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• RD Sharma Class 12 Solutions for Maths

# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.22

### Solution:

Let On dividing numerator and denominator by cos2x, we get Let us considered tan x = t

So, sec2x dx = dt Again, let us considered 3t = u

3dt = du = (3/2) × (1/2) × tan-1(u/2) + c

= (1/6)tan-1(3t/2) + c

Hence, I = (1/6)tan-1(3tanx/2) + c

### Solution:

Let On dividing numerator and denominator by cos2x, we get Now, let us considered tan x = t

So, sec2xdx = dt Again, let us considered 2t = u

2dt = du = (1/2) × (1/√5) × tan-1(u/√5) + c

= (1/2√5) × tan-1(2t/√5) + c

Hence, I = (1/2√5) × tan-1(2tanx/√5) + c

### Solution:

Let On dividing numerator and denominator by cos2x, we get Now, let us considered tan x = t

So, sec2x dx = dt ### Solution:

Let On dividing numerator and denominator by cos2x, we get Now, let us considered tan x = t

So, sec2x dx = dt Again, let us considered √3t = u

So, √3dt = du ### Solution:

Let On dividing numerator and denominator by cos2x, we get Now, let us assume 2tan x = t

So, 2sec2x dx = dt

I = 1/2 ∫dt/(1 + t2)

= 1/2 tan-1t + c

Hence, I = 1/2 tan-1(2tanx) + c

### Solution:

Let On dividing numerator and denominator by cos2x, we get Now, let us assume √3 tanx = t

So, √3 sec2x dx = dt Hence, I = (1/√15)tan-1(√3tanx/√5) + c

### Solution:

Let On dividing numerator and denominator by cos2x, we get Now, let us assume tanx = t

So, sec2x dx = dt ### Solution:

Let On dividing numerator and denominator by cos4x, we get Now, let us assume tan2x = t

So, 2tanx sec2x dx = dt

I = ∫dt/(t2 + 1)

= tan-1t + c

I = tan-1(tan2x) + c

### Solution:

Let On dividing numerator and denominator by cos2x, we get Now, let us assume 2 + tanx = t

So, sec2x dx = dt

I = ∫dt/t

= log|t| + c

I = log|2 + tanx| + c

### Solution:

Let On dividing numerator and denominator by cos2x, we get Now, let us assume tanx = t

So, sec2x dx = dt ### Solution:

Let On dividing numerator and denominator by cos2x, we get Now, let us assume √2tanx = t

So, √2sec2dx = dt

I = 1/√2 ∫1/(1 + t2)

= 1/√2 tan-1t + c

Hence, I = 1/√2 tan-1(√2tanx) + c

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