How do I solve these integrals involving trigonometric functions?

[BuBbLeS01]

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??

 

[Gib Z]

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?

 

[BuBbLeS01]

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?

 

[Gib Z]

SOrry sorry! I meant u=cos x, du = -sin x dx.

 

[BuBbLeS01]

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??

 

[Gib Z]

When you make the substitution,

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

You should be able to finish it off with the power rule, and replace back in u=cos x at the end.

 

[BuBbLeS01]

so if I leave it at...

1 - u^2 * sqrt(u) du

I could write it asx - cosx^2 * sqrt(cosx)

 

[Gib Z]

Did you follow my last post :( ? Do you know how to integrate those terms I had in the last post?

 

[BuBbLeS01]

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.

 

[Gib Z]

Well when I evaluate the integral I posted, I get $$2\cos^{3/2} x \left( \frac{1}{3} - \frac{\cos^2 x}{7}} \right)$$

 

[BuBbLeS01]

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?

 

[Gib Z]

Do you follow post 6? I get my answer from post 6 by directly using the power rule.

 

[BuBbLeS01]

I thought it would be...

u^3/2 - u^5/2

 

[rocomath]

$$\int\sin^3 x\sqrt{\cos x}dx$$

$$\int(1-\cos^2 x)\sqrt{\cos x}\sin xdx$$

$$u=\cos x$$
$$du=-\sin xdx$$

$$-\int(1-u^2)\sqrt udu$$

$$-\int(u^{\frac 1 2}-u^{\frac 5 2})du$$

 

[Gib Z]

I thought it would be...

u^3/2 - u^5/2

Check the power rule again.

$$\int\sin^3 x\sqrt{\cos x}dx$$

$$\int(1-\cos^2 x)\sqrt{\cos x}\sin xdx$$

$$u=\cos x$$
$$du=-\sin xdx$$

$$-\int(1-u^2)\sqrt udu$$

$$-\int(u^{\frac 1 2}-u^{\frac 5 2})du$$

I believe that was already well established in this thread.

 

[BuBbLeS01]

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?

 

[rocomath]

BuBbLeS ... what is 1.5 and 2.5 in fractions?

 

[Snazzy]

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?

$$x^a\cdot x^b = x^a^+^b$$

Now add 1/2 and 2.

 

[BuBbLeS01]

OMG I feel like such an IDIOT right now...thanks!