Evaluate ∫ d x 5 + 4 cos x .

Given that, We have to evaluate ∫ d x 5 + 4 cos x . ⇒ ∫ d x 5 + 4 cos x = ∫ d x 5 + 4 1 - tan 2 x 2 1 + tan 2 x 2 ∵ cos θ = 1 - tan 2 θ 2 1 + tan 2 θ 2 ⇒ ∫ d x 5 + 4 cos x = ∫ 1 + tan 2 x d x 5 1 + tan 2 x 2 + 4 1 - tan 2 x 2 ⇒ ∫ d x 5 + 4 cos x = ∫ s e c 2 x d x 5 1 + tan 2 x 2 + 4 1 - tan 2 x 2 ∵ 1 + tan 2 θ = s e c 2 θ ⇒ ∫ d x 5 + 4 cos x = ∫ s e c 2 x d x 9 + tan 2 x 2 Put tan x 2 = t ⇒ 1 2 s e c 2 x 2 d x = d t ⇒ ∫ s e c 2 x d x 9 + tan 2 x 2 = ∫ 2 d t 9 + t 2 = 2 ∫ d t 3 2 + t 2 = 2 3 tan - 1 t 3 + C ∵ ∫ d x a 2 + x 2 = 1 a tan - 1 x a Put t = tan x 2 , ⇒ 2 3 tan - 1 tan x 2 3 + C Therefore, ∫ d x 5 + 4 cos x = 2 3 tan - 1 tan x 2 3 + C .

Evaluate ∫ 2 x - 1 ( x - 1 ) ( x + 2 ) ( x - 3 ) d x

Given, ∫ 2 x - 1 ( x - 1 ) ( x + 2 ) ( x - 3 ) d x Since the denominator of the integrand contains three non-repeated linear factors. So, we can write the integrand into the sum of three partial fractions. Thus, we have 2 x - 1 ( x - 1 ) ( x + 2 ) ( x - 3 ) = A x - 1 + B x + 2 + C x - 3 . . . . i , where A , B , and C are real numbers. ⇒ 2 x - 1 ( x - 1 ) ( x + 2 ) ( x - 3 ) = A x + 2 x - 3 + B x - 1 x - 3 + C x - 1 x + 2 ( x - 1 ) ( x + 2 ) ( x - 3 ) ⇒ 2 x - 1 = A x + 2 x - 3 + B x - 1 x - 3 + C x - 1 x + 2 . . . . i i Substituting x = 3 in equation i i , we have C 2 5 = 2 3 - 1 = 5 ⇒ C = 1 2 Now, Substituting x = - 2 in equation i i , we have 2 - 2 - 1 = B - 3 - 5 ⇒ - 4 - 1 = 15 B ⇒ B = - 1 3 Now, Substituting x = 1 in equation i i , we have ⇒ 2 1 - 1 = A 1 + 2 1 - 3 ⇒ 1 = - 6 A ⇒ A = - 1 6 Thus, from equation i , we have 2 x - 1 ( x - 1 ) ( x + 2 ) ( x - 3 ) = - 1 6 x - 1 - 1 3 x + 2 + 1 2 x - 3 Thus, ∫ 2 x - 1 ( x - 1 ) ( x + 2 ) ( x - 3 ) d x = ∫ - 1 6 x - 1 - 1 3 x + 2 + 1 2 x - 3 d x = - 1 6 ∫ 1 x - 1 d x - 1 3 ∫ 1 x + 2 d x + 1 2 ∫ 1 x - 3 d x = - 1 6 log x - 1 - 1 3 log x + 2 + 1 2 log x - 3 + C , where C is an integral constant. Hence, ∫ 2 x - 1 ( x - 1 ) ( x + 2 ) ( x - 3 ) d x = = - 1 6 log x - 1 - 1 3 log x + 2 + 1 2 log x - 3 + C , where C is an integral constant.