∫(( sin θ + cos θ )/√ sin 2θ )dθ is equal to:

Question

∫(( sin θ + cos θ )/√ sin 2θ )dθ is equal to: (A)  log | cos θ - sin θ +√ sin 2θ |+c  (B)  log | sin θ - cos θ +√ sin 2θ |+c  (C)  sin -1( sin θ - cos θ )+c  (D)  sin -1( sin θ + cos θ )+c

Tardigrade Admin · Accepted Answer

Let I=∫( sin θ + cos θ /√1+ sin 2θ -1)dθ =∫( sin θ + cos θ /√1-( sin θ - cos θ )2)dθ Let sin θ - cos θ =t ⇒ ( cos θ + sin θ )dθ =dt ∴ I=∫(1/√1-t2)dt= sin -1( sin θ - cos θ )+c

Q. $\int \frac{( s i n θ + c o s θ )}{s i n 2 θ} d θ$ is equal to:

$lo g ∣ cos θ - sin θ + sin 2 θ ∣ + c$

$lo g ∣ sin θ - cos θ + sin 2 θ ∣ + c$

$sin^{- 1} (sin θ - cos θ) + c$

$sin^{- 1} (sin θ + cos θ) + c$

$sin^{- 1} (cos θ - sin θ) + c$

Let $I = \int \frac{s i n θ + c o s θ}{1 + s i n 2 θ - 1} d θ$
$= \int \frac{s i n θ + c o s θ}{1 - ( s i n θ - c o s θ ) ^{2}} d θ$
Let $sin θ - cos θ = t$
$\Rightarrow$ $(cos θ + sin θ) d θ = d t$
$∴$ $I = \int \frac{1}{1 - t ^{2}} d t = sin^{- 1} (sin θ - cos θ) + c$

Q. ∫sin2θ​(sinθ+cosθ)​dθ is equal to:

log∣cosθ−sinθ+sin2θ​∣+c

log∣sinθ−cosθ+sin2θ​∣+c

sin−1(sinθ−cosθ)+c

sin−1(sinθ+cosθ)+c

sin−1(cosθ−sinθ)+c

Let I=∫1+sin2θ−1​sinθ+cosθ​dθ =∫1−(sinθ−cosθ)2​sinθ+cosθ​dθ Let sinθ−cosθ=t ⇒ (cosθ+sinθ)dθ=dt ∴ I=∫1−t2​1​dt=sin−1(sinθ−cosθ)+c