Integration H2 Maths

Audi SG

Have difficulties with these 3 questions.

1. integrate xe^x^2 dx

2. integrate e^ sq rt x / sq rt x dx

3. integrate cos 3x (sin 3x)^0.5 dx

Yep, thats all. If you have any idea on how to solve these, please help me out. :) Appreciate it.

qdtimes2 SG

1. Differentiate e^(x^2), you get 2xe^(x^2). So, 0.5Int [2xe^(x^2)]dx = 0.5e^(x^2)

2. this is similar to qn 1..differentiate e^ sqrt x and see what you get first

3. this is a form after you differentiated [sin(3x)]^3/2.. i dunno how else we can do to solve this problem though

wee_ws SG

Hi,

These questions are of the form

- f'(x) e^{f(x)},

- f'(x) [ f(x) ]^n,

so be sure to recognise them.

Thanks.

Cheers,
Wen Shih

Han chiang SG

Hi, I just did a revision of calculus few days back. I hope I can help you out.

One important formula for f'(x).[f(x)]^n. When integrating, it becomes[f(x)]^n+1 / (n+1) . Means [f(x)] power plus 1, divide by the new power.

2) Integrate e^ sq rt x / sq rt x . This is the in the form of f'(x).[f(x)]^n, so rearrange (1/sq rt x) times e^ sq rt x. e^sq rt x is the [f(x)]^n. So f'(x) = 1/2sqrt x . And 1/sq rt x = 2(1/2sq rt x.) We have the f'(x) now. So integrate e^sq rt x / sq rt x = 2e^sq rt x.

3) Integrate cos3x(sin3x)^0.5. Same thing f'(x).[f(x)]^n. (sin3x)^0.5 is the [f(x)]^n. f'(x) = 3cos3x. And cos3x = 1/3 (3cos3x). We have the f'(x) now. So integrate cos3x(sin3x)^0.5 = 2/9 (sin3x)^3/2

Hope I didn't make any mistakes.