• Mar 18th 2010, 05:05 AM
joey1
$\displaystyle \int (3x-1)(x+2)dx$

b)$\displaystyle \int cos(7x+2)dx$

c)$\displaystyle \int \frac {y}{\sqrt {1+2y^2}} dy$

d)$\displaystyle \int \frac {sin2x}{1+sin^2x}dx$

e)$\displaystyle \int_{0}^\frac{\pi}{2}{sin2xcosx} dx$

f)$\displaystyle \int [\ln{4} - \frac {1}{\ln(e^{2x})}]dx$
• Mar 18th 2010, 05:38 AM
General
Quote:

Originally Posted by joey1
$\displaystyle \int (3x-1)(x+2)dx$

b)$\displaystyle \int cos(7x+2)dx$

c)$\displaystyle \int \frac {y}{\sqrt {1+2y^2}} dy$

d)$\displaystyle \int \frac {sin2x}{1+sin^2x}dx$

e)$\displaystyle \int_{0}^\frac{\pi}{2}{sin2xcosx} dx$

f)$\displaystyle \int [\ln{4} - \frac {1}{\ln(e^{2x})}]dx$

(a) : Expand.
(b) & (c) : Use a substitution.
(d) & (e) : Rewrite $\displaystyle sin(2x)$ as $\displaystyle 2sin(x)cos(x)$, then use a substitution.
(f) : Recall that : $\displaystyle \int ( \, f(x) - g(x) \, ) \, dx = \int f(x) \, dx \, - \int g(x) \, dx$ .. and Note that $\displaystyle ln(e^{f(x)}) = f(x)$.

Also, post only two question in each thread.