How do I solve these integrals involving trigonometric functions?

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In summary, the conversation involved finding the integral involving sin, cos, sec, and tan, with two parts A and B. In part A, the integrand was sin^3x * sqrt(cosx)dx and the solution involved using u-substitution with u = cosx. In part B, the integrand was sec^3(2x) * tan(2x)dx and the solution involved using the power rule and simplifying the resulting expression.
  • #1
BuBbLeS01
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Find the integral...Please HELP!

Homework Statement



Find the integral involving sin, cos, sec, and tan:
A. sin^3x * sqrt(cosx)dx
B. sec^3(2x) * tan(2x)dx

Homework Equations





The Attempt at a Solution


I will start with part A first...
A. sin^3x * sqrt(cosx) dx
sin^2*x * sqrt(cosx) * sinx dx
(1-cos^2*x) * sqrt(cosx) * sinx dx
sinx * sqrt(cosx) - sinx * cosx dx

but not I don't know what to do??
 
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  • #2
You did well, but just stop on the second last line, make a substitution u= sin x, then you have a simple integral in powers of u no?
 
  • #3
So I stop here...

(1-cos^2*x) * sqrt(cosx) * sinx dx

?

I am not understanding how to use u-substitution with this? u = sinx, du = cosx but that's not it?
 
  • #4
SOrry sorry! I meant u=cos x, du = -sin x dx.
 
  • #5
Umm but I am still not understanding lol I am sorry...

(1-cos^2*x) * sqrt(cosx) * sinx dx

u=cos x, du = -sin x dx

So if I do u^1/2 the integral is 2/3 * u^3/2

but that only gets me... sqrt(cosx) * -sinx dx

right??
 
  • #6
When you make the substitution,

[tex]\int (1- \cos^2 x) \sqrt{\cos x} \sin x dx = - \int (1-u^2) \sqrt{u} du = -\int ( u^{1/2} - u^{5/2}) du [/tex]

You should be able to finish it off with the power rule, and replace back in u=cos x at the end.
 
Last edited:
  • #7
so if I leave it at...

1 - u^2 * sqrt(u) du

I could write it asx - cosx^2 * sqrt(cosx)
 
  • #8
Did you follow my last post :( ? Do you know how to integrate those terms I had in the last post?
 
  • #9
woops that wasn't done right...

(x - 1/3*cosx^3) * 2/3*sqrt(cosx)^3/2

can I write it like this...I know its not completely reduced yet but I will do that.
 
  • #10
Well when I evaluate the integral I posted, I get [tex]2\cos^{3/2} x \left( \frac{1}{3} - \frac{\cos^2 x}{7}} \right) [/tex]
 
  • #11
Why am I not understanding this...

(1-u^2) * 2/3 u^3/2

so I plug u=cosx back in and integrate...

(x - 1/3*cosx^3) * 2/3*(cosx)^3/2

can you show me how you got your answer?
 
  • #12
Do you follow post 6? I get my answer from post 6 by directly using the power rule.
 
  • #13
I thought it would be...

u^3/2 - u^5/2
 
  • #14
[tex]\int\sin^3 x\sqrt{\cos x}dx[/tex]

[tex]\int(1-\cos^2 x)\sqrt{\cos x}\sin xdx[/tex]

[tex]u=\cos x[/tex]
[tex]du=-\sin xdx[/tex]

[tex]-\int(1-u^2)\sqrt udu[/tex]

[tex]-\int(u^{\frac 1 2}-u^{\frac 5 2})du[/tex]
 
  • #15
BuBbLeS01 said:
I thought it would be...

u^3/2 - u^5/2

Check the power rule again.

rocomath said:
[tex]\int\sin^3 x\sqrt{\cos x}dx[/tex]

[tex]\int(1-\cos^2 x)\sqrt{\cos x}\sin xdx[/tex]

[tex]u=\cos x[/tex]
[tex]du=-\sin xdx[/tex]

[tex]-\int(1-u^2)\sqrt udu[/tex]

[tex]-\int(u^{\frac 1 2}-u^{\frac 5 2})du[/tex]

I believe that was already well established in this thread.
 
  • #16
Okay I don't know why I am not seeing what you are doing but when you have...
(1-u^2) * sqrt u

don't you multiply u^1/2 by (1-u^2) ?

and get u^1/2 - u^3/2

so I don't understand why the second term is ^5/2?
 
  • #17
BuBbLeS ... what is 1.5 and 2.5 in fractions?
 
  • #18
BuBbLeS01 said:
Okay I don't know why I am not seeing what you are doing but when you have...
(1-u^2) * sqrt u

don't you multiply u^1/2 by (1-u^2) ?

and get u^1/2 - u^3/2

so I don't understand why the second term is ^5/2?

[tex]x^a\cdot x^b = x^a^+^b[/tex]

Now add 1/2 and 2.
 
  • #19
OMG I feel like such an IDIOT right now...thanks!
 

What is the purpose of finding the integral?

The purpose of finding the integral is to determine the area under a curve or the accumulation of a quantity over a given interval. It is an important concept in calculus and is used in various fields such as physics, engineering, and economics.

What is the process for finding the integral?

The process for finding the integral involves using integration techniques such as substitution, integration by parts, and partial fractions. The specific method used depends on the type of function being integrated.

What are the different types of integrals?

There are two types of integrals: indefinite integral and definite integral. The indefinite integral is used to find the general antiderivative of a function, while the definite integral is used to find the specific value of the area under a curve or the accumulation of a quantity over a given interval.

What are the common applications of finding the integral?

Finding the integral has many applications, including calculating displacement, velocity, and acceleration in physics, determining the total profit or loss in economics, and finding the center of mass in engineering. It is also used in probability and statistics to calculate the probability of a certain event occurring.

What are some tips for solving integrals?

Some tips for solving integrals include understanding the properties of integrals, such as linearity and the power rule, practicing different integration techniques, and being familiar with common trigonometric and exponential functions. It is also important to carefully check your work and use a graphing calculator or software to verify your answer.

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