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

Question

Tardigrade Admin · Accepted Answer

Let I=∫ ((1+x) ex/ cot (x ex)) d x Put x ex=t ⇒ (x ex+ex) d x=d t ⇒ (x+1) ex d x=d t ∴ I=∫ (d t/ cot t) =∫ tan t d t = log | sec t|+C Again, put t=x ex ⇒ I= log | sec (x ex)|+C

Q. $\int \frac{( 1 + x ) e ^{x}}{co t ( x e ^{x} )} d x$ is equal to

$l o g ∣ cos (x e^{x}) ∣ + C$

$l o g ∣ co t (x e^{- x}) ∣ + C$

$l o g ∣ sec (x e^{- x}) ∣ + C$

$l o g ∣ cos (x e^{- x}) ∣ + C$

$l o g ∣ sec (x e^{x}) ∣ + C$

Q. ∫cot(xex)(1+x)ex​dx is equal to

log∣cos(xex)∣+C

log∣cot(xe−x)∣+C

log∣sec(xe−x)∣+C

log∣cos(xe−x)∣+C

log∣sec(xex)∣+C

Let I=∫cot(xex)(1+x)ex​dx Put xex=t ⇒(xex+ex)dx=dt ⇒(x+1)exdx=dt ∴I=∫cottdt​ =∫tantdt =log∣sect∣+C Again, put t=xex ⇒I=log∣sec(xex)∣+C

Q. $\int \frac{( 1 + x ) e ^{x}}{co t ( x e ^{x} )} d x$ is equal to

$l o g ∣ cos (x e^{x}) ∣ + C$

$l o g ∣ co t (x e^{- x}) ∣ + C$

$l o g ∣ sec (x e^{- x}) ∣ + C$

$l o g ∣ cos (x e^{- x}) ∣ + C$

$l o g ∣ sec (x e^{x}) ∣ + C$