∫ e tan θ( sec θ- sin θ) d θ is equal to

Question

∫ e tan θ( sec θ- sin θ) d θ is equal to (A) - e tan θ sin θ+ c (B) e tan θ sin θ+ c (C) e tan θ sec θ+ c (D)  e tan θ cos θ+c

Tardigrade Admin · Accepted Answer

I=∫ e tan θ( sec θ- sin θ) d θ =∫ e tan θ(( sec θ- sin θ/ sec 2 θ)) sec 2 θ d θ I=∫ e tan θ((1/ sec θ)-( sin θ/ sec 2 θ)) sec 2 θ d θ =∫ e tan θ((1/√1+ tan 2 θ)-( tan θ/(1+ tan 2 θ)3 / 2)) sec 2 θ d θ = e tan θ ⋅ (1/√1+ tan 2 θ)= e tan θ ⋅ cos θ+C

Q. $\int e^{t a n θ} (sec θ - sin θ) d θ$ is equal to

$- e^{t a n θ} sin θ + c$

$e^{t a n θ} sin θ + c$

$e^{t a n θ} sec θ + c$

$e^{t a n θ} cos θ + c$

Q. ∫etanθ(secθ−sinθ)dθ is equal to

−etanθsinθ+c

etanθsinθ+c

etanθsecθ+c

etanθcosθ+c

I=∫etanθ(secθ−sinθ)dθ =∫etanθ(sec2θsecθ−sinθ​)sec2θdθ I=∫etanθ(secθ1​−sec2θsinθ​)sec2θdθ =∫etanθ(1+tan2θ​1​−(1+tan2θ)3/2tanθ​)sec2θdθ =etanθ⋅1+tan2θ​1​=etanθ⋅cosθ+C

Q. $\int e^{t a n θ} (sec θ - sin θ) d θ$ is equal to

$- e^{t a n θ} sin θ + c$

$e^{t a n θ} sin θ + c$

$e^{t a n θ} sec θ + c$

$e^{t a n θ} cos θ + c$