1) ∫ (x + 1)¯ᴺ dx
2) ∫ (cos³x sin2x) dx
3) ∫ t sqrt(1-t²) dt
2) ∫ (cos³x sin2x) dx
3) ∫ t sqrt(1-t²) dt
-
1) You can let u = x+1 and then dx = du, so you shifted the problem one unit to the right.
The integral of u^(-n) du is u^(-n+1)/(-n+1) by the Power Rule for integration.
Notice it does not work if n=1 because then you get ln, the natural logarithm.
2) sin 2x = 2 sin x cos x
so the integral there is equal to
integral 2 cos^4(x) sin x dx
Let u = cos x and then du = - sin x dx
so the integral is now equal to
integral -2 u^4 du
= -2 u^5/5
= -2 cos^5 x/5
You might wanna differentiate that to see if it checks out.
3) u = 1-t^2 du = -2 t dt
= integral t sqrt(u) du/(-2 t)
then the t's cancel out in the top and bottom
= -1/2 integral sqrt(u) du
then the power is 1/2
so it is 2/3 u^(3/2) times the -1/2
The integral of u^(-n) du is u^(-n+1)/(-n+1) by the Power Rule for integration.
Notice it does not work if n=1 because then you get ln, the natural logarithm.
2) sin 2x = 2 sin x cos x
so the integral there is equal to
integral 2 cos^4(x) sin x dx
Let u = cos x and then du = - sin x dx
so the integral is now equal to
integral -2 u^4 du
= -2 u^5/5
= -2 cos^5 x/5
You might wanna differentiate that to see if it checks out.
3) u = 1-t^2 du = -2 t dt
= integral t sqrt(u) du/(-2 t)
then the t's cancel out in the top and bottom
= -1/2 integral sqrt(u) du
then the power is 1/2
so it is 2/3 u^(3/2) times the -1/2