The correct option is D tanx cotx 4x+cc Now, ∫ (tan x x)^2 dx =∫ (tan^2x 2+^2x)dx =∫ (^2x 4+cos ec^2x)dx =tan x 4x x+c . [ Where ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Integration by Parts

∫tanx -cotx2d...

Question

$\int (t a n x - c o t x)^{2} d x =$

t a n x + x + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

t a n x - x + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

t a n x - c o t x + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

t a n x - c o t x - 4 x + c c

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $t a n x - c o t x - 4 x + c c$
Now,
$\int (tan x - cot x)^{2} d x$
$= \int ({tan}^{2} x - 2 + {cot}^{2} x) d x$
$= \int ({sec}^{2} x - 4 + cos e c^{2} x) d x$
$= tan x - 4 x - cot x + c$ . [ Where $c$ is integrating constant]

Suggest Corrections

Similar questions

\frac{d}{d x} [\frac{t a n x - c o t x}{t a n x + c o t x}] =

is equal to

A. tan x + cot x + C

B. tan x + cosec x + C

C. − tan x + cot x + C

D. tan x + sec x + C

\int \frac{\cos 2 x - 1}{\cos 2 x + 1} d x =

\frac{e^{x} - t a n x}{c o t x - x^{n}}

x

\int {1 + 2 t a n x (t a n x + s e c x)}^{1 / 2} d x =

Join BYJU'S Learning Program