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Grade: 12th pass
        
Please can someone solve this integral problem for me......................
3 months ago

Answers : (1)

Arun
23747 Points
							

Let I = ∫[SQRT (tan x) + SQRT (cot x)] dx = ∫(√tan x + √cot x) dx

= ∫(√tan x +1/√tan x) dx = ∫[( tan x + 1)/√tan x] dx

Put tan x = u²

Differentiating sec²x dx = 2u du

Since sec²x =1+tan²x =1+u⁴, dx=2u.du/(1+u⁴)

∴ I = ∫(u² + 1)/u x 2u.du/(1+u⁴)] = 2 ∫(u² + 1).du/(u⁴+1)

=2 ∫(u² + 1) du/(u² + i) (u² - i) [Factorizing the denominator]

=∫2 u² du/(u² + i) (u² - i) + ∫2du/(u² + i) (u² - i) [Splitting the numerator]

= ∫ [(u² + i)+(u² - i)]du/(u² + i)(u² - i) + ∫(1/i) [(u² + i)-(u² - i)]du/(u² + i) (u²-i)

= ∫(u² + i)du/(u² + i) (u² - i) + ∫(u² - i)du/(u² + i) (u² - i)

+ (1/i)∫(u² + i)du/(u² + i) (u² - i) + (1/i)∫(u² - i)du/(u² + i) (u² - i) [Splitting the numerator]

= ∫du/(u² - i) + (1/i)∫du/(u² - i) + ∫du/(u² + i) - (1/i)∫du/(u² + i)

= (1 +1/i) ∫du/(u² - i) + (1- 1/i)∫du/(u² + i)

= (i +1)/i . ∫du/(u² - i) + (i- 1)/i. ∫du/(u² + i)

Now (i +1)/i =i(i +1)/i²=(i² + i)/-1 =(-1 + i)/-1=1-i and (i- 1)/i=i(i- 1)/i²=(-1-i)/-1 = 1+i

∴ I = (1-i) ∫ du/[u² - (√i)²] + (1+i) ∫ du/[u² + (√i)²]

Both the integrals above are of standard form:

∫ dx/(x² - a²) = (1/2a). log (x-a)/(x+a) + c

∫ dx/(x² + a²) = 1/a. tan¯¹(x/a) + c

Here, x=u and a= √i

∴ I = (1-i)/2√i . log |(u-√i)/(u + √i)| + (1+i)/√i .tan¯¹ (u/√i)

Substituting back for u = √tan x

∫[SQRT (tan x) + SQRT (cot x)] dx

= (1-i)/2√i .log |(√tan x -√i)/(√tan x + √i)|+(1+i)/√i .tan¯¹(√tan x/√i)+ C (Answer)

3 months ago
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