Calculate the integral (limits)?

calculate:

int cos(x)/(sin(x)+2)(sin(x)+3) from -pi/6 to pi/6

Let u = sin(x) <==> du = cos(x) dx. Then, we have:

Lower bound: x = -π/6 ==> u = sin(-π/6) = -1/2

Upper bound: x = π/6 ==> u = sin(π/6) = 1/2.

∫ cos(x)/{[sin(x) + 2] * [sin(x) + 3]} dx (from x=-π/6 to π/6)

= ∫ 1/[(u + 2)(u + 3)] du (from u=-1/2 to 1/2).

By partial fractions, we have for some A and B:

1/[(u + 2)(u + 3)] = A/(u + 2) + B/(u + 3)

==> 1 = A(u + 3) + B(u + 2).

i) Letting u = -2 ==> A = 1

ii) Letting u = -3 ==> -B = 1 ==> B = -1.

Thus:

1/[(u + 2)(u + 3)] = 1/(u + 2) - 1/(u + 3).

Then, applying the FTC yields:

∫ 1/[(u + 2)(u + 3)] du (from u=-1/2 to 1/2)

= ∫ [1/(u + 2) - 1/(u + 3)] du (from u=-1/2 to 1/2)

= (ln|u + 2| - ln|u + 3|) (evaluated from u=-1/2 to 1/2)

= [ln(5/2) - ln(7/2)] - [ln(3/2) - ln(5/2)]

= ln(5/7) - ln(3/5)

= ln(25/21).

I hope this helps!

#1

∫(cos(x)) dx/(sin(x) + 2)(sin(x) + 3) dx

u = sin(x)

du = cos(x) dx

Change the limits in terms of u since we're integrating with respect to u now.

New limits:

a=sin(-pi/6) = -1/2

b = sin(pi/6) = 1/2

∫du/(u+2)(u+3) from -1/2 to 1/2

Use partial fractions:

A/(u+2) + B/(u+3) = 1/(u+2)(u+3)

(u+3)A + (u+2)B = 1

B = -1 and A = 1

∫(1/(u+2) - 1/(u+3)) du from -1/2 to 1/2

ln|u + 2| - ln|u + 3| eval. from -1/2 to 1/2

=ln(5/2) - ln(7/2) - ln(3/2) + ln(5/2) = ln(25/4) - ln(21/4) = ln(25/21)

#2