24. e sec x (1+ tanx) dx equals(A) e* cosx C(C) e sin x C(B) e secx C(D) e tan x C

Open in App

Solution

The given integration is,

∫ e x secx( 1+tanx )dx = ∫ e x ( secx+secxtanx )dx

The given function is in the form of,

∫ e x ( f( x )+ f ′ ( x ) )dx = e x f( x )+C

Assume, f( x )=secx and f ′ ( x )=secxtanx.

The integration of the given function is,

∫ e x ( secx+secxtanx )dx = e x secx+C

Therefore, option (B) is correct.

Suggest Corrections

Similar questions

Choose the correct answer in the given question.
$\int e^{x} s e c x (1 + t a n x) d x$ equals
$(a) e^{x} c o s x + C (b) e^{x} s e c x + C (c) e^{x} s i n x + C (d) e^{x} t a n x + C$

Q. If tan θ + sec θ =e^x, then cos θ equals

(a)

\frac{e^{x} + e^{- x}}{2}

(b)

\frac{2}{e^{x} + e^{- x}}

(c)

\frac{e^{x} - e^{- x}}{2}

(d)

\frac{e^{x} - e^{- x}}{e^{x} + e^{- x}}

Q. If

lim x \to 0 \frac{{(1 + x)}^{1 / x} - e + \frac{e x}{2}}{x^{2}} = \frac{a e}{24}

, the find

a

\int \frac{sec x (1 + tan x) d x}{(e^{- x} + sec x)} = f (x) + c

.
If

f (0) = ln (2)

, then

f (\frac{π}{4})

Q. Match the following.
I.

\int e^{x} (sin x + cos x) d x =

e^{x} tan x + c

II.

\int e^{x} (cos x - sin x) d x =

e^{x} log sec x + c

III.

\int e^{x} (tan x + {sec}^{2} x) d x =

e^{x} sin x + c

IV.

\int e^{x} (tan x + log sec x) d x =

e^{x} cos x + c

Join BYJU'S Learning Program