Find ∫ x - 3 e x x - 1 3 d x

Let I = ∫ x - 3 e x x - 1 3 d x ⇒ I = ∫ x - 1 - 2 e x x - 1 3 d x ⇒ I = ∫ e x x - 1 x - 1 3 - 2 x - 1 3 d x ⇒ I = ∫ e x 1 x - 1 2 - 2 x - 1 3 d x ⇒ I = e x 1 x - 1 2 + c , c = constant of integration ∵ ∫ e x f x + f ' x d x = e x f x + c

Find ∫ x - 3 e x x - 1 3 d x

Let I = ∫ x - 3 e x x - 1 3 d x ⇒ I = ∫ x - 1 - 2 e x x - 1 3 d x ⇒ I = ∫ e x x - 1 x - 1 3 - 2 x - 1 3 d x ⇒ I = ∫ e x 1 x - 1 2 - 2 x - 1 3 d x ⇒ I = e x 1 x - 1 2 + c , c = constant of integration ∵ ∫ e x f x + f ' x d x = e x f x + c

∫ 0 π 2 cos 2 x d x

Let I = ∫ 0 π 2 cos 2 x d x ⇒ I = ∫ 0 π 2 cos 2 x + 1 2 d x We know that ∫ a b k f x d x = k ∫ a b f x d x ⇒ I = 1 2 ∫ 0 π 2 cos 2 x + 1 d x We know that ∫ a b f x ± g x d x = ∫ a b f x d x ± ∫ a b g x d x ⇒ I = 1 2 ∫ 0 π 2 cos 2 x d x + ∫ 0 π 2 d x ⇒ I = 1 2 sin 2 x 2 + x 0 π 2 ⇒ I = 1 2 sinπ 2 + π 2 - 0 ⇒ I = 1 2 0 + π 2 ⇒ I = π 4

Find the integral of 1 x 2 + a 2 with respect to x and hence evaluate ∫ 1 x 2 + 2 x + 2 d x .

Let I = ∫ 1 x 2 + a 2 d x Substitute x = a tan θ ⇒ θ = tan - 1 x a Differentiating with respect to θ on both sides ⇒ d x = a sec 2 θ d θ ⇒ I = ∫ a sec 2 θ a 2 tan 2 θ + a 2 d θ ⇒ I = ∫ a sec 2 θ a 2 tan 2 θ + 1 d θ ⇒ I = 1 a ∫ sec 2 θ sec 2 θ d θ ∵ sec 2 x - tan 2 x = 1 ⇒ I = 1 a ∫ d θ ⇒ I = 1 a θ + c ⇒ I = 1 a tan - 1 x a + c Now, ∫ 1 x 2 + 2 x + 2 d x = ∫ 1 x 2 + 2 x + 1 + 1 d x ⇒ ∫ 1 x 2 + 2 x + 2 d x = ∫ 1 x + 1 2 + 1 2 d x ⇒ ∫ 1 x 2 + 2 x + 2 d x = tan - 1 x + 1 + c .

Prove that ∫ − a a f ( x ) d x = { 2 ∫ 0 a f ( x ) d x , if f ( x ) is an event function 0 , if f ( x ) is an odd function and hence evaluate ∫ − 1 1 sin 5 ⁡ x cos 4 ⁡ x d x .

