Question

what is answer of this photo
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Answer 1

Answer:

This photo is not clear bro

Answer 2

Step-by-step explanation:

1+sin2x

= sin²x+cos²x+2sinxcosx

= (sin x + cos x)²

1/√(1+sin2x)

= 1/√(sin x + cos x)²

= 1/(sin x + cos x)

∫ 1/√(1+sin2x)dx

= ∫ 1/(sin x + cos x) dx

= ∫ (1/√2)/(1/√2*sin x + 1/√2*cos x) dx

= ∫ (1/√2)/(sin x*cos(π/4) + sin(π/4)*cos x) dx

= 1/√2 ∫ 1/(sin {x+(π/4)}) dx

= 1/√2 ∫ cosec {x+(π/4)} dx

= 1/√2 log | tan [{x+(π/4)}/2] | +c

= 1/√2 log | tan {(x/2)+(π/8)} | +c

here 'c' is constant

using formula

sin²x+cos²x = 1

sin2x = 2sin x*cos x

sin A * cos B + cos A * sin B = sin (A+B)

1/sin x = cosec x

∫ cosec x dx = log | tan (x/2) | +c

= log | cosec x - cot x | +c

(x+y)² = x²+y²+2xy

(√x)² = x