Question

having trouble with trig substitution integrals.

Answer 1

What is the problem?

Answer 2

The question is meant to help me, but I've been having trouble understanding the lecture. Here is the question:

Answer 3

I'm sorry I couldn't type it out, but it's on a website :/

Answer 4

the dirst thing is that you should be thorough with you trigonometry then only the correct substitution will strike you

Answer 5

*first

Answer 6

I know a decent amount of trig.... I know that I need to use arctan for this substitution.

Answer 7

or tan... sorry.

Answer 8

g(t) = sqrt(1+(6t-7)^2)? Haven't taken this calculus in a while.

Answer 9

How did you get that? I had set x=tan(6t-7), but I was stuck after that.

Answer 10

you want to get the original equation to be simplified. so 1/g(t) where g(t) = another function which can be integrated?

Answer 11

...Where are we getting tan from?

Answer 12

so that u get 1+ tan^2 t in square root sign

Answer 13

so that u get sec t in denominator

Answer 14

so that u get cos t in numerator

Answer 15

Ah okay I see, your using an identity. Okay.

Answer 16

f(t) = cos t

Answer 17

I'm sorry, but I'm still a bit confused about what I'm doing :/. am I still using x=tan(6t-7)? My apologies, but it's all very new to me.

Answer 18

u want 1+tan^2 t under square root sign so that u can write it as sec^2 t ......ok?
now u had (6x-7)^2 to make it tan^2 t
u put 6x-7 = tan t
or
6x=tan t +7
or x = (tan t +7) / 6 = g(t)
got this ?

Answer 19

ah ok I think I understand now. Thank you.

Answer 20

glad to help ^_^