Indefinite integrals and Differentiate

**ADARSH** · February 9th 2009, 03:26 AM

1)

$ \frac{1}{4} = 2^ {-2} $
$ 16 = 2^4 $

so
$2^{-2x} * 2^{\frac{4x}{2}}$

$ =2^{2x-2x} =1 $

**CAIR** · February 9th 2009, 03:30 AM

thank you

**ADARSH** · February 9th 2009, 03:35 AM

b) If secod question is
$log_2 {x} + 5log_2{x} +\frac{1}{2}log_2{(x-1)}$
than
$=log_2 {x} + log_2{(x+1)^5} +log_2{(x-1)^\frac{1}{2}}$

$ =log_2 {[x(x+1)^5(x-1)^\frac{1}{2}]} $

which seems to be the answer

**ADARSH** · February 9th 2009, 03:50 AM

$ \int{\frac{7x+3}{\sqrt{x}}dx} $
$ = \int{ 7x * x^{\frac{-1}{2}}+3x^{\frac{-1}{2}}dx} $
$ = \int{ 7x^{\frac{1}{2}}dx}+\int{3x^{\frac{-1}{2}}dx} $

$ = \frac{14x^{\frac{3}{2}}}{3} + 6x^{\frac{1}{2}} + c $

**CAIR** · February 9th 2009, 04:17 AM

thanks

adarsh

**CAIR** · February 9th 2009, 02:11 PM

Im stuck on the last question can someone help?

**Jhevon** · February 9th 2009, 06:11 PM

Originally Posted by CAIR

Im stuck on the last question can someone help?

for $\int \frac {\sin^3 x + 3 \cos x + 2}{\sin^2 x}~dx$

divide the $\sin^2 x$ into each term in the numerator, you get

$\int \left( \sin x + 3 \csc x \cot x + 2 \sec^2 x \right) ~dx = \int \sin x ~dx + 3 \int \csc x \cot x ~dx + 2 \int \sec^2 x ~dx$

each of those integrals you should be able to do. Hint: think "what functions would I take the derivative of to get these?"

As for the derivative problems: please clarify what F(x) is. is it $\ln (x) \cdot \tan^{-1} (4x)$ or $\ln [x \cdot \tan^{-1}(4x)]$ . more than likely it's the first. in either case, you are going to have to use the chain rule and the product rule. do you remember these? what have you tried?

for $G(x) = \cos^{-1} ( \sin^{-1} x)$ you will also need the chain rule.

recall that the chain rule says: $\frac d{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$

here, $f(x) = \cos^{-1} x$ and $g(x) = \sin^{-1} x$

now try it.

by the way, the places where you put *, you shouldn't put *. students make mistakes with this all the time, it is WRONG to think of something like sin(x) as sin times x. sin by itself makes no sense, it is not a product!

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