Evaluate ∫ 1 x 3 + 1 d x

Let I = ∫ 1 x 3 + 1 d x ⇒ I = ∫ 1 ( 1 + x ) x 2 − x + 1 d x Consider, 1 ( 1 + x ) x 2 − x + 1 = A x + 1 + B x + C x 2 − x + 1 ⇒ 1 = A x 2 − x + 1 + B x + C x + 1 ⇒ 1 = A x 2 − A x + A + B x 2 + C x + B x + C ⇒ 1 = x 2 A + B + x - A + B + C + A + C On comparing the coefficients of like powers of x , we get: A + B = 0 , - A + B + C = 0 , A + C = 1 On solving this, we get: A = 1 3 , B = - 1 3 , C = 2 3 Thus, I = 1 3 ∫ 1 x + 1 d x + ∫ − x 3 + 2 3 x 2 − x + 1 d x = 1 3 log ⁡ ( x + 1 ) + 1 3 ∫ − x + 2 x 2 − x + 1 d x = 1 3 log ⁡ ( x + 1 ) + 1 6 ∫ − 2 x + 4 x 2 − x + 1 d x = 1 3 log ⁡ ( x + 1 ) + 1 6 ∫ − 2 x + 1 + 3 x 2 − x + 1 d x = 1 3 log ⁡ ( x + 1 ) + 1 6 − ∫ ( 2 x − 1 ) x 2 − x + 1 d x + ∫ 3 x 2 − x + 1 d x Consider, ∫ ( 2 x − 1 ) x 2 − x + 1 d x Putting x 2 - x + 1 = t ⇒ 2 x - 1 d x = d t Thus, ∫ ( 2 x − 1 ) x 2 − x + 1 d x = ∫ d t t = log t + c = log x 2 - x + 1 + c Now consider, ∫ 3 x 2 − x + 1 d x = ∫ 3 x 2 − x + 1 4 - 1 4 + 1 d x = ∫ 3 x - 1 2 2 + 3 4 d x = 3 ∫ d x x - 1 2 2 + 3 2 2 = 3 × 1 3 2 tan - 1 x - 1 2 3 2 + k = 2 3 tan - 1 2 x - 1 3 + k Thus, I = 1 3 log ⁡ ( x + 1 ) - 1 6 ∫ ( 2 x − 1 ) x 2 − x + 1 d x + 1 6 ∫ 3 x 2 − x + 1 d x ⇒ I = 1 3 log ⁡ ( x + 1 ) - 1 6 log ⁡ x 2 − x + 1 + 1 6 · 2 3 tan − 1 ⁡ 2 x − 1 3 C = c + k ⇒ I = 1 3 log ⁡ ( x + 1 ) - 1 6 log ⁡ x 2 − x + 1 + 1 3 tan − 1 ⁡ 2 x − 1 3 + C

Evaluate ∫ sin − 1 ⁡ x a + x d x

Let, I = ∫ sin - 1 x a + x d x Assume, x = a tan 2 θ ⇒ d x = 2 a tan θ sec 2 θ d θ ∴ I = 2 a ∫ sin - 1 ( a tan 2 θ a sec 2 θ ) tan θ sec 2 θ d θ I = 2 a ∫ θ × tan θ × sec 2 θ d θ ∵ sin - 1 sin θ = θ Assume, tan θ = t ⇒ sec 2 θ d θ = d t I = 2 a ∫ tan - 1 t × t d t ∵ θ = tan - 1 t Now, applying integration by parts, we get I = 2 a tan - 1 t ∫ t d t - ∫ d d t tan - 1 t × ∫ t d t d t ⇒ I = 2 a t 2 2 tan - 1 t - ∫ 1 1 + t 2 × t 2 2 d t ⇒ I = 2 a t 2 2 tan - 1 t - 1 2 ∫ t 2 + 1 - 1 1 + t 2 d t ⇒ I = 2 a t 2 2 tan - 1 t - 1 2 ∫ 1 + t 2 1 + t 2 - 1 1 + t 2 d t ⇒ I = 2 a t 2 2 tan - 1 t - 1 2 ∫ 1 - 1 1 + t 2 d t By using standard formulae of integration we get, I = a t 2 tan - 1 t - t + tan - 1 t + c Now, by putting value of t = x a , we get, a x a tan - 1 x a - x a + tan - 1 x a + c

Evaluate: ∫ 1 3 1 1 + x 2 d x

Let I = ∫ 1 3 1 1 + x 2 d x ⇒ I = tan - 1 x 1 3 ⇒ I = tan - 1 3 - tan - 1 1 ⇒ I = tan - 1 tan π 3 - tan - 1 tan π 4 ⇒ I = π 3 - π 4 ∵ tan - 1 tan A = A ⇒ I = π 12

