anti-derivative problems

• Nov 9th 2008, 04:53 PM
sess
anti-derivative problems
hi guys, i need help finding the anti-derivatives for these problems

Any help woud be appreciated

y' = tan(x)sec^2(x)

y' = 3x^(-1/2)

y' = -2cot^2(x)csc(x)
• Nov 9th 2008, 05:00 PM
Chris L T521
Quote:

Originally Posted by sess
hi guys, i need help finding the anti-derivatives for these problems

Any help woud be appreciated

y' = tan(x)sec^2(x)

Make the substitution $z=\sec x$

Quote:

y' = 3x^(-1/2)
Use the simple power rule...

--Chris
• Nov 9th 2008, 05:06 PM
sess
aight these are the answers that i got

y' = tan(x)sec^2(x) -> y = sec^2(x)tan(x)

y' = 3x^(-1/2) -> y = 1/(2x^(3/2))

y' = -2cot^2(x)csc(x) -> y = 2 (ln(abs(sin(x))))(csc(x))

did i get it?
• Nov 9th 2008, 05:16 PM
sess
pls help me out guys
• Nov 9th 2008, 05:18 PM
11rdc11

The answer for the 1st is

$\frac{\sec^2{x}}{2} + c$
• Nov 9th 2008, 05:29 PM
sess
can u show me the steps on how u got that answer please?
• Nov 9th 2008, 05:35 PM
11rdc11
Quote:

Originally Posted by sess
can u show me the steps on how u got that answer please?

$\int \tan{x}\sec^2{x}~dx$

Like chris said

$\sec{x}= u$

$\sec{x}\tan{x}dx=du$

$\int u^{n}~du = \frac{u^{n+1}}{n+1} + C$

Can you finish up?
• Nov 9th 2008, 05:42 PM
sess
ok Ive no idea wat u just said

i do know that the first derivative of sec(x) is sec(x)tan(x) right?

i also know that the first derivative of tan(x) is sec^2(x)

Where do i go from here? I really must appologize to you 11rdc11
I have not understood the concept of finding the anti-derivative at all

Would you please be so kind and explain? :(
• Nov 9th 2008, 06:02 PM
11rdc11
Quote:

Originally Posted by sess
ok Ive no idea wat u just said

i do know that the first derivative of sec(x) is sec(x)tan(x) right?

i also know that the first derivative of tan(x) is sec^2(x)

Where do i go from here? I really must appologize to you 11rdc11
I have not understood the concept of finding the anti-derivative at all

Would you please be so kind and explain? :(

Ok do you know the basic integral formula

$\int u^{n}~du = \frac{u^{n+1}}{n+1} + C$
• Nov 9th 2008, 06:04 PM
sess
Quote:

Originally Posted by 11rdc11
Ok do you know the basic integral formula

$\int u^{n}~du = \frac{u^{n+1}}{n+1} + C$

yep i understand that part
• Nov 9th 2008, 06:26 PM
11rdc11
Quote:

Originally Posted by sess
yep i understand that part

Ok n just represents the power.

So notice if we plug in the u of sub that I showed earlier the problem becomes the basic formula. Then just continue using the integral formula and back sub in the end.