## sat_chen Group Title Can someone explain me this trigonometric substitution one year ago one year ago

1. sat_chen Group Title
2. sat_chen Group Title

so what i am confused is on how someone went from

3. sat_chen Group Title

|dw:1359567855994:dw|

4. sat_chen Group Title

to the solution

5. sat_chen Group Title

i think you have to log in to wolfram before you can see the steps

6. sat_chen Group Title

we get sin taking the integral of cos?

7. shubhamsrg Group Title

Yes I can't see the steps.

8. shubhamsrg Group Title

Was the substitution x = 5 siny ?

9. sat_chen Group Title

yea you have to log in for that i think and it would take me a year to type the solutions here XD

10. sat_chen Group Title

so someone with wolfram account tell me how they went from that step to the solution

11. sat_chen Group Title

12. shubhamsrg Group Title

Leme try anyways, let x= 5secy => dx= 5 secy tany dy so we have inside integral 5 secy tany dy/( 125 sec^3 y ( -25 + 25sec^2 y)^1/2) => 5sec y tany dy/( 625 sec^3 y (tan^2 y)^1/2 ) => 1/125 ( dy/sec^2y)) or simply cos^2y dy Am I right so far ?

13. sat_chen Group Title

i got a better way for you to see the solutions this is part 2 of the solution

14. sat_chen Group Title

15. sat_chen Group Title

theres all the steps to the solution can you take me from that 0/250 + sin20/500 + c to the solution

16. sat_chen Group Title

and yes you are right with your steps

17. sat_chen Group Title

so from the integral of (1/125)(cos^20d0) you do the half angle formula to get integral of (1/125)(1+cos20)/2dp then you split and get 0/250 + sin2(theta)/500 + c so my question if how you get from here to the answer

18. sat_chen Group Title

|dw:1359570048133:dw|

19. sat_chen Group Title

i get the second part where it is the inverse of sec-1 (x/5) but i dont get where they got the 5sqrt(x^2) stuff from

20. sat_chen Group Title

|dw:1359570931896:dw| the stuff i dont get is in that box in case you are still confused on what i am confused about

21. phi Group Title

you are asking about $\frac{1}{2} \sin(2 u)$ with $u= \sec^{-1}\left(\frac{x}{5}\right)$ In terms of cosine, rather than secant, we can say $u= \cos^{-1}\left(\frac{5}{x}\right)$ If we take the hypotenuse to be x, in a triangle with side 5 |dw:1359572361980:dw| the other side is $$\sqrt{x^2-25}$$ Using these sides, we have $\cos u = \frac{5}{x}$ and $\sin u = \frac{\sqrt{x^2-25}}{x}$ sin(2u) = 2 sin(u) cos(u) sub in the above definitions wolfram then multiplied by sqrt(x^2)/x (why?) to get your boxed expression