Integration Q (1 Viewer)

[Rmd_1]

anyone can help me on this questions! thx
i can do it in another way just can get the same answer...........?

 

[Rmd_1]

sorry forgot type the questions :apig: :apig:

 

[CrashOveride]

it's just black ?

 

[redruM]

Originally posted by CrashOveride
it's just black ?

it worked for me.

 

[kimmeh]

it doesnt work for me

 

[CM_Tutor]

It's black for me too, and even if I use Paint to fill it white, i can't read the integration question that sort of appears.

 

[velox]

works for me, are u sure this is 4u?

 

[schmeichung]

well i think so..3u would b easier
and i didnt face too much tri sub in 3u

*use tri sub

 

[:: ck ::]

nerppp black for me too

 

[McLake]

I can't read bmp's. Can you try another pic.

 

[Rmd_1]

just download it and then u can see it!!!

 

[CM_Tutor]

Rmd_1, thanks for the advice. For all, the questions are:

1. Find ∫ (0 --> 1/2) dx / (1 - x²)^3/2, where the answer is 1 / √3

2. Find &int; (tan x --> cot x) dt / (1 + t²), for 0 < x < &pi; / 2, where the answer is (&pi; / 2) - 2x

3. Find &int; x / sqrt(1 + x²) dx, where the answer is sqrt(1 + x²) + C

 

[CM_Tutor]

Question 1 is a substitution problem, x = sin &theta;

Question 2 integrates to tan^-1t, so hinges on you being able to show that tan^-1cot x = (&pi; / 2) - x

Question 3 is a substitution problem, u = 1 + x²