Solved Examples on Indefinite Integral

Example 1: If ∫ xe^x cos x dx = f(x) dx + c, then f(x) is equal to

∫ xe^{(1 + i)x} / 1+i – ∫ e^(1+i)x / 1 + i dx = xe^(1+i)x / 1+ i – e^(1+i)x / (1+i)²

e^{(1 + i)x}[x(1 + i –1 / (1+i)²)]

e^x (cos x + i sin x) [(x–1) + ix / 1 + i – 1]

e^x/–2 [icos x – sin x][(x – 1) + ix]

I = e^x/2 [(1 – x) sin x – x cos x] + c.

Example 2: Let x² +1 ≠ nπ, n ∈ N, then

$\int x\sqrt{\frac{2 sin (x^{2}+1) - sin(2x^{2}+1)}{2 sin (x^{2}+1) + sin(2x^{2}+1)}} dx is$

(a) ln |1/2 sec (x² + 1) | + c

(b) ln | sec { ½ (x²+ 1) | + c

(d) ln | sec (x² + 1) | + c

Solution: Let

I = $\int x\sqrt{\frac{2 sin (x^{2}+1) - sin(2x^{2}+1)}{2 sin (x^{2}+1) + sin(2x^{2}+1)}}$

Put x² + 1= t

x dx = ½ dt

Then I = ½ ∫ √ (2 sin t – sin 2t)/(2 sin t + sin 2t) dt

= ½ ∫ √ (1 - cos t)/(1 + cos t) dt

= ½ ∫ tan (t/2) dt

= ½ .2. ln | sec (t/2)| + c

= ln | sec {(x²+1)/2} + c

= ln | sec {1/2 (x² + 1)} + c

Example 3: Let f(x) be a function such that f(0) = f’(0) = 0, f”(x) = sec⁴x + 4, then the function is

(a) ln |(sin x)| + 1/3 tan³x + x

(b) 2/3 ln |(sec x)| + 1/6 tan²x + 2x²

(d) none of these

Solution: Since f”(x) = sec⁴x + 4

So, f”(x) = (1 + tan²x) sec²x + 4

Therefore, f’(x) = tan x + tan³x/3 + 4x + c

Since, f’(0) = 0

So, 0 = c

Then, f’(x) = tan x + + tan3x/3 + 4x

= tan x + 1/3 tan x (sec²x – 1) + 4x

f’(x) = 2/3 tan x + 1/3 tan x sec²x + 4x

So, f(x) = 2/3 ln | sec x| + tan² x/6 + 2x² + d

But, f(0) = 0

So, d = 0

Then, f(x) = 2/3 ln | sec x | + 1/6 tan 2x + 2x².

Example 4: ∫ x^-2/3 (1 + x^1/2)^-5/3 dx is equal to

(a) 3(1 + x^-1/2)^-1/3 + c

(b) 3(1 + x^-1/2)^-2/3 + c

(a) none of these

More than one

Example 5: more than one

If $\int \frac{(4e^{x}+6e^{-x})}{()9e^{x}-4e^{-x})}dx =Ax + B log (9e^{2x}-4)+C, then$

(a) A = 3/2

(a) B = 35/36

(a) C is indefinite

(a) A + B = -19/36

Example 6: Fill in the blanks: ∫ √ x + √(x² + 1) dx is equal to

Example 7: True/false: The antiderivative of f(x) = ln(ln x) + (ln x)^-2 whose graph passes through (e, e) is x ln(ln x) – x(ln x)^-1.

Solution:

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