Evaluate ∫ 0 π 4 sin ⁡ 2 x sin 4 ⁡ x + cos 4 ⁡ x d x

Let I = ∫ 0 π 4 sin ⁡ 2 x sin 4 ⁡ x + cos 4 ⁡ x d x ⇒ I = ∫ 0 π 4 2 sin ⁡ x cos ⁡ x cos 4 ⁡ x + sin 4 ⁡ x d x ∵ sin 2 A = 2 sin A cos A Dividing numerator and denominator by cos 4 x , we get: ⇒ I = ∫ 0 π 4 2 tan ⁡ x sec 2 ⁡ x 1 + tan 4 ⁡ x d x Let tan 2 ⁡ x = t . Then, d tan 2 ⁡ x = d t ⇒ 2 tan ⁡ x sec 2 ⁡ x d x = d t Also, x = 0 ⇒ t = tan 2 ⁡ 0 = 0 and x = π 4 ⇒ t = tan 2 ⁡ π 4 = 1 Putting t = tan 2 ⁡ x and 2 tan ⁡ x sec 2 ⁡ x d x = d t , we get: I = ∫ 0 1 1 1 + t 2 d t ⇒ I = tan − 1 ⁡ t 0 1 = π 4 − 0 = π 4

Prove that ∫ 1 a 2 − x 2 d x = sin − 1 ⁡ x a + C

Let I = ∫ 1 a 2 − x 2 d x Putting x = a sin ⁡ θ and d x = a cos ⁡ θ d θ , we get: ⇒ I = ∫ a cos ⁡ θ d θ a 2 − a 2 sin 2 ⁡ θ ⇒ I = ∫ cos ⁡ θ d θ cos 2 ⁡ θ ∵ cos 2 A + sin 2 A = 1 ⇒ I = ∫ 1 ⋅ d θ ⇒ I = θ + C ⇒ I = sin − 1 ⁡ x a + C ∵ x = a sin ⁡ θ ⇒ sin ⁡ θ = x a ⇒ θ = sin − 1 ⁡ x a

Prove that: ∫ 0 π d x 1 − 2 a cos ⁡ x + a 2 = π 1 − a 2 ; ( a < 1 ) .

Let I = ∫ 0 π d x 1 - 2 a cos x + a 2 ⇒ I = ∫ 0 π d x 1 - 2 a 1 - tan 2 x 2 1 + tan 2 x 2 + a 2 ⇒ I = ∫ 0 π sec 2 x 2 d x 1 + tan 2 x 2 - 2 a + 2 a tan 2 x 2 + a 2 + a 2 tan 2 x 2 ∵ 1 + tan 2 A = sec 2 A ⇒ I = ∫ 0 π sec 2 x 2 d x a - 1 2 + a + 1 2 tan 2 x 2 Let, tan x 2 = t ⇒ 1 2 sec 2 x 2 d x = d t When x = 0 , t = 0 and x = π , t = ∞ ∴ I = 2 ∫ 0 ∞ d t a - 1 2 + a + 1 2 t 2 = 2 a + 1 2 ∫ 0 ∞ d t 1 - a a + 1 2 + t 2 = 2 a + 1 2 a + 1 1 - a tan - 1 t a + 1 1 - a 0 ∞ = 2 a + 1 2 a + 1 1 - a tan - 1 ∞ - tan - 1 0 = 2 1 - a 2 π 2 Thus, I = π 1 - a 2 Hence, proved.

