If -sec3 - d-1asec-tan-12ln-sec-tan-C - then the value of a is-

I=∫ sec θ(sec^2θ)d θ =sec θ ∫ sec^2 θ d θ ∫ tan nbsp;θ (sec θ tan nbsp;θ)d θ =sec θ tan nbsp;θ ∫ sec θ tan^2 nbsp;θ d nbsp;θ =sec θ tan θ ∫ (sec^2 ...

You visited us 4 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Other

Engineering Mathematics

Reduction Formulae and Trigonometric System

If ∫sec3 θ dθ...

Question

$If \int {sec}^{3} θ d θ = \frac{1}{a} (sec θ tan θ) + \frac{1}{2} ln|sec θ + tan θ | + C, then the value of a is____$ .
2

Open in App

Solution

The correct option is A 2
$I = \int s e c θ (s e c^{2} θ) d θ$
$= s e c θ \int s e c^{2} θ d θ - \int t a n θ (s e c θ t a n θ) d θ$
$= s e c θ t a n θ - \int s e c θ t a n^{2} θ d θ$
$= s e c θ t a n θ - \int (s e c^{2} θ - 1) s e c θ d θ$
$= s e c θ t a n θ - \int s e c^{3} θ d θ + \int s e c θ d θ$
$= s e c θ t a n θ - I + l n | s e c θ + t a n θ | + C_{1}$
$I = \frac{1}{2} [s e c θ t a n θ + l n | s e c θ + t a n θ |] + C$

on comparing

$a = 2$

Suggest Corrections

Similar questions

\frac{t a n θ + s e c θ - 1}{t a n θ - s e c θ + 1} =

_____.

If $(s e c θ - t a n θ) = \frac{1}{3}$ , the value of; $(s e c θ + t a n θ)$ is:

Q. Prove :

\frac{1}{s e c θ - t a n θ} + \frac{1}{s e c θ + t a n θ} = 2 s e c θ

Prove

t a n Θ + s e c θ = \frac{1}{s e c Θ - t a n Θ}

Q. If

t a n θ = m - \frac{1}{4 m}

, then find k such that

s e c θ - t a n θ = - \frac{1}{k m}

Join BYJU'S Learning Program

If ∫sec3θdθ=1a(secθtanθ)+12ln|secθ+tanθ|+C, then the value of a is____.2

$If \int {sec}^{3} θ d θ = \frac{1}{a} (sec θ tan θ) + \frac{1}{2} ln|sec θ + tan θ | + C, then the value of a is____$ .
2