Evaluate the following integral: ∫ ( log x ) 2 x d x

Let I = ∫ ( log x ) 2 x d x Here, d d x log x = 1 x We know that ∫ f ' ( x ) [ f ( x ) ] n d x = [ f ( x ) ] n + 1 n + 1 + c ∴ I = ( log x ) 2 + 1 2 + 1 + c ⇒ I = ( log x ) 3 3 + c ⇒ ∫ ( log x ) 2 x d x = ( log x ) 3 3 + c where c is an integration constant.

Evaluate the following integral: ∫ ( log x ) 2 x d x

Let I = ∫ ( log x ) 2 x d x Here, d d x log x = 1 x We know that ∫ f ' ( x ) [ f ( x ) ] n d x = [ f ( x ) ] n + 1 n + 1 + c ∴ I = ( log x ) 2 + 1 2 + 1 + c ⇒ I = ( log x ) 3 3 + c ⇒ ∫ ( log x ) 2 x d x = ( log x ) 3 3 + c where c is an integration constant.

Evaluate the integral: ∫ 3 + log ⁡ x x d x

Let I = ∫ 3 + log ⁡ x x d x Put 3 + log x = t ⇒ 1 x d x = d t ∴ I = ∫ t d t = t 3 2 3 2 + C = 2 3 t 3 2 + C = 2 3 ( 3 + log ⁡ x ) 3 2 + C

Evaluate the integral: ∫ 3 ( x + 1 ) ( x + log ⁡ x ) 2 x d x

Let I = ∫ 3 ( x + 1 ) ( x + log ⁡ x ) 2 x d x Put x + log ⁡ x = t ⇒ 1 + 1 x d x = d t ⇒ x + 1 x d x = d t ∴ I = ∫ 3 t 2 d t = 3 × t 3 3 + c = t 3 + c = ( x + log ⁡ x ) 3 + c

Evaluate the integral: I = ∫ tan − 1 ⁡ x 1 + x 2 d x

Let I = ∫ tan − 1 ⁡ x 1 + x 2 d x Put tan − 1 ⁡ x = t ⇒ 1 1 + x 2 d x = d t ∴ I = ∫ t d t = t 3 2 3 2 + C = 2 3 t 3 2 + C = 2 3 tan − 1 ⁡ x 3 2 + C

Evaluate the integral: ∫ sin − 1 ⁡ 2 x 3 1 − 4 x 2 d x

Let I = ∫ sin − 1 ⁡ 2 x 3 1 − 4 x 2 d x Put sin − 1 ⁡ 2 x = t ⇒ 2 1 − ( 2 x ) 2 d x = d t ⇒ 1 1 − 4 x 2 d x = d t 2 ∴ I = 1 2 ∫ t 3 d t = 1 2 × t 4 4 + c = 1 8 sin − 1 ⁡ 2 x 4 + c

Evaluate the integral: ∫ sin ⁡ x cos 5 ⁡ x d x

Let I = ∫ sin ⁡ x cos 5 ⁡ x d x Put cos ⁡ x = t ⇒ − sin ⁡ x d x = d t ∴ I = − ∫ t 5 d t = − t 6 6 + C = − cos 6 ⁡ x 6 + C

Evaluate the integral: ∫ cos 3 ⁡ ( a x + b ) sin ⁡ ( a x + b ) d x

Let I = ∫ cos 3 ⁡ ( a x + b ) sin ⁡ ( a x + b ) d x Put cos ⁡ ( a x + b ) = t ⇒ − a sin ⁡ ( a x + b ) d x = d t ⇒ sin ⁡ ( a x + b ) d x = − 1 a d t ∴ I = − 1 a ∫ t 3 d t = − 1 a × t 4 4 + c = − cos 4 ⁡ ( a x + b ) 4 a + c

Evaluate the integral: ∫ sec 3 ⁡ x tan ⁡ x d x

Let I = ∫ sec 3 ⁡ x tan ⁡ x d x Put sec x = t ⇒ sec ⁡ x tan ⁡ x d x = d t ∴ I = ∫ t 2 d t = t 3 3 + c = sec 3 ⁡ x 3 + c

Evaluate the integral : ∫ cot 3 x cosec 2 x d x

Let I = ∫ cot 3 x cosec 2 x d x Put cot x = t Differentiating both sides, we get ⇒ - cosec 2 x d x = d t ⇒ cosec 2 x d x = - d t ∴ I = ∫ t 3 - d t = - t 4 4 + c ∴ I = - 1 4 cot 4 x + c .