Let I = ∫ − a a f ( x ) d x ⇒ I = ∫ − a a f ( x ) d x = ∫ − a 0 f ( x ) d x + ∫ 0 a f ( x ) d x = I 1 + I 2 In I 1 , put x = - t ⇒ d x = - d t When x = - a , t = a and when x = 0 , t = 0 Therefore, I 1 = ∫ a 0 f ( − t ) ( − d t ) = − ∫ a 0 f ( − t ) d t = ∫ 0 a f ( − t ) d t ∵ ∫ a b f ( x ) d x = − ∫ b a f ( x ) d x = ∫ 0 a f ( − x ) d x ∵ ∫ a b f ( x ) d x = ∫ a b f ( t ) d t ∴ ∫ − a a f ( x ) d x = ∫ 0 a f ( − x ) d x + ∫ 0 a f ( x ) d x If f is an even function f - x = f x Therefore, ∫ − a a f ( x ) d x = ∫ 0 a f ( x ) d x + ∫ 0 a f ( x ) d x = 2 ∫ 0 a f ( x ) d x If f is an odd function f - x = - f x Therefore, ∫ − a a f ( x ) d x = - ∫ 0 a f ( x ) d x + ∫ 0 a f ( x ) d x = 0 Now, let I 3 = ∫ − 1 1 sin 5 ⁡ x cos 4 ⁡ x d x . Let f ( x ) = sin 5 ⁡ x cos 4 ⁡ x Then f ( − x ) = sin 5 ⁡ ( − x ) cos 4 ⁡ ( − x ) = − sin 5 ⁡ x cos 4 ⁡ x = − f ( x ) i.e. f is an odd function Hence, I 3 = 0 .

Prove that ∫ 0 a f ( x ) d x = ∫ 0 a f ( a - x ) d x and hence evaluate ∫ 0 π 2 cos 5 x sin 5 x + cos 5 x d x .

To prove ∫ 0 a f ( x ) d x = ∫ 0 a f ( a - x ) d x Let ( a - x ) = t ⇒ - d x = d t When x = 0 ⇒ t = a When x = a ⇒ t = 0 Now, ∫ 0 a f ( a - x ) d x = - ∫ a 0 f ( t ) d t = ∫ 0 a f ( t ) d t (Apply property ∫ b a f ( x ) d x = - ∫ a b f ( x ) d x ) ⇒ ∫ 0 a f ( a - x ) d x = ∫ 0 a f ( x ) d x (Apply property ∫ a b f ( x ) d x = ∫ a b f ( t ) d t ) Hence, proved. Now, we have to find ∫ 0 π 2 cos 5 x sin 5 x + cos 5 x d x Let I = ∫ 0 π 2 cos 5 x sin 5 x + cos 5 x d x I = ∫ 0 π 2 cos 5 π 2 - x sin 5 π 2 - x + cos 5 π 2 - x d x ∵ ∫ 0 a f ( x ) d x = ∫ 0 a f ( a - x ) d x = ∫ 0 π 2 sin 5 x cos 5 x + sin 5 x d x So, 2 I = ∫ 0 π 2 sin 5 x cos 5 x + sin 5 x d x + ∫ 0 π 2 cos 5 x sin 5 x + cos 5 x d x = ∫ 0 π 2 sin 5 x + cos 5 x cos 5 x + sin 5 x d x = ∫ 0 π 2 d x ⇒ 2 I = [ x ] 0 π 2 = π 2 - 0 ⇒ I = π 4 Therefore, ∫ 0 π 2 cos 5 x sin 5 x + cos 5 x d x = π 4 .

Prove that ∫ − a a f ( x ) d x = { 2 ∫ 0 a f ( x ) d x , if f ( x ) is an event function 0 , if f ( x ) is an odd function and hence evaluate ∫ − 1 1 sin 5 ⁡ x cos 4 ⁡ x d x .

