1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Nookie1986 [14]
3 years ago
15

Integrate this for me please

Mathematics
2 answers:
pashok25 [27]3 years ago
6 0
Let's do a variable substitution, by the formula \int u\,dv=uv-\int v\,du

I=\int\tan^2(x)\sec^4(x)dx=\int\underbrace{\tan(x)\sec^3(x)}_{u}\underbrace{\tan(x)\sec(x)dx}_{dv}\\ u=\tan(x)\sec^3(x)\\\\ du=(\sec^5(x)+3\sec^3(x)\tan^2(x))dx\\\\ du=\sec^3(x)(\underbrace{\sec^2(x)}_{\tan^2+1}+3\tan^2(x))dx\\\\ du=\sec^3(x)(4\tan^2(x)+1)dx\\\\\\ dv=\sec(x)\tan(x)dx\\\\ v=\sec(x)

So:

I=\tan(x)\sec^4(x)-\int\sec^4(x)(4\tan^2(x)+1)dx\\\\ I=\tan(x)\sec^4(x)-4\underbrace{\int\sec^4(x)\tan^2(x)dx}_{I}-\int\sec^4(x)dx\\\\ I=\tan(x)\sec^4(x)-4I-\int\sec^4(x)dx\\\\ 5I=\tan(x)\sec^4(x)-\underbrace{\int\sec^4(x)dx}_{I_2}\\\\

Solving I₂ using substitution, too:

I_2=\int\sec^4(x)dx=\int\underbrace{\sec^2(x)}_{u}\underbrace{\sec^2(x)dx}_{dv}\\\\\\ u=\sec^2(x)\\\\ du=2\sec^2(x)\tan(x)dx\\\\\\ dv=\sec^2(x)dx\\\\ v=\tan(x)

Then:

I_2=\tan(x)\sec^2(x)-\int 2\tan^2(x)\sec^2(x)dx\\\\ I_2=\tan(x)\sec^2(x)-2\int\tan^2(x)\sec^2(x)dx\\\\\\ y=\tan(x)\to dy=\sec^2(x)dx\to dx=\dfrac{dy}{\sec^2(x)}\\\\ \tan^2(x)\sec^2(x)dx=y^2\sec^2(x)\dfrac{dy}{\sec^2(x)}=y^2dy\\\\\\ I_2=\tan(x)\sec^2(x)-2\int y^2dy\\\\ I_2=\tan(x)\sec^2(x)-2\cdot\dfrac{y^3}{3}\\\\ I_2=\tan(x)\sec^2(x)-\frac{2}{3}\tan^3(x)

Hence, substituting I₂ in I:

5I=\tan(x)\sec^4(x)-I_2\\\\ 5I=\tan(x)\sec^4(x)-(\tan(x)\sec^2(x)-\frac{2}{3}\tan^3(x))\\\\ 5I=\tan(x)\sec^4(x)-\tan(x)\sec^2(x)+\frac{2}{3}\tan^3(x))\\\\ \boxed{I=\frac{1}{5}\tan(x)\sec^4(x)-\frac{1}{5}\tan(x)\sec^2(x)+\frac{2}{15}\tan^3(x)+C}

Now, using the limits of integration in the expression E of the statement:

E=\displaystyle\int^{\dfrac{\pi}{6}}_0\tan^2(x)\sec^4(x)dx\\\\\\ E=(\frac{1}{5}\tan(\frac{\pi}{6})\sec^4(\frac{\pi}{6})-\frac{1}{5}\tan(\frac{\pi}{6})\sec^2(\frac{\pi}{6})+\frac{2}{15}\tan^3(\frac{\pi}{6}))-\\\\ (\frac{1}{5}\tan(0)\sec^4(0)-\frac{1}{5}\tan(0)\sec^2(0)+\frac{2}{15}\tan^3(0))\\\\\\ E=(\frac{1}{5}\cdot\frac{1}{\sqrt3}\cdot(\frac{2}{\sqrt3})^4-\frac{1}{5}\cdot\frac{1}{\sqrt3}\cdot(\frac{2}{\sqrt3})^2+\frac{2}{15}(\frac{1}{\sqrt3})^3)-\\\\ (\frac{1}{5}\cdot0\cdot1^4-\frac{1}{5}\cdot0\cdot1^2+\frac{2}{15}\cdot0^3)


