ʃsin y dy/
√1 + cosy
please help..thanks..
that's integration of sin y dy divided by (over) square root of 1 + cosy.
Recall that
sin^2 (y/2) = (1 + cos(y))/2
so, √1 + cosy = √2 cos y/2
sin y = 2 sin y/2 cos y/2
So, the integrand becomes
2 siny/2 cosy/2 / √2cosy/2
= √2 sin(y/2)
Integral is just -2√2 cos(y/2)
oops:
cos^2(y/2) = (1 + cos(y))/2
cos^2(y/2)=(1+2cos(y/2))/2
but 2 enter cos(y/2)
hence=(1+cos(2y/2))/2
=(1+cosy)/2
To solve this integral, you can use a substitution method. Let's follow these steps:
Step 1: Begin by assigning a variable to the part inside the square root. Let's say u = 1 + cos(y).
Step 2: Calculate the derivative of u with respect to y. In this case, du/dy = -sin(y).
Step 3: Rearrange the equation from step 2 to solve for dy. We obtain dy = du/(-sin(y)).
Step 4: Substitute the expressions from steps 1 and 3 into the original integral:
∫(sin(y) dy) / √(1 + cos(y)) = ∫(sin(y) (du/-sin(y))) / √(u) = -∫(du/√u)
Since the numerator sin(y) cancels out with the denominator -sin(y), we are left with -∫(du/√u).
Step 5: Solve the simplified integral -∫(du/√u).
The integral of √u is 2/3 * u^(3/2), so -∫(du/√u) becomes -2/3 * u^(3/2) + C, where C is the constant of integration.
Step 6: Substitute back the original variable into the result. Replace u with 1 + cos(y):
= -2/3 * (1 + cos(y))^(3/2) + C
So, the solution to the given integral is -2/3 * (1 + cos(y))^(3/2) + C.