Let I = ∫ − a a f ( x ) d x ⇒ I = ∫ − a a f ( x ) d x = ∫ − a 0 f ( x ) d x + ∫ 0 a f ( x ) d x = I 1 + I 2 In I 1 , put x = - t ⇒ d x = - d t When x = - a , t = a and when x = 0 , t = 0 Therefore, I 1 = ∫ a 0 f ( − t ) ( − d t ) = − ∫ a 0 f ( − t ) d t = ∫ 0 a f ( − t ) d t ∵ ∫ a b f ( x ) d x = − ∫ b a f ( x ) d x = ∫ 0 a f ( − x ) d x ∵ ∫ a b f ( x ) d x = ∫ a b f ( t ) d t ∴ ∫ − a a f ( x ) d x = ∫ 0 a f ( − x ) d x + ∫ 0 a f ( x ) d x If f is an even function f - x = f x Therefore, ∫ − a a f ( x ) d x = ∫ 0 a f ( x ) d x + ∫ 0 a f ( x ) d x = 2 ∫ 0 a f ( x ) d x If f is an odd function f - x = - f x Therefore, ∫ − a a f ( x ) d x = - ∫ 0 a f ( x ) d x + ∫ 0 a f ( x ) d x = 0 Now, let I 3 = ∫ − 1 1 sin 5 ⁡ x cos 4 ⁡ x d x . Let f ( x ) = sin 5 ⁡ x cos 4 ⁡ x Then f ( − x ) = sin 5 ⁡ ( − x ) cos 4 ⁡ ( − x ) = − sin 5 ⁡ x cos 4 ⁡ x = − f ( x ) i.e. f is an odd function Hence, I 3 = 0 .

Prove that ∫ 0 a f ( x ) d x = ∫ 0 a f ( a - x ) d x and hence evaluate ∫ 0 a x x + a - x d x .

To prove: ∫ 0 a f ( x ) d x = ∫ 0 a f ( a - x ) d x Let ( a - x ) = t ⇒ - d x = d t When x = 0 ⇒ t = a and x = a ⇒ t = 0 Therefore, ∫ 0 a f ( a - x ) d x = - ∫ a 0 f ( t ) d t = ∫ 0 a f ( t ) d t (Apply property ∫ b a f ( x ) d x = - ∫ a b f ( x ) d x ) ⇒ ∫ 0 a f ( a - x ) d x = ∫ 0 a f ( x ) d x (Apply property ∫ a b f ( x ) d x = ∫ a b f ( t ) d t ) Hence, proved. Let I = ∫ 0 a x x + a - x d x → 1 ⇒ I = ∫ 0 a a - x a - x + a - a - x d x = ∫ 0 a a - x a - x + x d x → 2 (Apply ∫ 0 a f ( x ) d x = ∫ 0 a f ( a - x ) d x ) Adding 1 and 2 , we get: 2 I = ∫ 0 a a - x a - x + x d x + ∫ 0 a x x + a - x d x ⇒ 2 I = ∫ 0 a a - x + x a - x + x d x ⇒ 2 I = ∫ 0 a 1 · d x ⇒ 2 I = x 0 a ⇒ 2 I = a - 0 ⇒ 2 I = a ⇒ I = a 2 Hence, ∫ 0 a x x + a - x d x = a 2 .

Find the integral of 1 x 2 + a 2 with respect to x and hence find ∫ 1 x 2 - 6 x + 13 d x .

Let I = ∫ 1 x 2 + a 2 d x Substitute x = a tan θ ⇒ θ = tan - 1 x a Now, differentiating, then d x = a sec 2 θ · d θ Therefore, I = ∫ 1 a tan θ 2 + a 2 a sec 2 θ · d θ ⇒ I = ∫ 1 a 2 tan 2 θ + a 2 a sec 2 θ · d θ ⇒ I = ∫ 1 a 2 tan 2 θ + 1 a sec 2 θ · d θ ⇒ I = ∫ 1 a 2 sec 2 θ a sec 2 θ · d θ ⇒ I = 1 a ∫ 1 · d θ ⇒ I = 1 a θ + c ⇒ I = 1 a tan - 1 x a + c Now, I 1 = ∫ 1 x 2 - 6 x + 13 d x ⇒ I 1 = ∫ 1 x - 3 2 - 9 + 13 d x ⇒ I 1 = ∫ 1 x - 3 2 + 4 d x ⇒ I 1 = ∫ 1 x - 3 2 + 2 2 d x ⇒ I 1 = 1 2 tan - 1 x - 3 2 + c Hence, ∫ 1 x 2 - 6 x + 13 d x = 1 2 tan - 1 x - 3 2 + c .