Prove that ∫ 0 π x d x a 2 cos 2 ⁡ x + b 2 sin 2 ⁡ x = π 2 2 a b

Let I = ∫ 0 π x d x a 2 cos 2 ⁡ x + b 2 sin 2 ⁡ x . . . 1 ⇒ I = ∫ 0 π ( π − x ) a 2 cos 2 ⁡ ( π − x ) + b 2 sin 2 ⁡ ( π − x ) d x ∵ ∫ 0 a f ( x ) d x = ∫ 0 a f ( a − x ) d x ⇒ I = ∫ 0 π π − x a 2 cos 2 ⁡ x + b 2 sin 2 ⁡ x d x . . . 2 ∵ cos π - A = - cos A , sin π - A = sin A Adding 1 and 2 , we get: 2 I = ∫ 0 π x + π − x a 2 cos 2 ⁡ x + b 2 sin 2 ⁡ x d x ⇒ 2 I = π ∫ 0 π 1 a 2 cos 2 ⁡ x + b 2 sin 2 ⁡ x d x ⇒ 2 I = π ∫ 0 π 2 1 a 2 cos 2 ⁡ x + b 2 sin 2 ⁡ x + 1 a 2 cos 2 ⁡ ( π − x ) + b 2 sin 2 ⁡ ( π − x ) d x ∵ ∫ 0 2 a f ( x ) d x = ∫ 0 a { f ( x ) + f ( 2 a − x ) } d x ⇒ 2 I = 2 π ∫ 0 π 2 1 a 2 cos 2 ⁡ x + b 2 sin 2 ⁡ x d x ⇒ 2 I = 2 π ∫ 0 π 2 sec 2 ⁡ x a 2 + b 2 tan 2 ⁡ x d x [Dividing numerator and denominator by cos 2 x ] ⇒ I = π ∫ 0 π 2 sec 2 ⁡ x a 2 + b 2 tan 2 ⁡ x d x Let tan x = t . d ( tan ⁡ x ) = d t ⇒ sec 2 ⁡ x d x = d t Also, x = 0 ⇒ t = tan ⁡ 0 = 0 and x = π 2 ⇒ t = tan ⁡ π 2 = ∞ ∴ I = π ∫ 0 ∞ d t a 2 + b 2 t 2 = π b 2 ∫ 0 ∞ d t ( a b ) 2 + t 2 = π b 2 × 1 ( a b ) tan − 1 ⁡ t a b 0 ∞ ⇒ I = π a b tan − 1 ⁡ b t a 0 ∞ ⇒ I = π a b tan − 1 ⁡ ∞ − tan − 1 ⁡ 0 = π a b π 2 − 0 = π 2 2 a b

Evaluate ∫ 1 x 3 + 1 d x

Let I = ∫ 1 x 3 + 1 d x ⇒ I = ∫ 1 ( 1 + x ) x 2 − x + 1 d x Consider, 1 ( 1 + x ) x 2 − x + 1 = A x + 1 + B x + C x 2 − x + 1 ⇒ 1 = A x 2 − x + 1 + B x + C x + 1 ⇒ 1 = A x 2 − A x + A + B x 2 + C x + B x + C ⇒ 1 = x 2 A + B + x - A + B + C + A + C On comparing the coefficients of like powers of x , we get: A + B = 0 , - A + B + C = 0 , A + C = 1 On solving this, we get: A = 1 3 , B = - 1 3 , C = 2 3 Thus, I = 1 3 ∫ 1 x + 1 d x + ∫ − x 3 + 2 3 x 2 − x + 1 d x = 1 3 log ⁡ ( x + 1 ) + 1 3 ∫ − x + 2 x 2 − x + 1 d x = 1 3 log ⁡ ( x + 1 ) + 1 6 ∫ − 2 x + 4 x 2 − x + 1 d x = 1 3 log ⁡ ( x + 1 ) + 1 6 ∫ − 2 x + 1 + 3 x 2 − x + 1 d x = 1 3 log ⁡ ( x + 1 ) + 1 6 − ∫ ( 2 x − 1 ) x 2 − x + 1 d x + ∫ 3 x 2 − x + 1 d x Consider, ∫ ( 2 x − 1 ) x 2 − x + 1 d x Putting x 2 - x + 1 = t ⇒ 2 x - 1 d x = d t Thus, ∫ ( 2 x − 1 ) x 2 − x + 1 d x = ∫ d t t = log t + c = log x 2 - x + 1 + c Now consider, ∫ 3 x 2 − x + 1 d x = ∫ 3 x 2 − x + 1 4 - 1 4 + 1 d x = ∫ 3 x - 1 2 2 + 3 4 d x = 3 ∫ d x x - 1 2 2 + 3 2 2 = 3 × 1 3 2 tan - 1 x - 1 2 3 2 + k = 2 3 tan - 1 2 x - 1 3 + k Thus, I = 1 3 log ⁡ ( x + 1 ) - 1 6 ∫ ( 2 x − 1 ) x 2 − x + 1 d x + 1 6 ∫ 3 x 2 − x + 1 d x ⇒ I = 1 3 log ⁡ ( x + 1 ) - 1 6 log ⁡ x 2 − x + 1 + 1 6 · 2 3 tan − 1 ⁡ 2 x − 1 3 C = c + k ⇒ I = 1 3 log ⁡ ( x + 1 ) - 1 6 log ⁡ x 2 − x + 1 + 1 3 tan − 1 ⁡ 2 x − 1 3 + C

