Indefinite Integral Formula
INTEGRATION It is the inverse process of differentiation. If the derivative of F(x) is f(x) then we say that the antiderivative or integral of f(x) is F(x) and we write,
∫f(x) dx = F(x)
Thus,
d/dx[F(x)]=f(x) ⇒ ∫f(x) dx = F(x)
Example
Since
d/ dx(sin x) = cos x, we have ∫cos x dx = sin x.
Moreover, if C is any constant then d/dx(sin x +C) = cos x
So, in general, ∫cosx dx = (sin x +C)
Clearly, different values of C will give different integrals.
Thus, a given function may have an indefinite number of integral Because of this property, we call these integrals indefinite integrals.
Thus,
d/dx[F(x)]=f(x) ⇒ ∫f(x) dx = F(x)+C
, where C is constant called the constant of integration. Any function to be integrated is known as an integrand.
The following two results are a direct consequence of the definition integral.
Theorem –
∫xⁿdx={(xⁿ⁺¹)(n+1)}+C, when n ≠ -1
Proof
we have , d/dx (xⁿ⁺¹)(n+1)
=(n+1)xⁿ/n+1
=xⁿ
∴ ∫xⁿdx={(xⁿ⁺¹)\(n+1)}+Cfor e.g
∫x⁶dx =x⁶⁺¹/(6+1) + C
=x⁷/7+CTheorem –
∫(1/x)dx=log|x|+C, where x≠0
Proof Clearly,either x>0 or x<0
Case 1 When x>0
in this case, |x|=x
∴ d/dx[log|x|]=d/dx(logx) =1/x
so,we have ∫(1/x)dx=log|x|+C
Case 2 When x<0
in this case, |x|= -x
∴ d/dx[log|x|]=d/dx[log(-x)] =-1/-x =1/x
so,we have∫(1/x)dx=log|x|+C
Thus,from both the case ,
we have∫(1/x)dx=log|x|+C
Some Standard Results on Intergration
Theorem
d/dx{∫f(x)dx}=f(x)
Proof
Let ∫f(x)dx=F(x) ....(i)Then,
d/dx{F(x)}=f(x) [by def. of integral]
∴ d/dx{∫f(x)dx}=f(x) [using (i)]
Theorem
∫k.f(x)dx=k.∫f(x)dx, where k is constant
Proof
Let ∫f(x)dx=F(x) ....(i)
Then,d/dx{F(x)}=f(x) ....(ii) ∴ d/dx{k.F(x)}=k.d/dx{F(x)}
=k.f(x) [using (ii)]
so,by the definition of an intergral, we have
∫k.f(x)dx= k.F(x)=k.∫f(x)dx [using (i)]
Theorem
(i)∫{f₁(x)+f₂(x)}dx =∫f₁(x)dx+∫f₂(x)dx
Proof Let
∫{f₁(x)dx=F₁(x) and ∫f₂(x)dx=F₂(x) …….(i)
Then,
d/dx{F₁(x)}=f₁(x) and d/dx{F₂(x)}
= f₂(x) ……(ii)
Now, d/dx{F₁(x)+ F₂(x)}
=d/dx {F₁(x)} + d/dx{F₂(x)}
=f₁(x)+f₂(x) [using (ii)]∴ ∫{f₁(x)+f₂(x)}dx=F₁(x)+ F₂(x)
=∫f₁(x)dx+∫f₂(x)dx [using(i)]Evaluate
(i) ∫6ˣdx
solution
=∫6ˣdx
=6ˣ/log6 +C
Answer = 6ˣ/log6 +C
(ii) ∫(3sinx-4cosx+5sec²x-2cosec²x)dx
solution
=∫(3sinx-4cosx+5sec²x-2cosec²x)dx
= 3∫sinxdx-4∫cosxdx+5∫sec²xdx-2∫cosec²xdx
= 3(-cosx) – 4sinx + 5tanx – 2(-cotx)+C
= -3cosx – 4sinx+5tanx+2cotx+C
Answer= -3cosx – 4sinx+5tanx+2cotx+C
(iii) ∫tan²xdx
solution
=∫tan²xdx
=∫(sec²x-1)dx
=∫sec²xdx -∫dx
=tanx-x+C
Answer=tanx-x+C
(iv) ∫cot²xdx
solution
=∫cot²xdx
=∫(cosec²x-1)dx
= ∫cosec²xdx -∫dx
= – cotx- x +C
Answer = – cotx- x +C
(v) ∫{(1+sinx)/(1-sinx)}dx
solution
= ∫{(1+sinx)/(1-sinx)}dx
= ∫{(1+sinx)(1+sinx) /(1-sinx)(1+sinx)}dx
= ∫{(1+sinx)²/(1-sin²x)}dx
= ∫{(1+sin²x+2sinx)/cos²x}dx
= ∫(1/cos²x+sin²x/cos²x+2sinx/cos²x)dx
= ∫(sec²x+tan²x+2secxtanx)dx
= ∫(2sec²x-1+2secx tanx)dx
= 2∫(sec²x)dx-∫1dx+2∫(secx tanx)dx
= 2tanx-x+2secx+C
Answer = 2tanx-x+2secx+C
(vi) ∫{secx/(secx+tanx)}dx
solution
[∴sec²x+tanx²=1]
=∫{secx/(secx+tanx)}dx
∫{(secx)(secx-tanx)/(secx+tanx)(secx-tanx)}dx
=∫{(sec²x-secx tanx)/(sec²x+tan²x)}dx
=∫(sec²x-secx tanx)
=∫(sec²x)dx-∫(secx tanx)dx
=tanx – secx +C
Answer=tanx – secx +C
(vii) ∫{(4-5cos)/(sin²x)}dx
solution=∫{(4-5cos)/(sin²x)}dx
=∫{4/(sin²x) -(5cos)/(sin²x)}dx
=∫(4cosec²x-5cosecx cotx)dx
=∫(4cosec²x)dx-5∫(cosecx cotx)dx
= 4(-cotx)-5(-cosecx) +C
= -4cot+5cosecx+C
Answer = -4cot+5cosecx+C
(viii)∫(1/sin²xcos²x)dx
solution
= ∫(1/sin²xcos²x)dx [ ∴ sin²x+cos²x=1]
= ∫{(sin²x +cos²x)/sin²xcos²x)}dx
∫{(sin²x/ sin²xcos²x)dx+cos²x)/sin²xcos²x)}dx
=∫{(1/cos²x)+(1/sin²x)}dx
=∫(sec²x)dx +∫(cosec²x)dx
= tanx – cotx +C
Answer = tanx – cotx +C
(ix) ∫(cos2x/sin²xcos²x)dx
solution
=∫(cos2x/sin²xcos²x)dx
= ∫{(cos²x-sin²x)/(sin²xcos²x)}dx
= ∫{(1/sin²x)-(1/cos²x)}dx
= ∫(1/sin²x)dx-∫(1/cos²x)dx
= ∫cosec²xdx -∫sec²xdx
= -cotx-tanx+C
Answer = -cotx-tanx+C
Some Question for you
Evaluate
(i)∫sin⁻¹(cosx)dx
(ii)∫tan⁻¹(secx + tanx)dx
ये भी पढ़े….
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