I=∫sec3⁡xdx is equal to x∈0,π2

Let I=∫sec3⁡xdx=∫sec⁡xsec2⁡xdxI=∫1+tan2⁡xsec2⁡xdx Put tan⁡x=t⇒sec2⁡xdx=dt=∫1+t2dt=t21+t2+12log⁡t+t2+1+C=tan⁡x21+tan2⁡x+12log⁡tan⁡x+tan2⁡x+1+C=tan⁡xsec⁡x2+12log⁡|tan⁡x+sec⁡x|+C

I=∫sec3⁡xdx is equal to x∈0,π2

Let I=∫sec3⁡xdx=∫sec⁡xsec2⁡xdxI=∫1+tan2⁡xsec2⁡xdx Put tan⁡x=t⇒sec2⁡xdx=dt=∫1+t2dt=t21+t2+12log⁡t+t2+1+C=tan⁡x21+tan2⁡x+12log⁡tan⁡x+tan2⁡x+1+C=tan⁡xsec⁡x2+12log⁡|tan⁡x+sec⁡x|+C

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$I = \int \sec^{3} x d x is equal to [x \in (0, \frac{π}{2})]$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{\tan x \sec x}{2} + \frac{1}{2} \log |\sec x - \tan x| + C$

$\frac{\tan x \sec x}{2} + \frac{1}{2} \log |\sec x + \tan x| + C$

$\frac{\tan x \sec x}{2} + \frac{1}{2} \log |\sqrt{\sec x + \tan x}| + C$

$\frac{\tan x \sec x}{2} - \frac{1}{2} \log |\sqrt{\sec x + \tan x}| + C$

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$Let I = \int \sec^{3} x d x = \int \sec x \sec^{2} x d x$

$I = \int \sqrt{1 + \tan^{2} x} \sec^{2} x d x$

$Put \tan x = t \Rightarrow \sec^{2} x d x = d t$

$\begin{array}{l} = \int \sqrt{1 + t^{2}} d t = [\frac{t}{2} \sqrt{1 + t^{2}} + \frac{1}{2} \log |t + \sqrt{t^{2} + 1}|] + C \\ = \frac{\tan x}{2} \sqrt{1 + \tan^{2} x} + \frac{1}{2} \log |\tan x + \sqrt{\tan^{2} x + 1}| + C \\ = \frac{\tan x \sec x}{2} + \frac{1}{2} \log | \tan x + \sec x | + C \end{array}$