The integral I= displaystyle ∫ sin (2 θ )[(1 + (cos)2 â¡ θ /2 (sin)2 â¡ θ )]dθ simplifies to (where, c is the integration constant)

Question

Tardigrade Admin · Accepted Answer

I= displaystyle ∫ (cos θ (1 + (cos)2 â¡ θ )/sin â¡ θ )dθ Let sin θ =t ⇒ cos θ dθ =dt ⇒ I= displaystyle ∫ (2 - t2/t)dt =2ln |t|-(t2/2)+c =2ln |sin â¡ θ |-(sin2 â¡ θ /2)+c

Q. The integral $I = \int s in (2 θ) [\frac{1 + ( cos ) ^{2} θ}{2 ( s in ) ^{2} θ}] d θ$ simplifies to (where, $c$ is the integration constant)

$l n ∣ s in θ ∣ + cos θ + c$

$2 l n ∣ s in θ ∣ - \frac{s i n ^{2} θ}{2} + c$

$l n ∣ s in θ ∣ - s i n^{2} θ + c$

$l n ∣ cos θ ∣ + co s^{2} θ + c$

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

Q. The integral I=∫sin(2θ)[2(sin)2⁡θ1+(cos)2⁡θ​]dθ simplifies to (where, c is the integration constant)

ln∣sin⁡θ∣+cos⁡θ+c

2ln∣sin⁡θ∣−2sin2⁡θ​+c

ln∣sin⁡θ∣−sin2⁡θ+c

ln∣cos⁡θ∣+cos2⁡θ+c

I=∫sin⁡θcosθ(1+(cos)2⁡θ)​dθ Let sinθ=t ⇒cosθdθ=dt ⇒I=∫t2−t2​dt =2ln∣t∣−2t2​+c =2ln∣sin⁡θ∣−2sin2⁡θ​+c

Q. The integral $I = \int s in (2 θ) [\frac{1 + ( cos ) ^{2} θ}{2 ( s in ) ^{2} θ}] d θ$ simplifies to (where, $c$ is the integration constant)

$l n ∣ s in θ ∣ + cos θ + c$

$2 l n ∣ s in θ ∣ - \frac{s i n ^{2} θ}{2} + c$

$l n ∣ s in θ ∣ - s i n^{2} θ + c$

$l n ∣ cos θ ∣ + co s^{2} θ + c$

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