Evaluate: ∫ cos ⁡ 2 x − cos ⁡ 2 α cos ⁡ x − cos ⁡ α d x

Let I = ∫ cos 2 x - cos 2 α cos x - cos α d x As we know from trigonometric identities cos 2 γ = 2 cos 2 γ - 1 So, I = ∫ 2 cos 2 x - 1 - 2 cos 2 α - 1 cos x - cos α d x ⇒ I = 2 ∫ cos 2 x - cos 2 α cos x - cos α d x ⇒ I = 2 ∫ cos x - cos α cos x + cos α cos x - cos α d x ∵ β + γ β - γ = β 2 - γ 2 ⇒ I = 2 ∫ cos x + cos α d x ⇒ I = 2 ∫ cos x d x + 2 cos α ∫ 1 d x As we know, ∫ cos x d x = sin x + C 1 and ∫ 1 . d x = x + C 2 So, I = 2 sin x + 2 x cos α + C , where C = C 1 + C 2 ⇒ I = 2 sin x + x cos α + C .

Evaluate: ∫ 0 1 log ⁡ ( 1 + x ) 1 + x 2 d x

Let I = ∫ 0 1 log ⁡ ( 1 + x ) 1 + x 2 d x Put x = tan θ ⇒ d x = sec 2 ⁡ θ d θ When, x = 0 ⇒ θ = 0 When, x = 1 ⇒ θ = π 4 ∴ I = ∫ 0 π 4 log ⁡ ( 1 + tan ⁡ θ ) ⋅ sec 2 ⁡ θ d θ 1 + tan 2 ⁡ θ = ∫ 0 π 4 log ⁡ ( 1 + tan ⁡ θ ) ⋅ sec 2 ⁡ θ d θ sec 2 θ ∵ 1 + tan 2 A = sec 2 A = ∫ 0 π 4 log ⁡ ( 1 + tan ⁡ θ ) d θ . . . 1 = ∫ 0 π 4 log ⁡ 1 + tan ⁡ π 4 − θ d θ ∵ ∫ 0 a f ( x ) d x = ∫ 0 a f ( a − x ) d x = ∫ 0 π 4 log ⁡ 1 + tan ⁡ π 4 − tan ⁡ θ 1 + tan ⁡ π 4 ⋅ tan ⁡ θ d θ ∵ tan A - B = tan A - tan B 1 + tan A tan B = ∫ 0 π 4 log ⁡ 1 + 1 − tan ⁡ θ 1 + tan ⁡ θ d θ ∵ tan π 4 = 1 = ∫ 0 π 4 log ⁡ 1 + tan ⁡ θ + 1 − tan ⁡ θ 1 + tan ⁡ θ d θ = ∫ 0 π 4 log ⁡ 2 1 + tan ⁡ θ ⋅ d θ = ∫ 0 π 4 [ log ⁡ 2 − log ⁡ ( 1 + tan ⁡ θ ) ] d θ . . . 2 ∵ log m - log n = log m n Adding 1 and 2 , we get 2 I = ∫ 0 π 4 log 2 d θ ⇒ 2 I = log 2 × θ 0 π 4 ⇒ 2 I = log 2 × π 4 ⇒ I = π log 2 8

Prove that ∫ x 2 + a 2 d x = 1 2 x x 2 + a 2 + a 2 log ⁡ x + x 2 + a 2 + c .

Let I = ∫ x 2 + a 2 ⋅ d x ⇒ I = ∫ x 2 + a 2 · 1 d x Applying integration by parts by taking x 2 + a 2 as first function and 1 as second function. I = a 2 + x 2 ∫ 1 d x − ∫ d a 2 + x 2 d x ∫ 1 d x d x ⇒ I = a 2 + x 2 · x − ∫ 1 2 a 2 + x 2 − 1 2 ( 0 + 2 x ) x d x ⇒ I = x a 2 + x 2 − ∫ x 2 a 2 + x 2 d x ⇒ I = x a 2 + x 2 − ∫ a 2 + x 2 − a 2 a 2 + x 2 d x ⇒ I = x a 2 + x 2 − ∫ a 2 + x 2 d x + a 2 ∫ 1 a 2 + x 2 d x ⇒ I = x a 2 + x 2 − I + a 2 log ⁡ x + a 2 + x 2 ⇒ 2 I = x a 2 + x 2 + a 2 log x + a 2 + x 2 ⇒ I = x 2 a 2 + x 2 + a 2 2 log ⁡ x + a 2 + x 2 + c Hence proved.

