Integration Worksheet6 solution7

In this page integration worksheet6 solution7 we are going to see solution of some practice question from the worksheet of integration.

Question 16

Integrate the following with respect to x,  cosec³ x

Solution:

∫ cosec³ x dx = ∫ cosec x (cosec²x) dx

now we are going to apply partial differentiation

u = cosec x                     dv = cosec²x

du = - cosec x cot x           v = - cot x

∫ u dv = u v - ∫ v du

          = (cosec x)(- cot x) - ∫ - cot x (- cosec x cot x) dx

          = -cosec x cot x - ∫ cosec x cot²x dx

          = -cosec x cot x - ∫ cosec x (cosec²x - 1) dx

          = -cosec x cot x - ∫ cosec³x dx + ∫ cosec x dx

∫ cosec³x dx = -cosec x cot x - ∫ cosec³x dx + ∫ cosec x dx

∫ cosec³x dx + ∫ cosec³x dx = -cosec x cot x + log tan (x/2) + C

2∫ cosec³x dx = -cosec x cot x + log tan (x/2) + C

∫ cosec³x dx = (1/2)[-cosec x cot x + log tan (x/2)] + C

∫ cosec³x dx = (1/2)[-cosec x cot x] + (1/2)[log tan (x/2)] + C


Question 17

Integrate the following with respect to x,  e^ax cos bx

Solution:

∫ e^ax cos bx  dx

u = cos b x           dv = e^ax

du = - b sin bx        v = e^ax/a

∫ u dv = u v - ∫ v du

               = (cos b x)(e^ax/a) - ∫(e^ax/a) (- b sin bx) dx

               = (cos b x)(e^ax/a) + (b/a) ∫ e^ax (sin bx) dx--------(1)

now we are going to find the integration value of ∫ e^ax (sin bx) dx separately and then we can apply those values in the first equation.

∫ e^ax (sin bx) dx

 u = sin bx                 dv = e^ax

 du = b cos bx             v = e^ax/a

                = (sin bx)(e^ax/a) - ∫(e^ax/a)(b cos bx) dx

                 = (sin bx)(e^ax/a) - (b/a) ∫e^ax cos bx dx

   = (cos b x)(e^ax/a) + (b/a) [(sin bx)(e^ax/a) - (b/a) ∫e^ax cos bx dx]

∫e^ax cos bx dx+ (b²/a²) ∫e^ax cos bx dx

                           = (cos b x)(e^ax/a)+(b/a)(sin bx)(e^ax/a)

∫(1 + (b²/a²)) e^ax cos bx dx = (cos b x)(e^ax/a)+(b/a)(sin bx)(e^ax/a)

∫((a²+b²)/a²) e^ax cos bx dx = (cos b x)(e^ax/a)+(b/a)(sin bx)(e^ax/a)

∫e^ax cos bx dx = [a²/(a²+b²)](cos b x)(e^ax/a)+(b/a)(sin bx)(e^ax/a)

                      = [a²/(a²+b²)](e^ax/a)[(a cos b x+b sin bx)/a]

                      = [e^axa²/a²(a²+b²)][(a cos b x+b sin bx)]

                      = [e^ax/(a²+b²)][(a cos b x+b sin bx)]  

integration worksheet6 solution7 integration worksheet6 solution7