If \(\rm \displaystyle\int_{0}^{a} \frac{1}{1 + x^{2}}\: dx = \frac{\pi}{4}\) then, find the value of a ?

Concept:

Integration Formula

\(\rm \int \frac{dx}{{1 + x^{2}}} = tan^{-1} x + C\)

Trigonometry Formula

tan(\(\rm \frac{\pi}{4}\)) = 1

Calculation:

Given

⇒ \(\rm \int_{0}^{a} \frac{1}{1 + x^{2}}\: dx = \frac{\pi}{4}\)

⇒ \(\rm \left [tan^{-1}x \right ]_{0}^{a} = \frac{\pi }{4}\)

⇒ tan^-1(a) - tan^-1(0) = \(\rm \frac{\pi}{4}\)

⇒ tan-1(a) - 0 = \(\rm \frac{\pi}{4}\)

⇒ a = tan(\(\rm \frac{\pi}{4}\))

⇒ a = 1

The value of a is 1

Additional Information

Integral Formulas:

• ∫ dx = x + C
• ∫ a dx = ax+ C
• \(\rm \int x^{n} dx = \frac{x^{n + 1}}{n + 1} + C \: (n \neq -1)\)
• ∫ sin x dx = – cos x + C
• ∫ cos x dx = sin x + C
• ∫ sec2 x dx = tan x + C
• ∫ cosec2 x dx = - cot x + C
• ∫ sec x (tan x) dx = sec x + C
• ∫ cosec x ( cot x) dx = – cosec x + C
• ∫ (1/x) dx = ln |x| + C
• ∫ ex dx = ex + C
• \(\rm \int \frac{dx}{\sqrt{1 - x^{2}}} = sin^{-1} x + C\)
• \(\rm \int \frac{dx}{{1 + x^{2}}} = tan^{-1} x + C\)
• \(\rm \int \frac{dx}{x \sqrt{x^{2} - 1}} = sec^{-1} x + C\)