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Question Number 106570 by bobhans last updated on 06/Aug/20

(1) \underset{0}{\int^{a}} \frac{\sqrt{a - x}}{\sqrt{a - x} + \sqrt{x}} dx = ?

(a) 0 (b) \frac{a}{2} (c) a (d) 2 a (e) \frac{5}{2} a

(2) \underset{0}{\int^{π / 4}} \frac{1 - \tan x}{1 + \tan x} dx = ?

(a) 0 (b) \ln 2 (c) - \ln 2 (d) π \ln 2 (e) \frac{1}{2} \ln 2

(3) {(\sqrt{3} + 2)}^{x} > 7 - 4 \sqrt{3}, find the solution set

Answered by bemath last updated on 06/Aug/20

(2) \frac{1 - \tan x}{1 + \tan x} = \frac{\cos x - \sin x}{\cos x + \sin x}

= \frac{1 - \sin 2 x}{\cos 2 x} = \sec 2 x - \tan 2 x

\underset{0}{\int^{π / 4}} (\sec 2 x - \tan 2 x) dx =

{[\frac{1}{2} \ln ∣ \sec 2 x - \tan 2 x ∣ - \frac{1}{2} \ln ∣ \cos 2 x ∣]}_{0}^{\frac{π}{4}}

= \ln 2

Answered by john santu last updated on 06/Aug/20

(3) {(2 + \sqrt{3})}^{x} > 7 - 2 \sqrt{12}

{(2 + \sqrt{3})}^{x} > {(2 - \sqrt{3})}^{2}

{(2 + \sqrt{3})}^{x} > {(\frac{1}{2 + \sqrt{3}})}^{2}

{(2 + \sqrt{3})}^{x} > {(2 + \sqrt{3})}^{- 2}

x > - 2

Answered by Dwaipayan Shikari last updated on 06/Aug/20

\int_{0}^{a} \frac{\sqrt{a - x}}{\sqrt{a - x} + \sqrt{x}} d x = \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x} + \sqrt{a - x}} d x = I

2 I = \int_{0}^{a} \frac{\sqrt{a - x} + \sqrt{x}}{\sqrt{x} + \sqrt{a - x}} d x = \int_{0}^{a} 1 d x = a

I = \frac{a}{2}