If {2/1+ x} dx = 2 |A(x)-B(x)|+C and 0 ≤ x ≤ π/2 then B≤ft(π/4)= [AP EAPCET 20 May 2024 Shift 1]

Correct Answer is : 13+22{1}{{3+2{2}}}3+221

Correct Answer is : 13+22{1}{{3+2{2}}}3+221

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AP EAPCET 20 May 2024 Shift 1 Solved Paper

Section: Mathematics

Question : 72 of 160

Marks: +1, -0

\int \sqrt{\frac{2}{1+\sin x}} dx = 2 \log |A(x)-B(x)|+C

0 \leq x \leq \frac{\pi}{2}

B\left(\frac{\pi}{4}\right)=

Solution:

Use

1+\sin x=(\sin(x/2)+\cos(x/2))^2

. Then

\sqrt{2/(1+\sin x)}=\sqrt2/(\sin(x/2)+\cos(x/2))=\sec(x/2-\pi/4)

. Hence the integral is

2\log|\sec(x/2-\pi/4)+\tan(x/2-\pi/4)|+C

. Since

\sec

is even, this equals

2\log|\sec(\pi/4-x/2)-\tan(\pi/4-x/2)|+C

. Comparing with

2\log|A(x)-B(x)|+C

, we get

B(x)=\tan(\pi/4-x/2)

. Therefore

B(\pi/4)=\tan(\pi/8)=\sqrt2-1=1/\sqrt{3+2\sqrt2}

, which is option seq 1.

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