What is the integral of sin(x)tan(x)?

Question

I have spent ages trying to answer this question! Have tried a number of different methods but cannot seem to get the correct solution. ANy help would be really appreciated!

Answer 1

-cosx -ln|cos x| + C

Answer 2

int(sinxtanx)

=>int(sinx*sinx/cosx)

=>int(sin^2(x)/cosx)

=>int(1-cos^2x/cosx)

=>int(secx-cosx)

=log I secx+tanx I - sinx+C

Answer 3

ln ( | cos x | / | sin x - 1 | ) - sin x

Answer 4

sin(x) tan (x) =  sin(x) sin (x)/ cos (x)

=sin²(x)/cos(x)

= (1 - cos²(x))/cos(x)

=1/cos(x)   - cos²(x)/cos(x)

= sec(x) - cos(x)

The integral of the - cos (x) part is  -sin (x)

To do integral of sec x we must perform this trick

= sec(x) (sec x + tan x)/(sec x + tan x)  

= {1/(sec(x) + tan(x)}  {sec(x)sec(x) + sec(x)tan(x)}

={1/sec(x) + tan(x)} {sec²(x) + sec(x)tan(x)}

Now observe that derivative of sec(x) + tan (x)

is sec(x) tan(x) + sec²(x)  so

right side brackets expression is derivative of

denominator of left side brackets expression so

⌠{1/sec(x) + tan(x)} {sec²(x) + sec(x)tan(x)}dx

=⌠1/u du = ln|u| + C

so integral is ln|sec(x)+tan(x)| + C

The ENTIRE INTEGRAL is therefore

ln|sec(x)+tan(x)| - sin(x) + C

Answer 5

sin(x)tan(x) = sin^2(x) / cos(x)

= [1 - cos^2(x)] / cos(x)

= sec(x) - cos(x)

The integral of sec(x) is rather ... interesting.

integral of sec(x) = ln|sec(x) + tan(x)| + C

You can verify that by differentiating. So now we know your integral will be:

ln|sec(x) + tan(x)| - sin(x) + C

Answer 6

INT(sinx.tanx)

INT{(sin^2x)/cosx}

=>INT{(1-cos^2x)/cosx}

=>INT{secx-cosx}

=>Ln(Isecx+tanxI)-sinx+constant

[NOTE: int(secx)=Ln(Isecx+tanxI)   ]

u can also modify the answer in different forms by simplifying the answer....

also remember that indefinite integrals...can hav more than one answer..........