Trig Substitution for Integrating sin^3(x)/cos(x) - Homework Help

[nightshade123]

Homework Statement

$$\int$$ ((sin(x))^3/(cos(x)) )*dx

The Attempt at a Solution

alright i have been trying to use

u= cosx
-du = sinx

but it doesn't make sense bause there is still a sinx^2 to account for

so i know i need to make a trig substitution but i can't figure out the appropriate
substitution for sinx^2 even though i think the 1/2 ( 1- cos2x) is hte only one question... when u use a u substituion for u = cosx -du = sindx would u write

1/2 $$\int$$ ((sinx*( 1 - cos2x)) / cosx)*dx- 1/2 $$\int$$( (1 - 2u) / u ) *du
if u = cosx and u have to change a cos2x... what would you write 2u... that wouldn't make sensebecause it would come out 2*cosx which does not equatl cos2x

idk where I am lost... i can't logically follow through this problem for some reason, any advice would be amazing, thanks for your time.

 

[Mystic998]

Assuming your work is right, just write (1 - 2u)/u = 1/u - 2

 

[nightshade123]

according the calculator ti 89 the answer is -ln abs(cosx) - sinx^2 / 2accordign to what i ended up w/ i got

-1/2 ( ln abs(cosx) - 2*cosx )

which doesn't make sense my work is wrong it has to be lol

if u have a few mins to try to work through it on paper let me kno what u see

 

[Mystic998]

Oops, I misread. Just a second.

Edit: Your work is wrong. cos(2x) =/= 2u if u = cos(x).

I'd suggest instead of using $sin^{2}(x) = \frac{1 - cos(2x)}{2}$ that you use a more basic identity.

 

[nightshade123]

i know how to evaluate those easy integrals but the challenge is turning a complex integral into a easy integral

 

[nightshade123]

1/u du is

ln(u)

2 =2u

u = cosx

 

[nightshade123]

yea i have been trying to figure out a simpler substitution and it just isn't working

 

[Mystic998]

What's the very first trig identity you ever learned?

 

[Littlepig]

what about using $$sin(x)^{2}=1-cos(x)^{2}$$ ?

you get 2 simple integrals i guess...

 

[nightshade123]

sinx^2 + cosx^2 = 1

 

[nightshade123]

omg i made that way to hard >.< lol thanks
ill reevaluate it and see what happens

 

[fermio]

$$\int\frac{\sin^3 x}{\cos x}dx=\int\frac{\sin x\frac{1-\cos(2x)}{2}}{\cos x}dx=\frac{1}{2}\int\frac{\sin x(1-\cos(2x))}{\cos x}\frac{d(\cos x)}{-\sin x}=-\frac{1}{2}\int\frac{1-\cos(2x)}{\cos x}d(\cos x)=$$
$$=-\frac{1}{2}\int\frac{1-(2\cos^2 x-1)}{\cos x}d(\cos x)=-\frac{1}{2}\int\frac{2}{\cos x}-2\cos x d(\cos x)=-\ln\cos x+\frac{\cos^2 x}{2}+C$$

where $$d(\cos x)=-\sin x dx$$
$$dx=\frac{d(\cos x)}{-\sin x}$$
$$\cos(2x)=2\cos^2 x-1$$

Answer should be $$-\ln\cos x+\frac{1}{4}\cos (2x)+C$$ according to http://integrals.wolfram.com/index.jsp


$$\int\frac{\sin^3 x}{\cos x}dx=\int\frac{\sin x(1-\cos^2 x)}{\cos x}dx=\int\frac{\sin x(1-\cos^2 x)}{\cos x}\frac{d(\cos x)}{-\sin x}=-\int\frac{1-\cos^2 x}{\cos x}d(\cos x)=$$
$$=-\int\frac{1}{\cos x}-\cos x d(\cos x)=-(\ln\cos x-\frac{\cos^2 x}{2})+C$$
where $$d(\cos x)=-\sin x dx$$
$$dx=\frac{d(\cos x)}{-\sin x}$$

Does this http://integrals.wolfram.com/index.jsp integrator giving wrong answer? Becouse $$\frac{\cos^2 x}{2}$$ not equal to $$\frac{1}{4}\cos (2x)$$

 

[Big-T]

$$\frac{\cos^2(x)}{2}=\frac{\cos(2x)}{4}+C_1$$

 

[fermio]

Interesting
$$\frac{\cos^2 0.5}{2}\approx 0.385075576$$
$$\frac{\cos 1}{4}\approx 0.135075576$$
$$\frac{\cos^2 0.5}{2}-\frac{\cos 1}{4}=0.25$$
$$\frac{\cos^2 0.3}{2}-\frac{\cos 0.6}{4}=0.25$$
...
But why?

 

[awvvu]

$$\frac{\cos^2 0.5}{2}-\frac{\cos 1}{4}=0.25$$
$$\frac{\cos^2 0.3}{2}-\frac{\cos 0.6}{4}=0.25$$
...
But why?

cos^2(x) = 1/2(1 + cos(2x))

 

[nightshade123]

$$\int$$ 1 / (sinx - 1) dx

wth i can't figure this out, thinkin about it to hard again

i broke a more complex integral down into this final part, diffrent problemi know it equals

cosx / (sinx - 1)

 

[Big-T]

Try conjugation.

 

[nightshade123]

reduces back into itself

 

[Big-T]

What about integrating f(x)=1/sin(x)?

 

[nightshade123]

1/sinx does not equal 1 / (sinx -1 )

in that case u suggest f(x) = ln ( sinx / cosx + 1 )

 

[Big-T]

My mistake.
But:
$$\frac{1}{\sin x -1}=\frac{\sin x +1}{\sin^2 x -1}=\frac{\sin x +1}{-\cos^2 x}=-\frac{\sin x}{\cos^2 x}-\frac{1}{\cos^2 x}$$

Or am I mistaken again?

 

[nightshade123]

$$\frac{1}{\sin x -1}=\frac{\sin x +1}{\sin^2 x -1}$$thats false.. what are you trying to say?

 

[Big-T]

$$\frac{1}{\sin x -1}=\frac{\sin x +1}{\left(\sin x -1\right)\left(\sin x +1\right)}=\frac{\sin x +1}{\sin^2 x -1}$$
Isn't it?

 

[nightshade123]

yea that's right i read it wrong, integrate that throughout do u end up w/

cosx / (sinx -1)

i didnt, i got -(sinx + 1) / cosx

 

[nightshade123]

dont worry about it

 

[Big-T]

Those two functions are in fact equal.