Evaluate- -tan-1sin 2x-1 - cos 2x dx

The correct option is B x^22+c ∫tan^ 1(sin 2x/1 + cos 2x) dx=∫tan^ 1(2sin xcos x/2cos^2x) dx=∫tan^ 1(tan x) dx=∫ xdx=x^22+c ...

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

Byju's Answer

Standard XII

Mathematics

Existence of Limit

Evaluate: ∫...

Question

Evaluate: $\int {tan}^{- 1} (\frac{sin 2 x}{1 + cos 2 x}) d x$

x + c

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

\frac{x^{2}}{2} + c

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

x^{2} + c

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

\frac{x^{2}}{4} + c

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

Open in App

Solution

The correct option is B $\frac{x^{2}}{2} + c$
$\int {tan}^{- 1} (\frac{sin 2 x}{1 + cos 2 x}) d x = \int {tan}^{- 1} (\frac{2 sin x cos x}{2 {cos}^{2} x}) d x = \int {tan}^{- 1} (tan x) d x = \int x d x = \frac{x^{2}}{2} + c$

Suggest Corrections

Similar questions

\frac{x^{2} + 1}{(x^{2} + 4) (x - 2)} = \frac{A x + B}{x^{2} + 4} + \frac{C}{x - 2} \Rightarrow A + B + C =

Evaluate $\int \frac{1}{(1 - x^{2}) \sqrt{1 + x^{2}}} d x$

The solution of differential equation $\frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{(1 + x^{2})^{2}}$ is
(a) $y (1 + x^{2}) = C + t a n^{- 1} x$
(b) $\frac{y}{1 + x^{2}} = C + t a n^{- 1} x$
(c) $y l o g (1 + x^{2}) = C + t a n^{- 1} x$
(d) $y (1 + x^{2}) = C + s i n^{- 1} x$

Q. Evaluate:

\int \frac{d x}{x \sqrt{x^{2} + c}}

Solveis equal to

(A) (B) (C) (D)

Join BYJU'S Learning Program

Evaluate: ∫tan−1(sin2x1+cos2x)dx

The correct option is B x22+c∫tan−1(sin2x1+cos2x)dx=∫tan−1(2sinxcosx2cos2x)dx=∫tan−1(tanx)dx=∫xdx=x22+c

Evaluate: $\int {tan}^{- 1} (\frac{sin 2 x}{1 + cos 2 x}) d x$

The correct option is B $\frac{x^{2}}{2} + c$
$\int {tan}^{- 1} (\frac{sin 2 x}{1 + cos 2 x}) d x = \int {tan}^{- 1} (\frac{2 sin x cos x}{2 {cos}^{2} x}) d x = \int {tan}^{- 1} (tan x) d x = \int x d x = \frac{x^{2}}{2} + c$