Question

evalute the integral (sin^3 (x)/cosx)dx

Answer 1

Note that sin^3 = sin(1- cos^2)

Put that into your expression and it should look quite a bit easier.

Answer 2

\[\frac{\sin^3(x)}{\cos(x)}=\frac{\sin(x)\sin^2(x)}{\cos(x)}=\frac{\sin(x)(1-\cos^2(x))}{\cos(x)}=\cdots\]

Answer 3

can u continue plz, because i needed to check the answer

Answer 4

and is integral

Answer 5

Tell me what you think the answer is and I'll tell you if it's right.

First the integral of sin x/cos x.
And then the integral of sin x.cos^2 x/cos x

Answer 6

you cant do that i guess... do we put u= cosx ?

Answer 7

In the first one, yes.

Answer 8

and then is the - integral (1-u^2)/u du ??

Answer 9

No. if u = cos x, then du/dx = -sin x

so \[\frac{\sin x}{\cos x} dx = - \frac{du/dx}{u} dx = - \frac{du}{u}\]

Answer 10

Is the answer >

Answer 11

-ln|cosx| + (cos^2(x)/2) ?

Answer 12

And make sure you add a constant C.

Answer 13

Isnt the sign positive.. becuz the - would multiply in integral

Answer 14

yes, you're right.

Answer 15

put u=cosx
du=-sinxdx
so -(1-u^2)/u du
-(ln(u)-u^2/2)+c
ln(sec(x))+cosx^2/2+c