Integral of Trigonometric Functions 1

Topics

Try VividMath Premium to unlock full access

Start Free Trial

Lets see your results!

Points

Score

Time

Retry Quiz

Answered
Review

Question 1 of 4

Find the integral

\int {tan}^{2} x d x

$int tan^2 x dx$

1. $tan x + x + c$ $\text(tan) x+x+c$
2. $tan x - 2 x + c$ $\text(tan) x-2x+c$
3. ${tan}^{2} x - x + c$ $\text(tan)^2 x-x+c$
4. $tan x - x + c$ $\text(tan) x-x+c$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Integrals of Trigonometric Functions

\int cos = sin

$int \text(cos)=\text(sin)$

\int sin = - cos

$int \text(sin)=-\text(cos)$

\int {sec}^{2} = tan

$int \text(sec)^2=\text(tan)$

Integrating Trigonometric Functions

$\int f(\color{#004ec4}{g(x)}) dx=f(g(x))\cdot\frac{1}{\color{#004ec4}{g'(x)}} +c$

First, convert the function into a derivable function

Take note that

$\text(sec)^2=1+\text(tan)^2x$

$\text(sec)^2$	$=$	$1+\text(tan)^2x$
$\text(sec)^2$ $-1$	$=$	$1+\text(tan)^2x$ $-1$	Subtract $1$ from both sides
$\text(sec)^2-1$	$=$	$\text(tan)^2x$

Therefore, we can use

$\text(sec)^2-1$ as a derivable substitute

Finally, substitute the components into the formula

$\int f(\color{#004ec4}{g(x)}) dx$	$=$	$f(g(x))\cdot\frac{1}{\color{#004ec4}{g'(x)}} +c$

$\int \text{sec}^2\color{#004ec4}{x-1}\;dx$	$=$	$\text{tan}\;x-x +c$	Substitute known values and integrate

$\text(tan) x-x+c$

Question 2 of 4

2. Question
Find the integral

$int tan x dx$
- 1. $-ln (\text(cos) x)+c$
- 2. $ln (\text(cos) x)+c$
- 3. $-ln (\text(sin) x)+c$
- 4. $ln (\text(sin) x)+c$
Hint

Help Video
Correct

Keep Going!
Incorrect
Need Text
Play
Current Time 0:00
/
Duration Time 0:00
Remaining Time -0:00
Stream TypeLIVE
Loaded: 0%
Progress: 0%
0:00
Fullscreen
Video Acceleration:
On
Off

00:00
Mute
Playback Rate
1x
2x
1.5x
1.25x
1x
0.75x
0.5x
Subtitles
subtitles off
Captions
captions off
English
Chapters
Chapters

Integrals of Trigonometric Functions

$int \text(cos)=\text(sin)$

$int \text(sin)=-\text(cos)$

$int \text(sec)^2=\text(tan)$

Integrating Trigonometric Functions

$\int f(\color{#004ec4}{g(x)}) dx=f(g(x))\cdot\frac{1}{\color{#004ec4}{g'(x)}} +c$

First, convert the function into a derivable function

Take note that $\text(tan) x=(\text(sin) x)/(\text(cos) x)$

Therefore, we can use $(\text(sin) x)/(\text(cos) x)$ as a derivable substitute

Next, take note that $d/(dx)\text(cos) x=-\text(sin) x$

This means that the function satisfies the derivative of a natural logarithm $(f'(x))/f(x)$ , if the equation is balanced

We can use $-1$ as a constant to balance the function

$=$ $-int (-\text(sin) x)/(\text(cos) x)$

Finally, integrate the function into a natural logarithm

$-int (-\text(sin) x)/(\text(cos) x)$ $=$ $-ln (\text(cos) x)+c$

$-ln (\text(cos) x)+c$

Question 3 of 4

Find the integral

$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\;\text{cos}x\;dx$

1. $1/(sqrt2)$
2. $1-sqrt2$
3. $1-1/(sqrt2)$
4. $sqrt2$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Integrals of Trigonometric Functions

$int \text(cos)=\text(sin)$

$int \text(sin)=-\text(cos)$

$int \text(sec)^2=\text(tan)$

First, integrate the trigonometric function

$\int_{\color{#00880A}{\frac{\pi}{4}}}^{\color{#9a00c7}{\frac{\pi}{2}}}\;\text{cos}\;x\;dx$

$=$

$\bigg[\text{sin}\;x\bigg]_{\color{#00880A}{\frac{\pi}{4}}}^{\color{#9a00c7}{\frac{\pi}{2}}}$

Integrate

$\text(cos)x$

Finally, get the difference of the upper and lower limits substituted to the integral as

$x$ .

	$\bigg[\text{sin}\;x\bigg]_{\color{#00880A}{\frac{\pi}{4}}}^{\color{#9a00c7}{\frac{\pi}{2}}}$

$=$	$\sin{\color{#9a00c7}{\frac{\pi}{2}}}-\sin{\color{#00880A}{\frac{\pi}{4}}}$	Substitute the limits

$=$	$1-1/(sqrt2)$	Evaluate

$1-1/(sqrt2)$

Question 4 of 4

Find the integral

$\int_{0}^{\frac{\pi}{3}}\;3\;\text{sin}\frac{x}{2}\;dx$

1. $3sqrt3-6$
2. $-3sqrt3+6$
3. $3sqrt3+6$
4. $-3sqrt3-6$

Incorrect

Need Text

Play

Remaining Time -0:00

Stream TypeLIVE

Loaded: 0%

Progress: 0%

0:00

Fullscreen

00:00

Mute

Integrals of Trigonometric Functions

$int \text(cos)=\text(sin)$

$int \text(sin)=-\text(cos)$

$int \text(sec)^2=\text(tan)$

First, integrate the trigonometric function

$\int_{\color{#00880A}{0}}^{\color{#9a00c7}{\frac{\pi}{3}}}\;\text{sin}\;\frac{x}{2}\;dx$	$=$	$\bigg[3\left(-\text{cos}\;\frac{x}{2}\right)\bigg]_{\color{#00880A}{0}}^{\color{#9a00c7}{\frac{\pi}{3}}}$	Integrate $3\text(sin) x/2$

	$=$	$\bigg[-6\text{cos}\;\frac{x}{2}\bigg]_{\color{#00880A}{0}}^{\color{#9a00c7}{\frac{\pi}{3}}}$	Simplify

Finally, get the difference of the upper and lower limits substituted to the integral as

$x$ .

	$\bigg[-6\text{cos}\;\frac{x}{2}\bigg]_{\color{#00880A}{0}}^{\color{#9a00c7}{\frac{\pi}{3}}}$

$=$	$-6\;\cos{\frac{\color{#9a00c7}{\frac{\pi}{3}}}{2}}-[-6\;\cos{\frac{\color{#00880A}{0}}{2}}]$	Substitute the limits

$=$	$-6 \text(cos)(pi/6)+(6*1)$	Evaluate

$=$	$-6*(sqrt3)/2+6$	$\text(cos) pi/6=(sqrt3)/2$

$=$	$-3sqrt3+6$	Simplify

$-3sqrt3+6$

Integral of Trigonometric Functions 1

Try VividMath Premium to unlock full access

Quiz summary

Information

1. Question

Hint

Excellent!

Integrals of Trigonometric Functions

Integrating Trigonometric Functions

2. Question

Hint

Keep Going!

Integrals of Trigonometric Functions

Integrating Trigonometric Functions

3. Question

Hint

Fantastic!

Integrals of Trigonometric Functions

4. Question

Hint

Well Done!

Integrals of Trigonometric Functions

Quizzes