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Solve the integrals: (a)\int^{\frac{\pi}{6}}_{0} \frac{\sin(t)}{\cos^{2}(t)}dt (b)\int^{13}_{0}...

Question:

Solve the integrals:

(a){eq}\int^{\frac{\pi}{6}}_{0} \frac{\sin(t)}{\cos^{2}(t)}dt{/eq}

(b){eq}\int^{13}_{0} \frac{dx}{\sqrt[3]{(1+2x)^{2}}}{/eq}

(c){eq}\int^{\frac{1}{16}}_{\frac{1}{48}} \csc (8\pi) + \cot (8\pi)+dt{/eq}

Integration.

In integration, we are given of function x, and we have to evaluate the derivative of it.

The process of finding the answer is called integration.

In integration the function is integrated is known as integrands.

The formula of integration is:

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

Answer and Explanation:

Here we have to calculate the given integrals:

Part A.)

{eq}\displaystyle\int _0^{\frac{\pi}{6}}\dfrac{sin(t)}{cos^2(t)}dt\\\\ {/eq}

Substituting:

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

Substituting the values of limits:

When {eq}t=0\\\\ {/eq}

{eq}z=1\\\\ {/eq}

When {eq}t=\dfrac{\pi}{6}\\\\ {/eq}

{eq}z=cos(\dfrac{\pi}{6})\\\\ z=\dfrac{\sqrt 3}{2}\\\\ {/eq}

Therefore integrating:

{eq}=\displaystyle\int _1^{\frac{\sqrt 3}{2}}\dfrac{-dt}{t^2}\\\\ =-\left ( \dfrac{-1}{t} \right )_1^{\frac{\sqrt 3}{2}}\\\\ =\left [ \dfrac{1}{\frac{\sqrt 3}{2}}-1 \right ]\\\\ =\dfrac{2}{\sqrt 3}-1\\\\ {/eq}

Part 2.)

{eq}\displaystyle\int _0^{13}\dfrac{dx}{\sqrt[3]{(1+2x)^2}}\\\\ =\displaystyle\int _0^{13}\dfrac{dx}{(1+2x)^{\frac{2}{3}}}\\\\ {/eq}

Substituting:

{eq}1+2x=z\\\\ 2\ dx=dz\\\\ dx=\dfrac{dz}{2}\\\\ {/eq}

Substituting for the values of limits:

When {eq}x=0\\\\ {/eq}

{eq}z=1\\\\ {/eq}

When {eq}x=13\\\\ {/eq}

{eq}z=1+2\cdot 13\\\\ z=27\\\\ {/eq}

Hence integrating:

{eq}\displaystyle\int _1^{27}\dfrac{1}{z^{\frac{2}{3}}}\cdot \dfrac{dz}{2}\\\\ =\dfrac{1}{2}\left [ \dfrac{z^{\frac{1}{3}}}{\frac{1}{3}} \right ]_1^{27}\\\\ =\dfrac{3}{2}\left [ \left ( 27 \right )^{\frac{1}{3}}-1 \right ]\\\\ =3 {/eq}

Part C.)

{eq}\displaystyle\int _{\frac{1}{48}}^{\frac{1}{16}}\left [ csc(8\pi)+cot(8\pi) \right ]dt\\\\ =\displaystyle\int _{\frac{1}{48}}^{\frac{1}{16}}csc(8\pi)dt+\displaystyle\int _{\frac{1}{48}}^{\frac{1}{16}}cot(8\pi)dt\\\\ =csc(8\pi)\left [ t \right ]_{\frac{1}{48}}^{\frac{1}{16}}+cot(8\pi)\left [ t \right ]_{\frac{1}{48}}^{\frac{1}{16}}\\\\ =csc(8\pi)\left [ \dfrac{1}{16}-\dfrac{1}{48} \right ]+cot(8\pi)\left [ \dfrac{1}{16}-\dfrac{1}{48} \right ]\\\\ =\dfrac{1}{24}csc(8\pi)+\dfrac{1}{24}cot(8\pi)\\\\ {/eq}


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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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