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• RD Sharma Class 12 Solutions for Maths

Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.1

(i) ∫ x4 dx

Solution:

∫ x4 dx = x4+1/(4+1) + Constant

= x5/5 + C

(ii) ∫ x5/4 dx

Solution:

∫ x5/4 dx = x5/4 + 1/(5/4 +1) + Constant

= 4/9 x9/4 + C

(iii) ∫ 1/x5 dx

Solution:

∫ 1/x5 dx = ∫ x-5 dx

= x-5+1/(-5+1) + Constant

= x-4/(-4)+ C

= -1/(4x4) + C

(iv) ∫ 1/x3/2 dx

Solution:

∫ x-3/2 dx = x-3/2 + 1/(-3/2 +1) + Constant

= x-1/2/(-1/2) + C

= -2/(√x)+ C

(v) ∫ 3x dx

Solution:

∫ 3x dx = 3x/log3 + Constant

(vi) ∫ 1/x2/3 dx

Solution:

∫ 1/x2/3 dx = ∫ x-2/3 dx

= x-2/3 + 1/(-2/3+1) + Constant

= x1/3/(1/3) + C

= 3x1/3 + C

(vii) ∫ 32log3 x dx

Solution:

∫ 32log3 x dx =

= ∫ x2 dx

= x2+1/(2+1) + Constant

= x3/3 + C

(i)

Solution:

We know, cos 2x = 2cos2 x – 1

=

= ∫cos x dx

= sin x + Constant

(ii)

Solution:

We know, cos 2x = 1 – 2sin2 x

= ∫ sin x dx

= -cos x + Constant

Question 3. Evaluate

Solution:

dx

We know, e loge x = x

= ∫ x2 dx

= x2+1/2+1 + Constant

= x3/3 + C

Question 4. Evaluate:

Solution:

= ∫ a-x b-x dx

= ∫ (ab)-x dx

= (ab)-x/loge (ab)-1 + Constant

= -a-x b-x/loge (ab) + C

or

= -a-x b-x/ ln(ab) + C

(i)

Solution:

We know, cos 2x = 1 – 2sin2 x

= ∫ 1/sin2x dx = ∫ cosec2x dx

= -cot x + Constant

(ii)

Solution:

We know, cos 2x = 2cos2 x – 1

= ∫ 1/cos2 x dx = ∫ sec2 x dx

= tan x + Constant

Question 6. Evaluate: ∫ elog√x /x  dx

Solution:

∫ eloge √x /x dx = ∫√x/x dx

= ∫ x-1/2 dx = x-1/2 + 1/(-1/2 + 1) + Constant

= x1/2 /(1/2) + C

= 2√x + C

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