/eq} i.e degree of numerator is less than that of denominator and if not do division first.

2. Factor Q(x) as far as possible.

3. Write down the correct form for the partial fraction decomposition and solve.

## Answer and Explanation:

1 .{eq}\displaystyle{\int \cos \sqrt{x} dx \\[10pt] \text{substitute t = } \sqrt{x} \\[10pt] \frac{dt}{dx} = \frac{1}{2\sqrt{x}}\hspace{100pt}\left( \frac{d}{dx} = x^n = n x^{n - 1}\right)\\[10pt] dx = 2\sqrt{x} \ dt\\[10pt] \int 2t\cdot \cos t \ dt\\[10pt] 2\int t\cdot \cos t \ dt\\[10pt] 2\left[ t \int \cos t \ dt - \int \frac{d}{dx}(t) \left(\int \cos t \ dt\right) \ dt \right]\hspace{90pt}\left(\int uv\hspace{5pt}dx=u\int v\hspace{5pt}dx-\int \frac{d}{dx}(u)(\int v\hspace{5pt}dx)dx\right)\\ 2\left[ t \sin t - \int 1 \cdot \sin t \ dt\right]\\[10pt] 2\left[ t \sin t - (- \cos t)\right] + C\\[10pt] 2\left[ t \sin t + \cos t\right ] + C\\[10pt] 2\left[ \sqrt{x} \sin \sqrt{x} + \cos \sqrt{x}\right] + C\\[10pt] \boxed{2 \sqrt{x} \sin \sqrt{x} + 2 \cos \sqrt{x} +C}\\} {/eq}

2.Evaluate {eq}\displaystyle{\int x^3 \sqrt{x^2 + 1} d x \\[10pt] \text{substitute u = } x^2 + 1 \implies x^2 = u - 1\\[10pt] \frac{du}{dx} = 2x \\[10pt] dx = \frac{1}{2x} \ du\\[10pt] \frac{1}{2} \int (u - 1) \sqrt{u} \ du\\[10pt] \frac{1}{2} \int \left( u^\frac{3}{2} - u^\frac{1}{2}\right) \ du \\[10pt] \frac{1}{2} \left[ \frac {2 u^\frac{5}{2}}{5} - \frac{2 u^\frac{3}{2}}{3}\right] +C \hspace{80pt}\left( \int x^n \ dx = \frac{x^{n+1}}{n+1} + C\right)\\[10pt] \frac{1}{2} \left[ \frac {2 (x^2 + !)^\frac{5}{2}}{5} - \frac{2 (x^2 + 1)^\frac{3}{2}}{3}\right] +C\\[10pt] \left[ \frac { (x^2 + 1)^\frac{5}{2}}{5} - \frac{ (x^2 + 1)^\frac{3}{2}}{3}\right] +C\\[10pt] \frac{ 3(x^2 + 1)^\frac{5}{2} - 5 (x^2 + 1)^\frac{3}{2}}{15} +C \\[10pt] \frac{ (x^2 + 1)^\frac{3}{2}[ 3(x^2 +1) -5]}{15} + C\\[10pt] \frac{ (x^2 + 1)^\frac{3}{2}[ 3x^2 + 3 -5]}{15} + C\\[10pt] \boxed{\frac{ (x^2 + 1)^\frac{3}{2}(3x^2 - 2)}{15} + C}\\} {/eq}

3.Evaluate {eq}\displaystyle{ I = \int \frac{ 2 x + 1}{(x + 1)(x - 3)(x + 3)} dx\\} {/eq}.

First of all , we resolve into partial fractions.

{eq}\displaystyle{\frac{ 2 x + 1}{(x + 1)(x - 3)(x + 3)}\equiv \frac{A}{ x + 1} + \frac{B}{ x - 3} + \frac{C}{x + 3}\hspace{50pt}(1)\\} {/eq}

Multiplying both sides by (x + 1)(x - 3) (x + 3), we get

{eq}\displaystyle{2 x + 1 = A(x - 3)(x + 3) + B(x +1)(x + 3) + C (x +1)(x -3)\\[10pt] \text{ put }\quad x +1 = 0 \implies x = -1 \\[10pt] -2 + 1 = A(-4)(2)\\[10pt] - 1 = -8 A\\[10pt] A = \frac{1}{8}\\[10pt] \text{ put }\quad x - 3 = 0 \implies x = 3 \\[10pt] 6 + 1 = B(4)(6)\\[10pt] 7 = 24 B\\[10pt] B = \frac{7}{24}\\[10pt] \text{ put }\quad x + 3 = 0 \implies x = - 3 \\[10pt] -6 + 1 = C( - 2)(- 6)\\[10pt] - 5 = 12 C\\[10pt] C = \frac{-5}{12}\\} {/eq}

putting the values of A,B and C in (1), we get,

{eq}\displaystyle{\frac{ 2 x + 1}{(x + 1)(x - 3)(x + 3)}\equiv \frac{1}{8( x + 1)} + \frac{7}{ 24(x - 3)} + \frac{-5}{12(x + 3)}\\[10pt] \therefore I = \int\frac{1}{8( x + 1)} \ dx + \int \frac{7}{ 24(x - 3)} \ dx - \int \frac{5}{12(x + 3)} \ dx\\[10pt] =\frac{1}{8} \ln (x + 1) + \frac {7}{24} \ln (x - 3 ) - \frac {5 } {12} \ln (x + 3)\hspace{90pt}\left( \int \frac{1}{x} \ dx = \ln x + C\right)\\[10pt] = \frac{6 \ln (x + 1) + 14 \ln (x - 3) - 20 \ln (x + 3)}{48} + C\\[10pt] = \frac{2[3 \ln (x + 1) + 7 \ln (x - 3) - 10 \ln (x + 3)]}{48} + C\\[10pt] \boxed{= \frac{3 \ln (x + 1) + 7 \ln (x - 3) - 10 \ln (x + 3)}{24} + C}\\} {/eq}

#### Learn more about this topic:

How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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