Question

integral cos^(3)xsin^(4)x dx.

Need help breaking this down to integrate.

Answer 1

Problem will become easy, if we could change its form to something like
\[\int\sin^ n x\times \cos x\]
Now if we put sin x = t, cos x dx=dt
It'll become
\[\int t^n dt\]

Answer 2

Let's try that
\[\cos^3 x \sin ^ 4x\]
\[(1-\sin^2 x )\cos x \times \sin^4 x\]
\[\sin^4 x \times \cos x -\sin^6 x \times \cos x\]
Do you get this?

Answer 3

yes i do

Answer 4

Now, just substitute sin x = t and then integrate. Can you do that?

Answer 5

so you would be looking something like:\[t ^{4}*cosx-t ^{6}*cosx\]
is that what it would look like before i Integrate?

Answer 6

\[\int (t^4 -t^6 )dt\]

Answer 7

Have you studied substitution method?

Answer 8

no, it is an online class and they are jumping around.

Answer 9

Ok, I'll explain

Answer 10

Suppose we have the simple integral
\[\int \sin x\times\cos x\ dx\]
There are several ways of doing this, I'll use the substitution method
Let
\[\sin x = t
\]Now differentiate it, with respect to x
\[\cos x =\frac {dt}{dx}\]
or
\[\cos x\ dx= dt\]
Our integral is
\[\int \sin x\times\cos x\ dx\]
substituting
sin x =t
and cos x dx =dt
We get
\[\int t\ dt\]
Integration is easy
\[\frac {t^2} 2 +c\]
Put the value of t back
\[\frac{\sin^2x}{2}+C\]

Answer 11

wow that is simple

Answer 12

In you problem, after using substitution you'll get
\[\int (t^4-t^6 ) dt\]
Remember we have substituted sin x =t
Not sin^4 x =t or sin^6 x =t
If we do that, there are consequences :P

Answer 13

so i get the following.\[((5sinx)/5)-((7sinx)/7)+c\] does that appear correct?

Answer 14

\[\frac{\sin^5 x }{5}-\frac{\sin^7 x}{7}+C\]

Use latex to type equations, I've used the following code
\frac{\sin^5 x }{5}-\frac{\sin^7 x}{7}+C

Answer 15

ok so the only issue i have is the answer I am supposed to get is \[\frac{ 1 }{ 70 }\sin ^{5}(x)(5\cos(2x)+9)+c\]

Answer 16

We can get that as well. Pull out sin^ 5 x
you'll get a sin^2 x term, substitute
\[\sin^ 2 x =\frac{1-\cos 2 x}{2}\]
Then simplify

Answer 17

thank you, I think I got it.

Answer 18

Welcome :)