# Evaluate the following integrals a) \int t^2 \cos(2t) dt b) \int \frac {1}{x\sqrt {In^2(x)-1}}dx...

## Question:

Evaluate the following integrals

{eq}\displaystyle a) \int t^2 \cos(2t) dt {/eq}

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

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

## Integration.

Integration it defines the area on every point of the function.

It also uses a table of ant derivatives.

There is one of the special rule of integration by parts can also be used to find the integration.

It is a process of germinating the definite integrals.

The formula is:

{eq}\displaystyle\int sinx\ dx=-cosx+c\\\\ \displaystyle\int secx\ dx=ln\left | tanx+secx \right |+c\\\\ \displaystyle\int cosx\ dx=sinx+c\\\\ {/eq}

## Answer and Explanation:

We have to evaluate all the given integrals:

**Part A.)**

{eq}\displaystyle\int t^2cos(2t)dt\\\\ {/eq}

Applying the formula of integration by parts:

{eq}\displaystyle\int (uv)dx=u\displaystyle\int v\ dx-\displaystyle\int \left [ \dfrac{d}{dx}(u)\displaystyle\int v\ dx \right ]dx\\\\ =t^2\displaystyle\int cos(2t)dt-\displaystyle\int \left [ \dfrac{d}{dx}(t^2)\displaystyle\int cos(2t)dt \right ]dt\\\\ =t^2\left [ \dfrac{sin(2t)}{2} \right ]-\displaystyle\int \left [ 2t\cdot \dfrac{sin2t}{2} \right ]dt\\\\ =\dfrac{t^2sin(2t)}{2}-\displaystyle\int tsin(2t)dt\\\\ {/eq}

Again applying the formula of integration by parts we get:

{eq}=\dfrac{t^2sin(2t)}{2}-\left [ t\displaystyle\int sin(2t)dt-\displaystyle\int \left [ \dfrac{d}{dt}(t)\displaystyle\int sin2t\ dt \right ] \right ]dt\\\\ =\dfrac{t^2sin(2t)}{2}-\left [ t\left ( \dfrac{-cos(2t)}{2} \right )\displaystyle\int \left ( \dfrac{-cos(2t)}{2} \right )dt \right ]\\\\ =\dfrac{t^2sin(2t)}{2}+\dfrac{t\ cos(2t)}{2}-\dfrac{1}{2}\displaystyle\int cos(2t)dt\\\\ =\dfrac{t^2sin(2t)}{2}+\dfrac{tcos(2t)}{2}-\dfrac{1}{4}sin(2t)+c\\\\ {/eq}

Here c is integration constant.

**Part B.)**

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

Let:

{eq}lnx=z\\\\ \dfrac{1}{x}dx=dz\\\\ {/eq}

Hence integration will be:

{eq}=\displaystyle\int \dfrac{1}{\sqrt {z^2-1}}dz\\\\ {/eq}

Substituting trigonometric values:

{eq}z=sect\\\\ dz=sect\ tant\\\\ {/eq}

Or,

{eq}t=sec^{-1}(z)\\\\ {/eq}

Hence integration further:

{eq}=\displaystyle\int \dfrac{sect\ tant}{\sqrt {sec^2t-1}}dt\\\\ {/eq}

We know the formula of trigonometric is:

{eq}sec^2x-tan^2x=1\\\\ {/eq}

{eq}=\displaystyle\int \dfrac{sect\ tant}{tant}dt\\\\ =\displaystyle\int sect\ dt\\\\ =ln\left | tan(t)+sec(t) \right |+c\\\\ =ln\left | tan\left ( sec^{-1}(z) \right )+sec\left ( ec^{-1}(z) \right ) \right |+c\\\\ =ln\left | tan\left ( sec^{-1}(lnx) \right )+lnx \right |+c\\\\ {/eq}

Here c is integration constant.

**Part C.)**

{eq}\displaystyle\int \dfrac{2x-3}{(x^2-4x+8)^{\frac{3}{2}}}dx\\\\ {/eq}

We may write as:

{eq}x^2-4x+8=(x-2)^2+4\\\\ {/eq}

Hence integration will be:

{eq}=\displaystyle\int \dfrac{2x-3}{\left ( (x-2)^2+4 \right )^{\frac{3}{2}}}dx\\\\ {/eq}

Let:

{eq}x-2=z\\\\ dz=dx\\\\ {/eq}

Or,

{eq}x=z+2\\\\ {/eq}

Therefore:

{eq}=\displaystyle\int \dfrac{2z+4-3}{\left ( z^2+4 \right )^{\frac{3}{2}}}dz\\\\ =\displaystyle\int \dfrac{2z+1}{\left ( z^2+4 \right )^{\frac{3}{2}}}dz+\displaystyle\int \dfrac{dz}{(z^2+4)^{\frac{3}{2}}}dz\\\\ {/eq}

Substituting:

{eq}z^2+4=t\\\\ 2z\ dz=dt\\\\ {/eq}

Hence:

{eq}=\displaystyle\int \dfrac{dt}{t^{\frac{3}{2}}}+\displaystyle\int \dfrac{sec^2t}{(4tan^2t+4)^{\frac{3}{2}}}dt\\\\ =\left [ \dfrac{t^{\frac{-1}{2}}}{\frac{-1}{2}} \right ]+\displaystyle\int \dfrac{2sec^2t}{(4sec^2t)^{\frac{3}{2}}}dt\\\\ =\dfrac{-2}{t^{\frac{1}{2}}}+\displaystyle\int \dfrac{2}{8sect}dt\\\\ =\dfrac{-2}{(z^2+4)^{\frac{1}{2}}}+\dfrac{1}{4}\displaystyle\int cost\ dt\\\\ =\dfrac{-2}{\left ( z^2+4 \right )^{\frac{3}{2}}}+\dfrac{1}{4}sint+c\\\\ {/eq}

Here c is integration constant.

#### Learn more about this topic:

from AP Calculus AB & BC: Homework Help Resource

Chapter 13 / Lesson 13