∫ ((1+x)ex/ sin2 (xex)) dx is equal to

Question

∫ ((1+x)ex/ sin2 (xex)) dx is equal to (A) - cot (ex)+C (B)  tan (xex)+C (C)  tan (ex)+C (D)  cot (xex)+C

Tardigrade Admin · Accepted Answer

Let I=∫ ((1+x) ex/ sin 2(x ex)) d x Put x ex=t ⇒ (1 ⋅ ex+x ⋅ ex) d x=d t ⇒ (1+x) ex d x=d t ∴ I=∫ (d t/ sin 2 t)=∫ cose c2 t d t =- cot t+C =- cot (x ex)+C

Q. $\int \frac{( 1 + x ) e ^{x}}{s i n ^{2} ( x e ^{x} )} d x$ is equal to

$- cot (e^{x}) + C$

$tan (x e^{x}) + C$

$tan (e^{x}) + C$

$cot (x e^{x}) + C$

$- cot (x e^{x}) + C$

Let $I = \int \frac{( 1 + x ) e ^{x}}{s i n ^{2} ( x e ^{x} )} d x$
Put $x e^{x} = t \Rightarrow (1 \cdot e^{x} + x \cdot e^{x}) d x = d t$
$\Rightarrow (1 + x) e^{x} d x = d t$
$∴ I = \int \frac{d t}{s i n ^{2} t} = \int cose c^{2} t d t$
$= - cot t + C$
$= - cot (x e^{x}) + C$

Q. ∫sin2(xex)(1+x)ex​dx is equal to

−cot(ex)+C

tan(xex)+C

tan(ex)+C

cot(xex)+C

−cot(xex)+C

Let I=∫sin2(xex)(1+x)ex​dx Put xex=t⇒(1⋅ex+x⋅ex)dx=dt ⇒(1+x)exdx=dt ∴I=∫sin2tdt​=∫cosec2tdt =−cott+C =−cot(xex)+C