# integration problems - urgent

• Jun 20th 2006, 11:27 AM
lalitchugh
integration problems - urgent
Hi,

Attached document contain question rearding indefinite integrals.

Can someone help to provide solution for all three questions ASAP.

Apologise for improper way of defining the question as i don't have the software in which we can draw maths problem nicely.

Best Regards,
Lalit Chugh[B]
• Jun 20th 2006, 12:20 PM
Jameson
1. $\int \frac{\tan(x)}{\cot(x)}dx$

1/cot(x) is simply tan(x), so this integral is really $\int \tan^2(x)dx$

Use this identity and it should be easy.

$1+\tan^2(x)=\sec^2(x)$
• Jun 20th 2006, 12:22 PM
Jameson
2. $\int \sqrt{x}\left(x^5+\frac{3}{x}\right)dx$

Just distribute the square root of x and use basic power rules.

$\int x^{5+\frac{1}{2}}+\frac{3}{x}*x^{\frac{1}{2}}dx$
• Jun 21st 2006, 07:06 PM
distance
Quote:

Originally Posted by Jameson
1. $\int \frac{\tan(x)}{\cot(x)}dx$

1/cot(x) is simply tan(x), so this integral is really $\int \tan^2(x)dx$

Use this identity and it should be easy.

$1+\tan^2(x)=\sec^2(x)$

_____________

for this problem, i think just rewrite the function in term tanx = sinx/cosx
then, you will flip cosx into the denominator. It will come out like this, intergal of (sinx/cos^2x)dx
next step, using the sub methods

let u = cosx, du = - sinxdx

-du = sinxdx

then just jam that into the integral.

u will get something like du/u^2

bring u^2 on top : u-2

take the integral. u^-1/-1 or

• Jun 22nd 2006, 01:53 AM
nath_quam
Quote:

Originally Posted by distance
_____________

for this problem, i think just rewrite the function in term tanx = sinx/cosx
then, you will flip cosx into the denominator. It will come out like this, intergal of (sinx/cos^2x)dx
next step, using the sub methods

let u = cosx, du = - sinxdx

-du = sinxdx

then just jam that into the integral.

u will get something like du/u^2

bring u^2 on top : u-2

take the integral. u^-1/-1 or

Yes. This is the correct answer. Using the substitution I gave you get $\int \sec^2(x)-1dx$. And thus the answer follows.