Prove that ∫ 0 π x d x a 2 cos 2 ⁡ x + b 2 sin 2 ⁡ x = π 2 2 a b

Let I = ∫ 0 π x d x a 2 cos 2 ⁡ x + b 2 sin 2 ⁡ x . . . 1 ⇒ I = ∫ 0 π ( π − x ) a 2 cos 2 ⁡ ( π − x ) + b 2 sin 2 ⁡ ( π − x ) d x ∵ ∫ 0 a f ( x ) d x = ∫ 0 a f ( a − x ) d x ⇒ I = ∫ 0 π π − x a 2 cos 2 ⁡ x + b 2 sin 2 ⁡ x d x . . . 2 ∵ cos π - A = - cos A , sin π - A = sin A Adding 1 and 2 , we get: 2 I = ∫ 0 π x + π − x a 2 cos 2 ⁡ x + b 2 sin 2 ⁡ x d x ⇒ 2 I = π ∫ 0 π 1 a 2 cos 2 ⁡ x + b 2 sin 2 ⁡ x d x ⇒ 2 I = π ∫ 0 π 2 1 a 2 cos 2 ⁡ x + b 2 sin 2 ⁡ x + 1 a 2 cos 2 ⁡ ( π − x ) + b 2 sin 2 ⁡ ( π − x ) d x ∵ ∫ 0 2 a f ( x ) d x = ∫ 0 a { f ( x ) + f ( 2 a − x ) } d x ⇒ 2 I = 2 π ∫ 0 π 2 1 a 2 cos 2 ⁡ x + b 2 sin 2 ⁡ x d x ⇒ 2 I = 2 π ∫ 0 π 2 sec 2 ⁡ x a 2 + b 2 tan 2 ⁡ x d x [Dividing numerator and denominator by cos 2 x ] ⇒ I = π ∫ 0 π 2 sec 2 ⁡ x a 2 + b 2 tan 2 ⁡ x d x Let tan x = t . d ( tan ⁡ x ) = d t ⇒ sec 2 ⁡ x d x = d t Also, x = 0 ⇒ t = tan ⁡ 0 = 0 and x = π 2 ⇒ t = tan ⁡ π 2 = ∞ ∴ I = π ∫ 0 ∞ d t a 2 + b 2 t 2 = π b 2 ∫ 0 ∞ d t ( a b ) 2 + t 2 = π b 2 × 1 ( a b ) tan − 1 ⁡ t a b 0 ∞ ⇒ I = π a b tan − 1 ⁡ b t a 0 ∞ ⇒ I = π a b tan − 1 ⁡ ∞ − tan − 1 ⁡ 0 = π a b π 2 − 0 = π 2 2 a b

Prove that ∫ 1 a 2 − x 2 d x = sin − 1 ⁡ x a + C

Let I = ∫ 1 a 2 − x 2 d x Putting x = a sin ⁡ θ and d x = a cos ⁡ θ d θ , we get: ⇒ I = ∫ a cos ⁡ θ d θ a 2 − a 2 sin 2 ⁡ θ ⇒ I = ∫ cos ⁡ θ d θ cos 2 ⁡ θ ∵ cos 2 A + sin 2 A = 1 ⇒ I = ∫ 1 ⋅ d θ ⇒ I = θ + C ⇒ I = sin − 1 ⁡ x a + C ∵ x = a sin ⁡ θ ⇒ sin ⁡ θ = x a ⇒ θ = sin − 1 ⁡ x a

Evaluate ∫ 0 π 4 sin ⁡ 2 x sin 4 ⁡ x + cos 4 ⁡ x d x

Let I = ∫ 0 π 4 sin ⁡ 2 x sin 4 ⁡ x + cos 4 ⁡ x d x ⇒ I = ∫ 0 π 4 2 sin ⁡ x cos ⁡ x cos 4 ⁡ x + sin 4 ⁡ x d x ∵ sin 2 A = 2 sin A cos A Dividing numerator and denominator by cos 4 x , we get: ⇒ I = ∫ 0 π 4 2 tan ⁡ x sec 2 ⁡ x 1 + tan 4 ⁡ x d x Let tan 2 ⁡ x = t . Then, d tan 2 ⁡ x = d t ⇒ 2 tan ⁡ x sec 2 ⁡ x d x = d t Also, x = 0 ⇒ t = tan 2 ⁡ 0 = 0 and x = π 4 ⇒ t = tan 2 ⁡ π 4 = 1 Putting t = tan 2 ⁡ x and 2 tan ⁡ x sec 2 ⁡ x d x = d t , we get: I = ∫ 0 1 1 1 + t 2 d t ⇒ I = tan − 1 ⁡ t 0 1 = π 4 − 0 = π 4