Maths

Students can find the formula of integration for a particular function in this article. It is important to memorise these formulas to solve the integral problem easily. Below is the integration of a particular function formula.

The formula of a particular function:

1. Power Rule: 

$\int x^{n}dx=\frac{x^{n+1}}{n+1}+C(n\ne -1)$

2. Reciprocal Function:

$\int \frac{1}{x}dx=ln\left|x\right|+C$

3. Trigonometric Function: 

• $\int sinxdx=-cosx+C$
• $\int cosxdx=sinx+C$
• $\int {sec}^{2}xdx=tanx+C$
• $\int {csc}^{2}xdx=-cotx+C$
• $\int secxtanxdx=secx+C$
• $\int cscxcotxdx=-cscx+C$

4. Inverse Trigonometric Function

• $\int \frac{1}{1+x^2}dx={tan}^{-1}x+C$
• $\int \frac{1}{\sqrt{1-x^2}}dx={sin}^{-1}x+C$
• $\int \frac{-1}{\sqrt{1-x^2}}dx={cos}^{-1}x+C$

? 2 adj (3 A adj (2A)|
= 23.? 3 A adj (2A)|
|2
= 23 ⋅ (33)2 ⋅ | A|2 ⋅ |adj (2 A)|2
= 23 ⋅ 36 ⋅ | A|2 ⋅ (|2 A|2)2
= 23 ⋅ 36 ⋅ | A|2 [ (23)2 ⋅ | A|2]2
= 23. 36. |A|2. 212. |A|4
= 215. 36. |A|6
= 215 ⋅ 36 ⋅ 56 = 2? ⋅ 3? ⋅ 5?
? ? = 15? = 6? = 6
? +? +? = 27

( (1+i)/ (1-i) )^ (m/2) = ( (1+i)/ (i-1) )^ (n/3) = 1
⇒ ( (1+i)²/2 )^ (m/2) = ( (1+i)²/ (-2) )^ (n/3) = 1
⇒ (i)^ (m/2) = (-i)^ (n/3) = 1
⇒ m/2 = 4k? and n/3 = 4k?
⇒ m = 8k? and n = 12k?
Least value of m = 8 and n = 12
∴ GCD = 4
∴ GCD = 4

(0.16)^log? (1/3 + 1/3² + . to ∞)
= (4/25)^log? (1/2)
= ( (5/2)? ² )^log? /? (1/2) = (5/2)^ (-2 log? /? (1/2)
= (5/2)^ (log? /? ( (1/2)? ² ) = 4

lim_ (x→0) (1/x? ) {1 - cos (x²/2) - cos (x²/4) + cos (x²/2)cos (x²/4)} = 2?
⇒ lim_ (x→0) ( (1 - cos (x²/2) (1 - cos (x²/4) / x? ) = 2?
⇒ 2? = 2? ⇒ k = 8

∴ center lies on x + y = 2 and in 1st quadrant center = (a, 2-a) where a > 0 and 2-a > 0 ⇒ 0 < a < 2
∴ circle touches x = 3 and y = 2
⇒ |3-a| = |2 - (2-a)| = radius
⇒ |3-a| = |a| ⇒ a = 3/2
∴ radius = a
⇒ Diameter = 2a = 3.

A = [ x 1 ]
[ 1 0 ]
A² = [ x 1 ] [ x 1 ] = [ x²+1 x ]
[ 1 0 ] [ 1 0 ] [ x 1 ]
A? = [ x²+1 x ] [ x²+1 x ]
[ x 1 ] [ x 1 ]
= [ (x²+1)²+x² x (x²+1)+x ]
[ x (x²+1)+x²+1 ]
a? = (x² + 1)² + x² = 109
⇒ x = ±3
a? = x² + 1 = 10

S = (2 . ¹P? - 3 . ²P? + 4 . ³P? upto 51 terms) + (1! - 2! + 3! - . upto 51 terms)
∴ [? ? P_ (n-1) = n!]
= (2! - 3! + 4! + 52!) + (1! - 2! + 3! - 4! + . . + (51)!)
= 1! + 52!

0 ≤ y ≤ x² + 1, 0 ≤ y ≤ x + 1, 1/2 ≤ x ≤ 2
Required area
= 19/24 + 5/2 = 79/24

LHL : lim_ (x→0? ) |1-x-x|/|λ-x-1| = 1/|λ-1|
RHL: lim_ (x→0? ) |1-x+x|/|λ-x+0| = 1/|λ|
For existence of limit
LHL = RHL
⇒ 1/|λ-1| = 1/|λ| ⇒ λ = 1/2
∴ L = 1/|λ| = 2