Find the integral of 1 x 2 - a 2 with respect to x and hence evaluate ∫ 1 x 2 - 16 d x .

Let I = ∫ 1 x 2 - a 2 d x Consider 1 x 2 - a 2 = 1 x - a x + a = A x - a + B x + a ⇒ 1 = A x + a + B x - a If x = a , then 1 = A 2 a + B 0 ⇒ A = 1 2 a If x = - a , then 1 = A 0 + B - 2 a ⇒ B = - 1 2 a So, 1 x 2 - a 2 = 1 2 a x - a - 1 2 a x + a Now, ∫ 1 x 2 - a 2 d x = ∫ 1 2 a x - a d x - ∫ 1 2 a x + a d x = 1 2 a ∫ 1 x - a d x - ∫ 1 x + a d x = 1 2 a log x - a - log x + a + C ∴ ∫ 1 x 2 - a 2 d x = 1 2 a log x - a x + a + C Now, ∫ 1 x 2 - 16 d x = ∫ 1 x 2 - 4 2 d x ∴ ∫ 1 x 2 - 16 d x = 1 8 log x - 4 x + 4 + C

Find the integral of 1 x 2 + a 2 with respect to x and hence evaluate ∫ 1 x 2 + 2 x + 2 d x .

Let I = ∫ 1 x 2 + a 2 d x Substitute x = a tan θ ⇒ θ = tan - 1 x a Differentiating with respect to θ on both sides ⇒ d x = a sec 2 θ d θ ⇒ I = ∫ a sec 2 θ a 2 tan 2 θ + a 2 d θ ⇒ I = ∫ a sec 2 θ a 2 tan 2 θ + 1 d θ ⇒ I = 1 a ∫ sec 2 θ sec 2 θ d θ ∵ sec 2 x - tan 2 x = 1 ⇒ I = 1 a ∫ d θ ⇒ I = 1 a θ + c ⇒ I = 1 a tan - 1 x a + c Now, ∫ 1 x 2 + 2 x + 2 d x = ∫ 1 x 2 + 2 x + 1 + 1 d x ⇒ ∫ 1 x 2 + 2 x + 2 d x = ∫ 1 x + 1 2 + 1 2 d x ⇒ ∫ 1 x 2 + 2 x + 2 d x = tan - 1 x + 1 + c .

∫ 0 π 2 cos 2 x d x

Let I = ∫ 0 π 2 cos 2 x d x ⇒ I = ∫ 0 π 2 cos 2 x + 1 2 d x We know that ∫ a b k f x d x = k ∫ a b f x d x ⇒ I = 1 2 ∫ 0 π 2 cos 2 x + 1 d x We know that ∫ a b f x ± g x d x = ∫ a b f x d x ± ∫ a b g x d x ⇒ I = 1 2 ∫ 0 π 2 cos 2 x d x + ∫ 0 π 2 d x ⇒ I = 1 2 sin 2 x 2 + x 0 π 2 ⇒ I = 1 2 sinπ 2 + π 2 - 0 ⇒ I = 1 2 0 + π 2 ⇒ I = π 4

Evaluate : ∫ 0 π 2 sin x 1 + cos 2 x d x

Given, ∫ 0 π 2 sin x 1 + cos 2 x d x Let cos x = t ⇒ - sin x d x = d t When x = π 2 , t = cos π 2 = 0 When x = 0 , t = cos 0 = 1 So, ∫ 0 π 2 sin x 1 + cos 2 x d x = ∫ 1 0 - d t 1 + t 2 = - tan - 1 t 1 0 = - tan - 1 0 - - tan - 1 1 = - 0 - - π 4 ∴ ∫ 0 π 2 sin x 1 + cos 2 x d x = π 4