BASIC INTEGRAL FORMULA
Integral Formulas
∫ 1 dx = x + C
∫ a dx = ax+ C
∫ xn dx = ((xn+1)/(n+1))+C ; n≠1
∫ (1/x) dx = ln |x| + C
∫ ex dx = ex+ C
∫ ax dx = (ax/ln a) + C ; a>0, a≠1
Trigonometric integration formula
∫ sin x dx = – cos x + C
∫ cos x dx = sin x + C
∫ sec2x dx = tan x + C
∫ csc2x dx = -cot x + C
∫ sec x (tan x) dx = sec x + C
∫ cosec x ( cot x) dx = – cosec x + c
∫ tan x = log |sec x| + C
∫ cot x = log |sin x| + c
∫ sec x = log |sec x + tan x| + C
∫ cosec x = log |cosec x - cot x| + C
Inverse trigonometric integration formula
∫1/√(1 - x2) dx = sin-1x + C
∫ 1/√(1 - x2) dx = -cos-1x + C
∫1/(1 + x2) dx = tan-1x + C
∫ 1/(1 + x2 ) dx = -cot-1x + C
∫ 1/x√(x2 - 1) dx = sec-1x + C
∫ 1/x√(x2 - 1) dx = -cosec-1 x + C
Some important integral formula
∫1/(x2 - a2) dx = 1/2a log|(x - a)(x + a| + C
∫ 1/(a2 - x2) dx =1/2a log|(a + x)(a - x)| + C
∫1/(x2 + a2) dx = 1/a tan-1x/a + C
∫1/√(x2 - a2)dx = log |x +√(x2 - a2)| + C
∫ √(x2 - a2) dx = x/2 √(x2 - a2) -a2/2 log |x + √(x2 - a2)| + C
∫1/√(a2 - x2) dx = sin-1 x/a + C
∫√(a2 - x2) dx = x/2 √(a2 - x2) dx + a2/2 sin-1 x/a + C
∫1/√(x2 + a2 ) dx = log |x + √(x2 + a2)| + C
∫ √(x2 + a2 ) dx = x/2 √(x2 + a2 )+ a2/2 log |x + √(x2 + a2)| + C
Example ; Compute ∫ (5x +2)4 dx
solution
Let u = (5x +2)
du = 5 dx
Thus ∫ (5x +2)4 dx = 1/5∫(u)4 du
Solving this equation, we get,
A (x + 2) + B (x + 1) = 1
Or, Ax + 2A + Bx + B = 1
x (A + B) + (2A + B) = 1
To make LHS and RHS to be equal,
we have ,A + B = 0 and 2A + B = 1.
On solving these two equations, we get
A = 1 and B = – 1.
Therefore, 1 / [(x + 1) (x + 2)] = 1 / (x + 1) – 1 / (x + 2)
Hence, ∫ dx / [(x + 1) (x + 2)] = ∫ dx / (x + 1) – ∫ dx / (x + 2)
Solution:
Given: ∫tan8x sec4 x dx
Let I = ∫tan8x sec4 x dx — (1)
sec4x = (sec2x) (sec2x)
Now, substitute in (1)
I = ∫tan8x (sec2x) (sec2x) dx
= ∫tan8x (tan2 x +1) (sec2x) dx
= ∫tan10x sec2 x dx + ∫tan8x sec2 x dx
=( tan11 x /11) + ( tan9 x /9) + C
Hence, ∫tan8x sec4 x dx = ( tan11 x /11) + ( tan9 x /9) + C
4. Write the anti-derivative of the following function: 3x2+4x3
Solution:
Given: 3x2+4x3
The antiderivative of the given function is written as:
∫3x2+4x3 dx = 3(x3/3) + 4(x4/4)
= x3 + x4
Thus, the antiderivative of 3x2+4x3 = x3 + x4
5. Integrate 2x sin(x2+ 1) with respect to x:by substitution method
Solution:
Given function: 2x sin(x2+ 1)
d/dx ( x2 + 1) = 2x.
By substitution method, we get
x2 + 1 = t, so that 2x dx = dt.
Hence, we get ∫ 2x sin ( x2 +1) dx = ∫ sin t dt
= – cos t + C
= – cos (x2 + 1) + C
Therefore, ∫ 2x sin(x2+ 1) = – cos (x2 + 1) + C by substitution method
6. Integrate: ∫ sin3 x cos2x dx
Solution:
Given that, ∫ sin3 x cos2x dx
∫ sin3 x cos2x dx = ∫ sin2 x cos2x (sin x) dx
=∫(1 – cos2x ) cos2x (sin x) dx —(1)
substitute t = cos x,
Then dt = -sin x dx
∫ sin3 x cos2x dx = – ∫ (1-t2)t2 dt
= – ∫ (t2-t4) dt (by multiplying t2
= – [(t3/3) – (t5/5)] + C —(2)
Now, substitute t = cos x in (2)
= -(⅓)cos3x +(1/5)cos5x + C
Hence, ∫ sin3 x cos2x dx = -(⅓)cos3x +(1/5)cos5x + C
7. Prove
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