E=\frac{1}{5\sqrt3}\cdot\frac{16}{9}-\frac{1}{5\sqrt3}\cdot\frac{4}{3}+\frac{2}{15}\cdot\frac{1}{3\sqrt3}-0+0-0\\\\\ E=\frac{1}{5\sqrt3}(\frac{16}{9}-\frac{4}{3}+\frac{2}{9})\\\\ E=\frac{1}{5\sqrt3}\cdot\frac{16-12+2}{9}=\frac{1}{5\sqrt3}\cdot\frac{6}{9}=\frac{1}{5\sqrt3}\cdot\frac{2}{3}\\\\ \boxed{E=\dfrac{2}{15\sqrt3}}
mel-nik [20]3 years ago
5 0
\int\limits_{0}^{\frac{\pi }{6}}tan^2(x)sec^2(x)\cdot dx
\\------------------\\
u=tan(x)\implies \frac{du}{dx}=sec^2(x)\implies \frac{du}{sec^2(x)}=dx
\\------------------\\
\int\limits_{0}^{\frac{\pi }{6}}u^2sec^2(x)\cdot \cfrac{du}{sec^2(x)}\implies 
\int\limits_{0}^{\frac{\pi }{6}}u^2\cdot du
\\ \quad \\


\textit{now, we need to change the bounds as well, so}
\\------------------\\
u(0)=tan(0)\implies 0
\\ \quad \\

u\left( \frac{\pi }{6} \right)=tan\left( \frac{\pi }{6} \right)\implies \frac{1}{\sqrt{3}}
\\------------------\\
thus\implies \int\limits_{0}^{\frac{1 }{\sqrt{3}}}u^2\cdot du

and surely you can take it from there,
recall, that, since we changed the bounds, with the u(x),
you don't need to change the variable "u", and simply,
get the integral of it, simple enough, and apply those bounds
You might be interested in
The vertices of a hyperbola are located at (−4, 1) and (4, 1). The foci of the same hyperbola are located at (−5, 1) and (5, 1).
topjm [15]

Answer:

x^2\ 16 - y^2/9 = 1

step-by-step explanation:

soooo the equation for a hyperbola that is horizontal is x^2/a^2 - y^2/b^2 = 1

a hyperbola should always equal one so dont forget that when writing your equation becuase it is easy to forget.

it helps to graph this so you can see it better

to find a it is the distance from the center to the vertices which is 4 so in the equation you will write 16 becuase it is a^2

then you need to find b. to get b you have to figure out that c is the distance from the center to the foci which is 5 and it is all related to the plythagorm theorum becuase it forms a right triangle. so you do c^2 - a^2 = b^2

you get 9 for b^2 because 25-16=9 and so you put that in the equation

3 0
3 years ago
Read 2 more answers
Help! Use trigonometric ratios to determine whether or not it is possible to construct triangle ABC if b = 3, c = 5, and B = 45°
elena-s [515]
No its not posible because the sine is 1.5 and it cant be higher then 1
7 0
3 years ago
Read 2 more answers
Dont send me pdfs or links please answer correctly for brainliest
Yuki888 [10]

Answer:

C. Sample: All names are written on pieces of paper and dropped into a container. Three hundred names are drawn.

Step-by-step explanation:

please mark this answer as brainlest

7 0
2 years ago
Read 2 more answers
The perimeter of a triangle is 76cm side of the triangle is twice as long as side b .side c is 1 cm longer than a side a .find t
stepladder [879]
P= a+b+c = 76
side a = 2b = 30
side b = b = 15
side c = 2b + 1 = 31
7 0
3 years ago
Max <br> p=3x+2y <br> subject to <br> 5x+y&lt;16 <br> 2x+3y&lt;22 <br> x&gt;0 <br> y&gt;0
garri49 [273]

Answer:For all the corner points, the maximum is at point (1, 3)

From the graph of the constraints, the corner points of the feasibility region are (0, 0), (0, 10/3), (1, 3), (2, 0)

For (0, 0): p = 0 + 2(0) = 0

For (0, 10/3): p = 0 + 2(10/3) = 20/3 = 6.67

For (1, 3): p = 1 + 2(3) = 1 + 6 = 7

For (2, 0): p = 2 + 2(0) = 2

Therefore, solution = (1, 3)

3 0
3 years ago
Other questions:
  • The polynomial X3+ 8 is equal to
    14·1 answer
  • Please help
    6·1 answer
  • 14, 15 and 16 please
    10·1 answer
  • If inflation from last year to this year was 2% and a pair of designer jeans sold for $75 last year, what would you expect to pa
    12·1 answer
  • What is a differential equation
    7·1 answer
  • Sam invested $7500, some at 7% interest and the rest at 10%. How much did he invest at each rate if he received $570.00 in inter
    8·1 answer
  • Please explain to me how to find y. I’m so confused.
    14·1 answer
  • In February, 1985, the cost of a gallon of milk was $2.26. In November of 2007, how much would that gallon of milk cost?
    15·1 answer
  • I need help on this please
    8·2 answers
  • A rectangular building for a gym is three times as long as it is wide. Just inside the walls of the building, there is a 6ft rec
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!