evaluate the intergral { 1) 4x^2 (^4\sqrt {(8+4x^3)}) dx\\ 2) cosec^2 (4 \theta + 6 ) d \theta...

Question:

evaluate the intergral

{eq}1) 4x^2 (^4\sqrt {(8+4x^3)}) dx\\ 2) cosec^2 (4 \theta + 6 ) d \theta \\3) \frac{(sin t)}{ (10+cos t)^4}\\ 4) x^2 cos (x^3+3) \sqrt {(sin (x^3 +3))} dx {/eq}

Integration.

If we apply the partial fraction method in integration the primary step involves the factorization of denominator.

If the numerator is not less than denominator then we have to apply the second steps that are to apply the long division rule of the polynomials.

If we want to calculate the area under the line then simply we can integrate it.

The formula is:

{eq}\displaystyle\int cosec^2x\ dx=-cotx+c\\\\ \displaystyle\int x^n\ dx=\dfrac{x^{n+1}}{n+1}+c\\\\ {/eq}

Answer and Explanation:

Here we have to evaluate the given integrals:

Part 1.)

{eq}4x^2\left ( \sqrt[4]{(8+4x^3)} \right )dx\\\\ =\displaystyle\int 4x^2\left ( \sqrt[4]{(8+4x^3)} \right )dx\\\\ =\displaystyle\int 4x^2\left ( 8+4x^3 \right )^{\frac{1}{4}}dx\\\\ {/eq}

Let:

{eq}8+4x^3=z\\\\ 4\cdot 3x^2\ dx=dz\\\\ 4x^2\ dx=\dfrac{dz}{3}\\\\ {/eq}

Hence:

{eq}\displaystyle\int z^{\frac{1}{4}}\cdot \dfrac{dz}{3}\\\\ =\dfrac{1}{3}\left [ \dfrac{z^{\frac{5}{4}}}{\frac{5}{4}} \right ]+c\\\\ =\dfrac{1}{3}\cdot \dfrac{4}{5}\left [ \left ( 8+4x^3 \right )^{\frac{5}{4}} \right ]+c\\\\ =\dfrac{4}{15}\dfrac{4}{5}\left [ \left ( 8+4x^3 \right )^{\frac{5}{4}} \right ]+c\\\\ {/eq}

Here c is integration constant.

Part 2.)

{eq}\displaystyle\int cosec^2(4\theta+6)d\theta\\\\ {/eq}

Let:

{eq}4\theta+6=z\\\\ 4\ d\theta=\dfrac{dz}{4}\\\\ {/eq}

Hence:

{eq}\displaystyle\int cosec^2z\cdot \dfrac{dz}{4}\\\\ =-\dfrac{cotz}{4}+c\\\\ =\dfrac{-cot(4\theta+6)}{4}+c\\\\ {/eq}

Here c is integration constant.

Part 3.)

{eq}\displaystyle\int \dfrac{sint}{(10+cost)^4}dt\\\\ {/eq}

Let:

{eq}10+cost=z\\\\ -sint\ dt=dz\\\\ {/eq}

Therefore:

{eq}\displaystyle\int \dfrac{-dz}{z^4}\\\\ =-\displaystyle\int z^{-4}dz\\\\ =-\left [ \dfrac{z^{-3}}{-3} \right ]\\\\ =\dfrac{1}{3}\left [ \dfrac{1}{z^3} \right ]+c\\\\ =\dfrac{1}{3}\left [ \dfrac{1}{(10+cost)^3} \right ]+c\\\\ {/eq}

Here c is integration constant.

Part 4.)

{eq}\displaystyle\int x^2cos(x^3+3)\sqrt {sin(x^3+3)}dx\\\\ {/eq}

Let:

{eq}sin(x^3+3)=z\\\\ cos(x^3+3)\cdot 3x^2\ dx=dz\\\\ x^2cos(x^3+3)dx=\dfrac{dz}{3}\\\\ {/eq}

Therefore integrating:

{eq}\displaystyle\int \sqrt z\cdot \dfrac{dz}{3}\\\\ =\dfrac{1}{3}\left [ \dfrac{z^{\frac{3}{2}}}{\frac{3}{2}} \right ]+c\\\\ =\dfrac{2}{9}\left [ sin(x^3+3) \right ]^{\frac{3}{2}}+c\\\\ {/eq}

Here c is integration 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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