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# MATH LOVERS OUT THERE.. please, help!!!?

my AP Calculus is one of the biggest pains i've ever gone through, please, help with some of these integrals (i couldn' t find the integral sign, so i'll be creative: S is "my integral" sign).

Thanks!

1) S(t^2 * sint dt)

2) S(x^3 * e^-2 dx)

3) S(tan^-1 y dy)

Solve the defferential equation:

4) dy/dx= x^2 * e^(4x)

even if you solve one of these problems, i'll be mooooore than happy!

for the first 3 problems just evaluate the integral :D

### 2 Answers

- 1 decade agoFavorite Answer
for the first two, use the equation:

S(u*dv) = u*v - S(v*du), where u and v are functions of t (in number one) , or x (in number two). you should do that twice for number one, and three times for number two, to get rid of the polynomial. Two is actually really easy as you typed it, but i'm guessing there's another x in there somewhere (to make it S(x^3 * e^-2x dx) ), so do what i said.

Number four, multiply both sides by dx, then integrate, same method as above.

If you don't recognize "u" and "v," another way to write that is

S( f(x)g'(x) ) = f(x)g(x) - S ( f'(x)g(x) ).

Have fun.

- Anonymous1 decade ago
The fourth one also looks for an integral... it's just written differently.

Except (3), they look like integration by parts exercises. Remember S(u dv) = uv - S(v du).

For (1) you have u=tÂ², dv=sint dt â du=2t dt, v=-cost. So...

S(u dv) = tÂ²(-cost) - S(-cost * 2t dt).

You will need to do integration by parts again on this one. Good luck!

For (3), try substituting u = tan^-1 y â y=tanu, dy = ... (look up the derivative of tan), etc.

- sahsjingLv 71 decade ago
All these problems can be done with integration by parts. Just be more patient. Also, try to use mental substitution to save time and space.

1)

â«t^2 * sint dt

= -t^2cost+â«2t dsint, integration by parts

= -t^2cost+2t sint + 2cost, integration by parts

= 2tsint-(t^2-2)cost

3)

â«tan^-1 y dy

= ytan^-1 y - â«y/(1+y^2) dy, integration by parts

= ytan^-1y - (1/2)ln(2+y^2), mental substitution